1. An Announcement
2. Patterns of Digits
3. Digit Patterns in Music
4. Introduction to the 11-Limit
5. A New 11-Limit Scale
6. More Information on the New 11-Limit Scale
7. Patterns in the Easter Date
8. Coding the Easter Song in 28EDO
9. More Information on 28EDO
10. Conclusion
An Announcement
Today is Easter Sunday -- the day after the blue Paschal full moon. Earlier, I wrote that I'd write a music post for Easter. Well, I decided to take it one step further.
As of today, this will no longer be a Geometry blog. Instead, I'm devoting this blog completely to music, including new songs and scales. Don't worry -- there will still be plenty of math in this post -- there just won't be any Geometry, in this post or in any post ever again.
Patterns of Digits
Back on Pi Day, I posted several songs based on the digits of pi. This idea clearly isn't original to me, since Michael Blake and others have also created such digital songs. In today's post, I want to make songs based on the digits of other real numbers.
I don't remember how old I was when I first learned about pi. It's possible that it could have been in first grade -- in 1988, around the time of the first ever Pi Day. Back then, all I knew about pi was that it was approximately 3.1416. I had no way of finding out more digits of pi -- indeed, I once guessed that maybe pi was 3.1416182022242628... and so on. Of course, now I know that the 6 isn't actually a digit of pi, since it really begins 3.14159....
In the end, the digits of pi didn't fascinate the young me since pi is an irrational number -- more digits are difficult to discover. Instead, I liked the digits of rational numbers, which are much easier to find.
The number 1/2 is simply 0.5. One third is the first repeating decimal, 0.333..., followed by the simple decimals 1/4 = 0.25 and 1/5 = 0.2. One sixth is similar to 1/3, 0.1666...., but then comes the first full repetend prime, 1/7 = 0.142857,142857....
Back in my September 20th post, I quoted Stanley Ogilvy as he wrote about full repetend primes:
- 1 * 142857 = 142857
- 2 * 142857 = 285714
- 3 * 142857 = 428571
- 4 * 142857 = 571428
- 5 * 142857 = 714285
- 6 * 142857 = 857142
- 7 * 142857 = 999999.
1/7 = 0.142857,14...
He writes out the remainders and show that they repeat after six digits (at the comma). The multiples of 142857 are the periods of the fractions 2/7 to 6/7, while 7/7 is just 0.999999,999999... = 1.
Ogilvy tells us that 1/7 is a "full repetend" prime -- one whose period (or repetend) is of the maximum length, which is one less than the prime itself. In long division, all the remainders appear except zero. He informs us that the next full repetend prime is 17:
1/17 = 0.588235294117647,0588...
But unfortunately, no one knows how to determine whether a prime is full repetend. The author does connect the length of the period to Fermat's Little Theorem:
10^6 == 1 (mod 7)
This tells us that 7 divides a string of six 9's (with the quotient equal to the period), but it doesn't tell us whether 6 is the smallest exponent such that:
10^e - 1 == 0 (mod 7)
With 7 it happens to be. But for instance,
10^10 - 1 == 0 (mod 11),
and we find that
1/11 = .090909...
Thus 11 is not a full repetend prime, since its repetend has only two digits, 09.
Returning to today, let me explain how I found repeating decimals when I was young. My first scientific calculator was a TI-34. I still have a TI-34 today -- one I found in the classroom where I taught last year. But here's one thing I noticed about this calculator as a young child:
5/3 = 1.666666667
2/3 = 0.666666666
So the TI rounds off numbers greater than 1, but truncates numbers between -1 and 1. This means that I can avoid the 3.1416 problem -- as long as the number isn't greater than 1, I know that the digits are correct and not rounded.
Let's work together to determine the digits of 1/19 using the TI-34. To start, we divide 1 by 19:
1/29 = 0.034482758
Now we find multiples of 1/29 -- which after all, repeat the same digits. On the TI-34, we constant add by pressing the + key, then the = key repeatedly. This produces 2/29, 3/29, 4/29, and so on.
We wish to find out which digit follows the last 8 above. We could find multiples of 1/29 until we get past 0.8 -- but there might be more than one fraction n/29 starting with 8. Indeed, we might eliminate 0.82 since we already have an 8 followed by a 2 showing on the calculator. But then there could be a third fraction n/29 starting with 0.8.
So we take the last two digits and look for a fraction n/29 starting with 0.58. Now we know that there can't be a second fraction starting with 0.58 -- after all, two decimals starting with 0.58 must differ by less than 1/100, which is less than 1/29. Just hit + and then keep pressing = until 0.58 appears:
0.586206896 (digits so far: 1/29 = 0.0344827586206896)
Now the last two digits are 96, so we return to pressing = until 0.96 shows up:
0.965517241 (digits so far: 1/29 = 0.03448275862068965517241)
But now we must reach 0.41, but pressing the = key only increases the fraction. There are two things we can do at this point. The TI-34 will let us press the 0 key and then the = key -- the calculator still remembers that 1/29 is the constant to be added. The other choice is to press the +/- key to switch to negative numbers. Then adding 1/29 actually decreases the absolute value of the number -- that is, it's equivalent to subtracting 1/29.
Since 0.41 is about half of 0.96, pressing 0 or +/- at this point is equally good. The one thing that we don't do is keep adding 1/29 until we reach 1.41. This is because once we pass 1, the calculator will start rounding, and then we might choose an incorrect digit as in 3.1416.
0.413793103 (digits so far: 1/29 = 0.0344827586206896551724137931,03)
Now we would need to reach 0.03 -- but this would take us back to the starting value 1/29. So we know this is where the digits must start to repeat.
Back when I was young, I used to have memorized all the digits of the full repetend primes up to an including the one we just found, 1/29 -- the others are 1/7, 1/17, 1/19, and 1/23:
1/7 = 0.142857,1
1/17 = 0.0588235294117647,05
1/19 = 0.052631578947368421,05
1/23 = 0.0434782608695652173913,04
1/29 = 0.0344827586206896551724137931,03
In fact, I noticed that 1/23 and 1/29 started with the same digits in a different order -- 1/23 starts out 0.0434782 while 1/29 goes 0.0344827. The next full repetend prime is 47 -- with such a large gap between 29 and 47, I never bothered to memorize the digits of 1/47. I also skipped ahead to some of my favorite composite decimal expansions, the squares:
1/36 = 0.027,7
1/49 = 0.020408163265306122448979591836734693877551,02
(notice the pattern 02, 04, 08, 16, 32, ...)
1/64 = 0.015625
1/81 = 0.012345679,01
1/100 = 0.01
And to this day, I still remember all the digits of 1/7, 1/17, and 1/19. It's ironic to hear about "drens" who say that because we have calculators, we don't need to memorize the multiplication table. For when I was armed with my first calculator, I used it to memorize more, not less -- I probably would never have memorized the digits of 1/29 if I had to long-divide 1 by 29. Of course, it's easy to argue that knowledge of the digits of numbers like 1/19 and 1/29 is useless.
By the way, the reason I was thinking about full repetend primes and cyclic numbers because of the website FiveThirtyEight, which I mentioned in a recent post as being a stats website:
https://fivethirtyeight.com/tag/the-riddler/
Every week there is a Riddler, or puzzle, posted at this website. And as it turns out, during the last two weeks in a row, the Riddler puzzle is related to full repetend primes and cyclic numbers:
https://fivethirtyeight.com/features/can-you-shuffle-numbers-can-you-find-all-the-world-cup-results/
This links to the answer for March 16th and the question for March 23rd. Here is the question and answer for March 16th:
Last week I posed a seemingly simple question: How many decimal numbers are there whose value equals the average of their digits? The digits of the number 2.3, for example, are 2 and 3, and their average is 2.5 — so that doesn’t quite work. But there are at least a few that do.
In fact, it appears there are an infinite number of such numbers. (I’ll call them Bollier numbers, after this puzzle’s submitter, Sam Bollier.)
It’s not too hard to find your first Bollier number. Single-digit whole numbers are obvious, if trivial, solutions. The average of the digits of the number 7, for example, is just 7. The number 4.5 works well, too — its digits are 4 and 5, and their average is 4.5. Bingo. The number 3.83333 does, too. The average of its digits is (3+8+3+3+3+3)/6 = 3.833333…, which we are allowed to truncate according to the rules of the riddle. There are more complicated numbers too. 4.428571 works, as does 4.571429. And so does 4.5294117647058824.
And that last number should look familiar -- compare to the digits of 1/17 we found above. It makes sense why we see these digits here -- the digit average is found by dividing the digit sum by the digit count, hence it's rational. So we expect the digits of rational numbers to appear. The rules of the riddle allow us to round (not truncate, as stated here), which is why we see 4.571429 when we know that the repetend of 4/7 is .571428.According to the link, here are the steps to finding a Bollier number:
- Pick a prime number, N.
- Compute the number 4.5 – 0.5*(1/N).
- Truncate this number to N digits.
http://mathworld.wolfram.com/MidysTheorem.html
The theorem states that if we cut the repetend into two equal parts and add them, then the sum consists of all 9's. The prime doesn't necessarily need to be a full repetend prime -- the only requirement is for the length of the repetend to be even (so we may cut it in half). So all full repetend primes are Midy primes (since if N is odd, then N - 1, the length of the full repetend, is even), but not all Midy primes are full repetend. The first Midy primes that aren't full repetend are 11 and 13:
1/11 = 0.09,09, (see that 0 + 9 = 9)
1/13 = 0.076923,07 (see that 076 + 923 = 999)
Once we have Midy's Theorem, the proof that the algorithm leads to Bollier numbers follows. We see that the digits of the repetend fit in pairs that add up to 9. Their sum must therefore be 9/2 * (N - 1), hence the mean of all N - 1 digits in the repetend is 9/2 = 4.5.
Now we append a 4 to the left of the decimal point. There are now N digits, and since we added a 4 to a set of digits with mean 4.5, the average drops slightly. The sum is now 4 + 9/2 * (N - 1), and dividing this by N does work out to be 4.5 - 0.5 * (1/N), which makes the rounded number a Bollier.
Some of the Bollier numbers are given as 4.5 + 0.5 * (1/N), instead of 4.5 - 0.5 * (1/N). This is also easily explained -- again we have a repetend of N - 1 digits with average 4.5, and placing a 4 in front gives N digits with a sum of 4 + 9/2 * (N - 1). But this time, the first digit of the repetend is a 5, which means that the first digit to repeat (after the last digit of the repetend) is also a 5. This means by the 4/5 rounding rules, the last digit must be rounded up. So the sum is now 4 + 9/2 * (N - 1) + 1, and dividing this by N now gives 4.5 + 0.5 * (1/N), which also makes the rounded number a Bollier.
The author of the link claims that there are infinitely many Bollier numbers. This is true provided we know that there are infinitely many full repetend primes. As it turns out, even though we suspect that there are infinitely many full repetend primes, no one has proven this. In fact, no one has proved the weaker statement that there are infinitely many Midy primes. Thus full repetend primes and Midy primes are just like twin primes (March 2nd post) -- we think, but can't prove, that they are infinite.
But here's a proof that there really are infinitely many Bollier numbers. Consider the following:
1/11 = 0.09,09,
1/101 = 0.0099,0099,
1/10001 = 0.00009999,00009999,
These are called generalized Fermat numbers -- 10^2^n + 1. Not all of these numbers are prime -- yes, 11 and 101 are prime, but 10001 = 73 * 137. Nonetheless, they all clearly have the Midy property, since a string of 0's plus a string of 9's equals a string of 9's.
And since these numbers all have the Midy property, they all lead to Bollier numbers:
4 + 5/11 = 4.4545454545
4 + 50/101 = 4.4950,4950,4950,4950,...,4950 (the block "4950" repeated 25 times)
4 + 5000/10001 = 4.499950000,...,499950000 (the block "49995000" repeated 1250 times)
There are infinitely many generalized Fermat numbers (not necessarily primes -- 10^2^n + 1 for every whole number n). Therefore there exist infinitely many Bollier numbers. QED
Actually, there might be one caveat, though -- the decimals for all of the generalized Fermat primes all end in 0. This gets me wondering whether decimals ending in 0 are allowed. One participant submitted hundreds of Bollier numbers, without single one ending in 0. This means that either his computer didn't consider that possibility, or else decimals ending in 0 are forbidden. If it's the latter, then I've proved nothing at all, and the infinitude of Bollier numbers remains an open problem.
By the way, let's look at the Riddler problem for March 23rd:
Imagine taking a number and moving its last digit to the front. For example, 1,234 would become 4,123. What is the smallest positive integer such that when you do this, the result is exactly double the original number? (For bonus points, solve this one without a computer.)
Hmmm -- when we multiply the number by 2, the digits are the same but in a different order. This sounds just like a cyclic number -- and where there are cyclic numbers, there are full repetend primes!
Let's work backwards, starting from the ones place. Suppose the units digit is a 1. When we double this number, our new number ends in 2, hence the tens digit is a 2. Now doubling a number ending in -21 gives a number ending in -42, so the hundreds digit is a 4. Now doubling a number ending in -421 gives a number ending in -842, so the thousands digit is an 8.
So far, our number ends in -8421. Let's look at the list of full repetend primes again:
1/19 = 0.052631578947368421,05
And voila -- there's our cyclic number, 052631578947368421. When we double this number, we indeed obtain the number 105263157894736842, so this should be our answer.
But this number begins with 0, so we might have another problem with zeros. But this is OK, since the number is cyclic. When we double 105263157894736842, we get 210526315789473684, and since the number doesn't start with 0, this should be a legal answer.
In fact, the any repetend from 2/19 to 9/19 has the desired property. But 10/19 doesn't have the property, since doubling this number adds an extra digit 1 in front, destroying the desired property.
I believe that this is the smallest numbers that works (at least in base 10). No number with fewer digits works. To find a number with more digits, just repeat the number:
2 * 105263157894736842105263157894736842 = 210526315789473684210526315789473684
By the way, the digits of 1/29 have the property that moving the last digit to the front triples, rather than doubles, the number:
1/29 = 0.0344827586206896551724137931,03
hence 0344827586206896551724137931 works -- or 1034482758620689655172413793 if starting with 0 is forbidden.
And the digits of 1/39 have the property that moving the last digit to the front quadruples, rather than triples, the number. But notice that 39 isn't a prime, much less a full reptend prime. It works anyway:
1/39 = 0.025641,02
4 * 025641 = 102564
4 * 102564 = 410256
It goes without saying that 1/49 -- one of my favorite repetends -- allows multiplication by 5 via a digit shift. It's easy to prove that the digits of 1/(ab - 1) in base b allows you to multiply by a by shifting the last digit to the front. The proof uses infinite geometric series.
By the way, I solved this problem without a computer -- it was easy since I still remember the digits of the fraction 1/19. Hey, perhaps knowledge of the digits of 1/19 can pay off after all!
Digit Patterns in Music
And so let's try coding some new songs using the digits of the full repetend primes. We begin with a simple example, the digits of 1/7:
http://www.haplessgenius.com/mocha/
10 DIM S(9)
20 FOR X=0 TO 9
30 READ S(X)
40 NEXT X
50 DATA 192,180,160,144,135
60 DATA 120,108,96,90,80
70 N=1
80 INPUT V
90 B=10
100 D=INT(B/V)
110 SOUND 261-N*S(D),4
120 M=D*V
130 S=B-M
140 B=S*10
150 GOTO 100
Notice that while there's a DATA line to code the scale, the notes don't require a DATA line. Instead, I simply have the computer calculate the digits by following the standard algorithm for long division (variables D, M, S, B should look familiar). Here V is used for the number we divide by -- I decided to use INPUT so that the user can choose a song. Our first song will be V=7 -- unless you really want to enter V=3 or V=9 and have a single note repeat.
When we choose V=7, the digits 142857 play -- do, fa, re, (high) do, sol, ti (or Bb-Eb-C-Bb-F-A). As these are repeating digits, the song plays forever until we press the BREAK (or Esc) key. If you wish, you can add the following line to make the digits print:
105 PRINT D;
Notice that 7 is the only full repetend prime that doesn't contain a zero. So unlike pi, where we can play 32 digits before we reach our first zero, the big zero issue comes up right away with our full repetend primes.
Because of Midy's Theorem, these songs will be symmetrical, just like the Fibonacci songs we looked at back in my Black Friday post. The difference is that opposite digits in Fibonacci songs add up to 10, while opposite digits in full repetend songs add up to 9. For Fibonacci, we made zero a rest, since the symmetry centers around digits 1-9. But for full repetend, zero is an integral part of the symmetry, which centers around digits 0-9.
Thus not only do we not want to make zero a rest, we need it to represent the lowest note. In the program above, I code Note 0 as Degree 192, which is the low A below the Bb at Degree 180.
Try playing some other songs. Here's our list of full repetend primes under 300 -- the larger primes take longer for the song to repeat:
7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, 223, 229, 233, 257, 263, 269
Right now, the song codes Notes 0-9 as low A up to high C, with a Bb major scale in between. The symmetry of the song makes opposing intervals appear in each half of the song. So where the second (C) appears in the first half, the seventh (A) appears in the second half. Where the third (D) appears in the first half, the sixth (G) appears in the second half, and so on.
It might also be desirable to code Notes 0-9 as low Bb up to high D -- it's just like the major tenth we used for earlier songs, except coded as Notes 0-9 rather than 1-0. If we code 0-9 as Bb-D, then a C scale appears in between:
C-D-Eb-F-G-A-Bb-C
This is the Dorian mode on C. The Dorian mode is a symmetric mode, so perhaps it's good to have it appear in a song based on symmetry. But this song isn't really symmetrical in just intonation. The major (9/8) and minor (10/9) tones appear in the correct places for Bb major, not a symmetric C Dorian scale. For example, C-D is 10/9, while Bb-C is 9/8.
It's possible to code a just Symmetric Dorian scale in Mocha. As it turns out, we need this symmetric Dorian scale to start on Bb:
Bb-C-Db-Eb-F-G-Ab-Bb
50 DATA 200,180,162,150,135
60 DATA 120,108,100,90,81
So notes 0-9 code as low Ab-high C. Notes 0-7 do sound like an Ab major scale, but this time the major and minor tones are set up for Bb Symmetric Dorian, not Ab major.
For both the major and Dorian scales, I like the song to end on Note 1, the tonic (Bb). Notice that for our primes that end in 9, like 19 and 29, the repetends themselves end on 1 -- we saw this when we were building the repetends to solve the March 23rd Riddler. Let's isolate the full repetend primes ending in 9:
19, 29, 59, 109, 149, 179, 229, 269
Notice that the first half of a song ending in 1 must end in its inversion 8 -- the octave. Thus there is a high Bb at the midpoint of each song.
Here's the New 7-Limit Scale coded in Mocha:
50 DATA 105,96,90,84,81
60 DATA 72,70,63,60,54
Here I decided to code 0-9 as low F-high E. This means that Note 1 is white G (as opposed to greenish G as I've posted it in the past.) It's difficult to make a 7-limit symmetrical scale in Mocha -- we have 3-limit Waldorf and 5-limit Symmetric Dorian, but making any symmetrical scale beyond the 5-limit is tricky.
But setting Note 0 to low F and Note 1 to greenish G, the scale has a different sort of symmetry. The inversion of a green note in this scale is red and vice versa (for example, Note 4 is green Bb, while Note 5 is red B). The inverse of a greenish note is white (for example, Note 2 is greenish G, while Note 7 is white D). The inverse of a white note is either greenish or also white (for example, Note 3 is white A, while Note 6 is white C.)
Of course, by "inversion" here I mean inversion in the New 7-Limit Scale. I already know that in Kite's color notation, the inverse of a green interval is yellow (for example, green 6/5 inverts to yellow 5/3) while the inverse of a red interval is blue (for example, red 9/7 inverts to blue 14/9).
Introduction to the 11-Limit
I wrote earlier that with my obsession with the number 11 lately, it's time for us to venture one step beyond the 7-limit, to 11, the next prime limit.
In Kite's color notation, the otonal 11-limit is called "jade," while the utonal 11-limit is "amber." We know that Mocha's SOUND command is based on utonal harmonics, so we use the color "amber."
Here are all the amber notes playable in Mocha:
Sound Degree Color Note Name
250 11 amber B
239 22 amber B
228 33 amber E
217 44 amber B
206 55 amber-green G
195 66 amber E
184 77 amber-red C#
173 88 amber B
162 99 amber A
151 110 amber-green G
140 121 deep amber F#
129 132 amber E
107 154 amber-red C#
96 165 amber-green C
85 176 amber B
63 198 amber A
41 220 amber-green G
30 231 amber-red F#
19 242 deep amber F#
As we see in the chart above, the fundamental amber note is amber B, at Degree 11 (Sound 250). But it's important to note the amber E, at Degree 33 (Sound 238). We compare this to the white E at Degree 32 (Sound 239). Thus amber E lies below white E at an interval of 33/32 -- this is also known as a quarter-tone, since at 53 cents, it's about half of a semitone. We can call it the undecimal (that is, 11-limit) quarter-tone, to distinguish it from the septimal (7-limit) quarter-tone 36/35 (about 49 cents, or the difference between white G and greenish G).
The note name Eb implies that it's a semitone below E. Since our amber E is a quarter-tone below E, we need to give is a name like "E half-flat." The half-flat symbol is a mirror image of the flat symbol, and since we render the latter as "b" in ASCII, we therefore write the half-flat as "d." In other words, another way to say "amber E" is "Ed." (Sometimes "half-flat" is called "quarter-flat," with "quarter" matching "quarter-tone.")
Notice that Degree 242 (Sound 19) is deep (that is, double) amber F#, which we write as F#dd. As its name implies, F# lowered by two quarter-tones ought to be F -- and in fact, white F appears one step below, at Degree 243 (Sound 18). Thus F and F#dd differ by the small subcomma 243/242 (only about 7 cents).
Triads involving amber notes are often called "neutral," since they lie a quarter-tone below major and a quarter-tone above minor. (In the past, I've often used the term "neutral" to refer to certain intervals involving the septimal quarter-tone as well.) Here are the two simplest neutral triads:
Degree 27-22-18 (Sound 234-239-243): G-Bd-D
Degree 33-27-22 (Sound 228-234-239): Ed-G-Bd
(Don't forget that Mocha degrees are based on undertones, not overtones. Using overtones, we have 18:22:27 as a jade triad, but using undertones it is amber.)
These two neutral triads sound very similar. The first triad goes white-amber-white, while the second goes amber-white-amber. It's probably easiest just to use Kite's color names to distinguish them -- the first is the amber triad (if the root and fifth are white, the third names the triad). The second is the jade triad (if the third is white and the root and fifth are the same, the opposite color names the triad).
A New 11-Limit Scale
So let's try to build a New 11-Limit Scale, similar to our New 7-Limit Scale. Recall that in building our New 7-Limit Scale, I wanted to maximize the number of triads playable in Mocha. And so we'll do the same with our New 11-Limit Scale.
Here's a list of all the playable amber and jade triads:
Amber Triads (by Degree):
27-22-18
54-44-36
81-66-54
108-88-72
135-110-90
162-132-108
189-154-126
216-176-144
243-198-162
Jade Triads (by Degree):
33-27-22
66-54-44
99-81-66
132-108-88
165-135-110
198-162-132
231-189-154
The root of the lowest amber triad is Degree 243, while the root of the lowest jade triad is 231. But odd degrees are unacceptable as the first note of a scale, since we can't build an octave on them.
So we start our scale on Degree 216 -- which, as we saw earlier in this thread, is a white G. It is the root of the amber triad G-Bd-D. One octave above this is the next higher white G, Degree 108.
At first, I naively included all the degrees in the above list between 216 and 108. But the intervals look better if we choose amber A as the base note instead of white G:
Key Sound Degree Note Ratio Color
1. 63 198 Ad 1/1 amber A
2. 72 189 A7 22/21 red A
3. 85 176 Bd 9/8 amber B
4. 96 165 C+d 6/5 amber-green C
5. 99 162 C 11/9 white C
6. 107 154 C#7d 9/7 amber-red C#
7. 117 144 D 11/8 white D
8. 126 135 Eb+ 22/15 green Eb
9. 129 132 Ed 3/2 amber E
10. 135 126 E7 11/7 red E
11. 151 110 G+d 9/5 amber-green G
0. 153 108 G 11/6 white G
1. 162 99 Ad 2/1 amber A
Here are all the triads playable using these notes:
Subminor (blue) triad: A7-C-E7
Minor (green) triads: Ad-C+d-Ed / C-Eb+-G / Ed-G+d-Bd
Jade triads: Ad-C-Ed / C+d-Eb+-G+d / Ed-G-Bd
Amber triads: G-Bd-D / A7-C#7d-E7 / C-Ed-G
Major (yellow) triad: C+d-Ed-G+d
Supermajor (red) triads: Ad-C#7d-Ed / C-E7-G
Suppose we want there to be only ten notes instead of twelve -- so that we can convert the digit-based songs from earlier to the new scale. So which two notes should we throw out?
We might try throwing out the two 22's in the list above. Just as with the New 7-Limit Scale, we might add the red A and green Eb back in to make an Extended 11-Limit Scale.
Key Sound Degree Note Ratio Color
1. 63 198 Ad 1/1 amber A
2. 85 176 Bd 9/8 amber B
3. 96 165 C+d 6/5 amber-green C
4. 99 162 C 11/9 white C
5. 107 154 C#7d 9/7 amber-red C#
6. 117 144 D 11/8 white D
7. 129 132 Ed 3/2 amber E
8. 135 126 E7 11/7 red E
9. 151 110 G+d 9/5 amber-green G
0. 153 108 G 11/6 white G
1. 162 99 Ad 2/1 amber A
Here are the remaining playable triads using these ten notes:
Minor (green) triads: Ad-C+d-Ed / Ed-G+d-Bd
Jade triads: Ad-C-Ed / Ed-G-Bd
Amber triads: G-Bd-D / C-Ed-G
Major (yellow) triad: C+d-Ed-G+d
Supermajor (red) triad: Ad-C#7d-Ed
Unfortunately, we lose our only subminor triad, since it used one of the skipped notes.
This scale still looks strange. The small interval 55/54 (about 32 cents) appears between two pairs of notes, the two C's and the two G's. But throwing out one note from each pair hurts the triads, since Degrees 162 and 108 appear in several triads, while Degrees 165 and 110 appear in the only playable major triad.
Of course, all that really matters is what the scale sounds like. We see that the first three notes of the scale sound like a just minor scale, before the new undecimal notes kick in. The last note, white G, sounds almost like a leading tone to the tonic Ad. Oh, and there's no symmetry in this scale.
To really hear what it sounds like, let's program it in Mocha. Again, we play Degree 216 as Note 0 so that the tonic, Degree 198, can be Note 1.
50 DATA 216,198,176,165,162
60 DATA 154,144,132,126,110
By the way, if you want to try playing the scale straight through, enter 81 at the prompt. This is because 1/81 = 0.012345679,01... -- admittedly 8 is skipped, but at least you'll hear the rest of it.
More Information on the New 11-Limit Scale
Let's try adding solfege to the new scale, just as we did for the 7-limit:
Key Sound Degree Note Ratio Solfege
1. 63 198 Ad 1/1 do
2. 85 176 Bd 9/8 re
3. 96 165 C+d 6/5 me
4. 99 162 C 11/9 mu
5. 107 154 C#7d 9/7 mo
6. 117 144 D 11/8 fu
7. 129 132 Ed 3/2 sol
8. 135 126 E7 11/7 lo
9. 151 110 G+d 9/5 te
0. 153 108 G 11/6 tu
1. 162 99 Ad 2/1 do
This is Andrew Heathwaite's solfege system. Notice that many neutral intervals contain the vowel "u," as in "mu," "fu," and "tu." According the Heathwaite, the letter "u" stands for undecimal. Of course, 11/7 is "lo" even though it's undecimal, since 11/7 isn't a "neutral" interval. (It's either a superfifth or a subsixth.)
I chose 198 as the tonic of this scale. This number factors as 2 * 3^2 * 11. It allows us to play some otonal undecimal intervals (11/9, 11/8, 11/7, 11/6) on the tonic, but since there is no factor of 5 or 7, all intervals involving those primes must be utonal (6/5, 9/7, and so on).
I might adjust this scale in the future as needed to fit actual songs. We might try fitting some scales to the notes of this scale. For example, recall the blues scale that I heard the middle school jazz players practice over a week ago:
F-Ab-Bb-B-C-Eb-F
We might try playing the interval F-B as 11/8, for example. This wouldn't quite fit the above scale, since there is no perfect fourth or minor seventh. (Then again, I know that blues scales are typically played as 7-limit, not 11-limit.)
Another example where 11-limit intervals might come into play are eleventh chords. For example, a F major 11th chord should be played as:
F-A-C-Eb-G-Bb
But more often than not, the last note appears as B rather than Bb. I wonder whether the B note is supposed to represent 11/8, and so the chord is intended to be a harmonic 11th chord 4:5:6:7:9:11.
Finally, many international scales often contain quarter-tones, so they could represent 11-limit. For example, one Middle Eastern scale, called "sigah," is played as:
Ed-F-G-A-Bd-C-D-Ed
Some (but not all) of those notes appear in the New 11-Limit Scale. Again, I might make a further adjustment if I desire to make my New 11-Limit Scale fit these other known scales.
Patterns in the Easter Date
Today is Easter. Last year, Easter was on April 16th. Next year, the holiday will fall on April 21st. A long time ago when I was young, I often thought of the changing Easter date like a bunny (the Easter bunny, of course) hopping his way around the calendar months of March and April.
I once noticed that the Easter dates follow a sort of pattern. For example, Easter fell on April 6th in 1980, the year I was born. When I had a part-time job at UCLA, two girls worked with me, and both were born on April 6th, 1980 (so there was a party of both of them the same day). As it turns out, Easter hasn't fallen on April 6th since -- and it won't again until their 62nd birthday!
I'm writing about patterns in the date because I'm wondering whether, just like the digits of repeating decimals, we could create a new song -- an Easter song, if you will -- out of the pattern in the dates.
Let's create a very simple Calendar Reform on the fly here. Actually, this calendar will be identical to the Gregorian Calendar, except it has a four-day week rather than a seven-day week. We'll simply call the four days of the week Oneday, Twoday, Threeday, and Fourday. This is a perpetual calendar in that March 1st will always fall on a Oneday -- which is convenient not only because Easter always falls in either March or April, but it allows us to make February 28th and 29th both blank days (and these are needed since 4 * 91 = 364).
We know that Easter always falls on a Sunday using the seven-day week. But what day of the week does Easter fall using the four-day calendar? For example, if March 1st is Oneday, then yesterday, March 31st, would be Threeday, and so today is Fourday. To make it simple, we might think of today as March 32nd, so we must reduce 32 mod 4.
Of course, Easter is defined to be Sunday of the seven-day week, not the four-day week. So we might expect the holiday to fall on each of the four days of the week equally. Well, let's find out by calculating a few Easters, starting with the first spring after the creation of the blog and extending through the 2020's:
April 5th, 2015 (March 36th = Fourday)
March 27th, 2016 (Threeday)
April 16th, 2017 (March 47th = Threeday)
April 1st, 2018 (March 32nd = Fourday)
April 21st, 2019 (March 52nd = Fourday)
April 12th, 2020 (March 43rd = Threeday)
April 4th, 2021 (March 35th = Threeday)
April 17th, 2022 (March 48th = Fourday)
April 9th, 2023 (March 40th = Fourday)
March 31, 2024 (Threeday)
April 20th, 2025 (March 51st = Threeday)
April 5th, 2026 (March 36th = Fourday)
March 28th, 2027 (Fourday)
April 16th, 2028 (March 47th = Threeday)
April 1st, 2029 (March 32rd = Fourday)
Of 15 Easters, seven fall on Threeday and eight on Fourday, with none on Oneday or Twoday. That we can invent a four-day week on the fly and then have a decade and a half's worth of Easters land on only two of the four days seems more like a pattern than a coincidence.
In fact, I discovered the pattern a few years earlier, when I look at the Easters during the decade before I created this blog:
March 27th, 2005
April 16th, 2006
April 8th, 2007
March 23rd, 2008
April 12th, 2009
April 4th, 2010
April 24th, 2011
April 8th, 2012
March 31st, 2013
April 20th, 2014
Of 10 Easters, seven are in April -- and notice that the April dates are 4, 8, 12, 16, 20, and 24. So all of these are multiples of four. We can consider March 31st to be April 0th -- and zero is also a multiple of four. The other two March dates, the 27th and 23rd, are four and eight days before April 0th, so the pattern continues.
Thus all ten Easters are on the same day of the four-day week! In order to avoid negative April dates, we've been converting these dates to positive March dates. The latest April date, the 24th, is converted to March 55th, and all ten Easters fall on Threeday. So during this stretch, we don't even have any Fourday Easters, much less Oneday or Twoday Easters. And we have an entire quarter century without any Oneday or Twoday Easters.
Does this mean that Oneday and Twoday Easters are impossible, just as Monday and Tuesday Easters are impossible in the seven-day week? Well, let's look at a chart:
1967-1984: Oneday and Twoday Easters
1985-1990: All Twoday Easters
1991-2004: Twoday and Threeday Easters
2005-2014: All Threeday Easters
2015-2028: Threeday and Fourday Easters
2029-2034: All Fourday Easters
2035-2048: Fourday and Oneday Easters
2049-2058: All Oneday Easters
So all four days are possible, but they appear in clusters. There are stretches with many Twoday Easters and stretches without them. My friends' Easter birth on April 6th, a Oneday, occurred just before a long stretch without any Oneday Easters. This is why they still have 24 more years left before they can celebrate another Easter birthday.
The following link provides us an explanation for this phenomenon:
http://www.webexhibits.org/calendars/calendar-christian-easter.html
Suppose you know the Easter date of the current year, can you easily find the Easter date in the next year? No, but you can make a qualified guess. If Easter Sunday in the current year falls on day X and the next year is not a leap year, Easter Sunday of next year will fall on one of the following days: X-15, X-8, X+13 (rare), or X+20. If Easter Sunday in the current year falls on day X and the next year is a leap year, Easter Sunday of next year will fall on one of the following days: X-16, X-9, X+12 (extremely rare), or X+19. (The jump X+12 occurs only once in the period 1800-2200, namely when going from 2075 to 2076.) If you combine this knowledge with the fact that Easter Sunday never falls before 22 March and never falls after 25 April, you can narrow the possibilities down to two or three dates.
Let's look at those lists again:
Non-leap: X - 15, X - 8, X + 13, X + 20
Leap: X - 16, X - 9, X + 12, X + 19
Of these eight possible "jumps" (bunny hops?), four of them are multiples of four, and none of them are two away from a multiple of four. Since most years don't have February 29th and the +13 jump is given as "rare," the three jumps -15, -8, +20 are the most common -- and two of these three jumps are multiples of four. During the stretch from 2005-2014, every time the -15 jump would have occurred, a February 29th conspired to make it a -16 jump, which is a multiple of four. And since the most common non-multiple of four jump is -15 (which is equivalent to +1 mod 4, not -1), the Easters tend to progress forward through the four-day week rather than backward.
It's also possible to see why these eight are the only possible jumps. Recall the definition of Easter as given at the above link:
Easter Sunday is the first Sunday after the first full moon after vernal equinox.
This means that we are taking a solar year (from equinox to equinox) and rounding it to an integer multiple of lunar months (from full moon to full moon), and then rounding the result off to an integer multiple of weeks (from Sunday to Sunday). There are three binary choices to make:
- Do we round the year up or down to an integer multiple of months?
- Do we round the result up or down to an integer multiple of weeks?
- Is there or is there not a Leap Day?
With three binary choices, there are only 2^3 or 8 possibilities -- and all of them are listed.
A solar year is either 12 or 13 months. (The 13th month or Leap Month often appears in lunisolar calendars such as the Hebrew Calendar -- the ultimate inspiration for Passover and Easter.) A year with 12 lunar months is about 354 days, which rounds to either 50 weeks or 51 weeks. As a solar year has 365 days, 50 weeks is 15 days short of a year (-15 jump) and 51 weeks is 8 days short (-8). A year with 13 lunar months is about 384 days, which rounds to either 54 weeks or 55 weeks. Then 54 weeks is 13 days long of a year (+13 jump) and 55 weeks is 20 days long of a year (+20). Also, notice that 354 days is about halfway between 50 and 51 weeks, while 384 days is much closer to 55 weeks than to 54 weeks. This is why the 54-week jump (+13) is "rare." The calculations for Leap Day are based on taking the year to be 366 days rather than 365.
Now that we finally know the pattern of the Easter dates, let's make them into an Easter song. When I first discovered the Easter pattern years ago, I wanted to make the common jumps correspond to consonant intervals. For example, maybe -8 would be a major third, while -15 is a perfect fifth. But then -16, another bunny hop, would be an augmented fifth (two major thirds) -- although this might sound as a minor sixth, which is consonant, depending on the situation. (Recall what I wrote about train whistles in an earlier post -- they sound like dissonant augmented triads.)
In the end, I wasn't able to make the Easter dates into a coherent song. But I came up with this idea years ago, back before I knew about scales other than 12EDO. Perhaps it's possible to make an Easter song after all, just in another scale besides 12EDO.
So what might be a good EDO for the Easter song? We notice that there are 35 possible Easters (from March 22nd to April 25th), so maybe 35EDO would work. Or we could use 34EDO instead, so that the notes for March 22nd and April 25th are an octave apart. Another possibility is 31EDO -- this is a commonly used EDO for 11-limit music (as explained in an earlier post). Also, since March has 31 days, it means that Easters on the same day of the month (March 22nd and April 22nd, up to March 25th and April 25th) are an octave apart.
But the Easter date patterns we found above involve two key numbers -- 7 and 4. And so the best EDO for an Easter song would be 7 * 4, or 28EDO. Indeed, 28EDO is the only EDO such that the inversion of a possible bunny hop is another bunny hop (such as -15 and +13, -9 and +19). And so 28EDO it is.
Coding the Easter Song in 28EDO
We know that music on Mocha is based on EDL (equal divisions of length) rather than EDO (equal division of the octave). Thus trying to play an EDO on Mocha is unnatural -- and indeed, we can only approximate 28EDO on the emulator.
I wrote earlier that the EDO's up to 12 (the "macrotonal EDO's") sound quite well on Mocha, but the accuracy drops off quickly past 12EDO. The multiples of four (16EDO, 20EDO, 24EDO, 28EDO) are slightly better than non-multiples of four. Once we reach 31EDO, the odd EDO's are marginally better than the even EDO's -- in reality, all of them are very inaccurate. (This is another situation where 16-bit Atari music shines -- although Atari music is also based on EDL, we'd be able to approximate EDO's better on Atari than on Mocha.)
As it turns out, Degree 210 (Sound 51) -- the root note of the New 7-Limit Scale -- is also a good root note for a 28EDO scale. To create the scale start with Degree 210 and divide by the 28th root of two for each step until we reach 105, one octave above Degree 210. Here's the resulting scale:
Step Degree Sound
0 210 51
1 205 56
2 200 61
3 195 66
4 190 71
5 186 75
6 181 80
7 177 84
8 172 89
9 168 93
10 164 97
11 160 101
12 156 105
13 152 109
14 148 113
15 145 116
16 141 120
17 138 123
18 134 127
19 131 130
20 128 133
21 125 136
22 122 139
23 119 142
24 116 145
25 113 148
26 110 151
27 108 153
28 105 156
Some of these are more accurate than others. These include:
Step Degree Sound
0 210 51
3 195 66
6 181 80
9 168 93
10 164 97
11 160 101
12 156 105
20 128 133
24 116 145
25 113 148
28 105 156
These can indicate how close some of the steps of 28EDO are to just intervals. For example, that Step 3 corresponds to Degree 195 tells us that this interval from Steps 0 to 3 represents the ratio 210/195, which reduces to 14/13 (a 13-limit interval).
One important interval is that from Steps 0 to 9. The ratio 210/168 reduces to 5/4, a major third. In fact, 28EDO approximates a just major third to within one cent -- better than any simpler EDO. And furthermore, -9 is one of the Easter bunny hops, so here I achieve my original goal of making the valid Easter jumps correspond to consonant intervals.
The reason that we see EDO's like 24 and 31 more often than 28 -- despite its accurate major third -- is that 28EDO perfect fifth is inaccurate (and fifths are more important than thirds). In fact, the perfect fifth of 28EDO is the same as that of 7EDO -- Step 16. This is confirmed by the presence of Degree 141 in the above chart for Step 16 -- had Step 16 been closer to a just 3/2, it would have been listed as Degree 140 (210/140 = 3/2), not Degree 141.
I've mentioned 7EDO in previous posts -- there was apparently an ancient Chinese scale (Qingyu) based on five notes of a 7EDO scale. We found out that this scale fails to distinguish between major and minor intervals.
In fact, the Xenharmonic website often gives 7EDO a special name -- whitewood. It refers to the idea of removing all of the black keys on a piano, leaving only the white keys. Without black keys, the interval from C-D is the same as that from E-F -- seven equal intervals add up to the octave.
Another name given at Xenharmonic for EDO's like 7 and 28 is "perfect EDO." This is because there are no "major" or "minor" intervals, only "perfect" intervals.
Of course, 28EDO does have a major third (Step 9) in addition to 7EDO's "perfect third." In other words, 28EDO is what we get if we cross a perfect EDO with a just major third.
This indicates what our Easter song will sound like. We'll hear essentially 7EDO music during the stretches when Easter falls on the same day of the four-week (like 2005-2014), then slowly more and more major thirds (and minor sixths) appear. Eventually we'll hit another stretch (like 2029-2034) with the same 7EDO scale transposed up a 28EDO step.
Let's program this song in Mocha. We'll begin by coding the 28EDO scale:
NEW
10 DIM S(35)
20 FOR X=1 TO 35
30 S(X)=INT(210/2^((X-4)/28)+.5)
40 NEXT X
Don't forget to use the up arrow for the exponentiation ^ symbol. This sets up a 28EDO scale, where Note 4 is Degree 210 and Note 28 is Degree 105. The idea is that Note 1 will correspond to March 22nd and Note 35 is April 25th, so that the earliest Easters play the lowest notes (which are the largest degrees).
Now for the trickiest part -- calculating an Easter date. We begin by asking the user for a year:
50 INPUT Y
Notice the definition of the Remainder or mod function given:
Remainder(x|y) (or x mod y) means the remainder when you divide x by y. It is never negative, and is defined in terms of the [] operation as follows:
- Remainder(x|y) = x - y[x/y]
In Mocha, [] is the INT operation, so we write:
R=X-Y*INT(X/Y)
This is actually the same remainder function we used in the repeating decimals song earlier. It would be easier if the mod function had a single symbol, like % in C and C++. But unfortunately, BASIC doesn't have such a symbol built in!
Let's proceed with the algorithm mentioned at the above link. For example, at the above link we see:
G = year mod 19
In the program, we write this as
60 G=Y-19*INT(Y/19)
Here is the rest of the algorithm:
70 C=INT(Y/100)
80 HH=C-INT(C/4)-INT((8*C-13)/25)+19*G+15
90 H=HH-30*INT(HH/30)
100 I=H-INT(H/28)*(1-INT(29/(H+1))*INT((21-G)/11))
110 JJ=Y+INT(Y/4)+I+2-C-INT(C/4)
120 J=JJ-7*INT(JJ/7)
130 L=I-J+7
You might notice that the link above gives L = I - J, not L = I - J + 7. But let's see why I included it:
L is the number of days from 21 March to the Sunday on or before the Paschal full moon (a number between -6 and 28)
In other words, L is the number of days from March 21st to Palm Sunday. We might as well add 7 to make this the number of days from March 21st to Easter Sunday. And then this becomes a number between 1 and 35, which we can enter directly into Mocha:
140 SOUND 261-S(L),4
The song is supposed to continue on to the following year to computer the next Easter:
150 Y=Y+1
160 GOTO 60
We can begin this song at any year, such as 2018, and hear a different part of the song. I suppose that purists should begin the song with 1583 -- the first spring of the Gregorian Calendar.
Just as with the repeating decimals, this song repeats -- but according to the link above, Easter repeats only after 5.7 million years!
The link also states that Julian (Orthodox) Easter repeats only after 532 years. The link doesn't state this, but the first lines (calculating I and J that we skipped over) are actually for Julian Easter. (But the way, this year Orthodox Easter is a week after Gregorian Easter on April 8th.) This we can code in Mocha by deleting some lines:
DEL 70-80
90 II=19*G+15
100 I=II-30*INT(II/30)
110 JJ=Y-INT(Y/4)+I
This song thus repeats after 532 notes. (Notice that 533 isn't a prime, much less a full reptend prime, but 541 is indeed a full reptend prime. So the Orthodox Easter song is similar in length to the song for the repeating decimal 1/541.)
More Information About 28EDO
Here's a link to the 28EDO page at the Xenharmonic website:
Basic properties
28edo, a multiple of both 7edo and 14edo (and of course 2edo and 4edo), has a step size of 42.857 cents. It shares three intervals with 12edo: the 300 cent minor third, the 600 cent tritone, and the 900 cent major sixth. Thus it tempers out the greater diesis648:625. It does not however temper out the 128:125 lesser diesis, as its major third is less than 1 cent flat (and its inversion the minor sixth less than 1 cent sharp). It has the same perfect fourth and fifth as 7edo. It also has decent approximations of several septimal intervals, of which 9/7 and its inversion 14/9 are also found in 14edo.Subgroups
28edo can approximate the 7-limit subgroup 2.27.5.21 quite well, and on this subgroup it has the same commas and tunings as 84edo. The temperament corresponding to orwell temperament now has a major third as generator, though as before 225/224, 1728/1715 and 6144/6125 are tempered out. The 225/224-tempered version of the augmented triad has a very low complexity, so many of them appear in the MOS scales for this temperament, which have sizes 7, 10, 13, 16, 19, 22, 25.Another subgroup for which 28edo works quite well is 2.5.11.19.21.27.29.39.
The chart at the link doesn't do this, but I like the idea of calling the 7EDO intervals "perfect," one step above perfect "major," and one step below perfect "minor." In other words, "major" is what the chart calls "up," and "minor" is what the chart calls "down." We might as use the name "augmented" for "double-up" and "diminished" for "double-down."
Then we can call Step 9 a "major third," which is what we expect for a interval near 5/4. Then Step 7, the "minor third," is the same as the minor third of 12EDO.
Meanwhile, many "augmented" and "diminished" represent septimal intervals -- for example, Step 2 is 21/20, Step 6 is 7/6, Step 10 is 9/7, and Step 14 is 7/5. Some of the "perfect" intervals correspond to undecimal intervals -- Step 4 is the perfect (neutral) second 11/10, while Step 8 is the perfect (neutral) third 11/9.
By this definition, two perfect fifths don't add up to a major second (instead of an octave). In other words, 9/8 is inconsistent. This happens because the whitewood perfect fifth isn't accurate, so two different perfect fifths are needed to reach 9/8. The circle of fifths clearly doesn't work in 28EDO the way it does in 12EDO.
In fact, this inconsistency results in some strange sounds when one tries to play a chord progression, such as in the Eagles' Hotel California:
In the Easter song, this chord drift is a feature, not a bug.
Conclusion
Yes, I did say that from now on, this would be a music blog, not a Geometry blog, right. Well....
APRIL FOOL!
Oh yeah, that's right -- today, Easter, is also April Fool's Day. Ha, ha, so you really thought that I wasn't going to post Geometry anymore! And I've made April Fool's Day posts in the past, but not last year. Combined with labeling this as an Easter post, I really got you there!
I really will continue to make music posts on the blog, especially as the Xenharmonic website is hosted by Wikispaces -- a disappearing host. So soon, this blog might be the only site where 28EDO and all these other alternate scales are recorded for prosperity.
By the way, the last time Easter fell on April Fool's Day was in 1956 (the last time Easter rotated in a Fourday/Oneday stretch), and the next April Fool's Easter will be in 1929 (the start of six straight Fourday Easters). This is mentioned in a Newsweek article:
http://www.newsweek.com/easter-april-fools-day-year-11-years-862141
If you think you’re seeing a pattern here, you’re right; many Easter Sunday dates move in an 11-year cycle. That’s not to say that Easter falls on April Fools’ Day, or any other date, every 11 years. Rather, a certain number of years will pass between instances of Easter occurring on a specific date, and an 11-year gap is by far the most common span.
That pattern applies to both Gregorian and Julian calendar dates. Of the 12 times western Easter fell on April 1 between 1700 and 2018, 10 of those instances were 11 years apart from another. Of the nine times it fell on March 30 in that same period, all but one were 11 years removed from another occurrence.
By the way, the last time Easter fell on April Fool's Day was in 1956 (the last time Easter rotated in a Fourday/Oneday stretch), and the next April Fool's Easter will be in 1929 (the start of six straight Fourday Easters). This is mentioned in a Newsweek article:
http://www.newsweek.com/easter-april-fools-day-year-11-years-862141
If you think you’re seeing a pattern here, you’re right; many Easter Sunday dates move in an 11-year cycle. That’s not to say that Easter falls on April Fools’ Day, or any other date, every 11 years. Rather, a certain number of years will pass between instances of Easter occurring on a specific date, and an 11-year gap is by far the most common span.
That pattern applies to both Gregorian and Julian calendar dates. Of the 12 times western Easter fell on April 1 between 1700 and 2018, 10 of those instances were 11 years apart from another. Of the nine times it fell on March 30 in that same period, all but one were 11 years removed from another occurrence.
The explanation for this 11-year phenomenon is that eleven solar years, rounded off to the nearest integer day, is 4018 days -- and 4018 is both a whole number of weeks (4018 = 7 * 574) and nearly a whole number of lunar months (136 months, since 4018/136 is about 29.54 days).
So if you want to pay tribute to math or be particularly clever this Easter and April Fools’ Day, you could find a way to incorporate this special number.
And that's exactly what I did in today's post. I paid tribute to math, I was clever for Easter and April Fool's Day, and I incorporated the special number 11 (as in 11-limit music).
Recall that my school district is always closed on Easter Monday. Therefore my next post -- and yes, it will be a Geometry post -- will be on Tuesday.
So if you want to pay tribute to math or be particularly clever this Easter and April Fools’ Day, you could find a way to incorporate this special number.
And that's exactly what I did in today's post. I paid tribute to math, I was clever for Easter and April Fool's Day, and I incorporated the special number 11 (as in 11-limit music).
Recall that my school district is always closed on Easter Monday. Therefore my next post -- and yes, it will be a Geometry post -- will be on Tuesday.
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