Tag Archives: recreational mathematics

The Rite of Sling

by in Base Behavior, Digit-sums, Mathematics and tagged , , , , , , , , , , , , , , , , , , , , | Leave a comment

Duels are interesting things. Flashman made his name in one and earnt an impressive scar in another. Maupassant explored their psychology and so did his imitator Maugham. Game theory might be a good guide on how to fight one, but I’d like to look at something simpler: the concept of duelling numbers.

How would two numbers fight? One way is to use digit-sums. Find the digit-sum of each number, then take it away from the other number. Repeat until one or both numbers <= 0, like this:

function duel(n1,n2){
print(n1," <-> ",n2);
do{
s1=digitsum(n1);
s2=digitsum(n2);
n1 -= s2;
n2 -= s1;
print(” -> ",n1," <-> ",n2);
}while(n1>0 && n2>0);
}

Suppose n1 = 23 and n2 = 22. At the first step, s1 = digitsum(23) = 5 and s2 = digitsum(22) = 4. So n1 = 23 – 4 = 19 and n2 = 22 – 5 = 17. And what happens in the end?

23 ↔ 22 ➔ 19 ↔ 17 ➔ 11 ↔ 7 ➔ 4 ↔ 5 ➔ -1 ↔ 1

So 23 loses the duel with 22. Now try 23 vs 24:

23 ↔ 24 ➔ 17 ↔ 19 ➔ 7 ↔ 11 ➔ 5 ↔ 4 ➔ 1 ↔ -1

23 wins the duel with 24. The gap can be bigger. For example, 85 and 100 are what might be called David and Goliath numbers, because the David of 85 beats the Goliath of 100:

85 ↔ 100 ➔ 84 ↔ 87 ➔ 69 ↔ 75 ➔ 57 ↔ 60 ➔ 51 ↔ 48 ➔ 39 ↔ 42 ➔ 33 ↔ 30 ➔ 30 ↔ 24 ➔ 24 ↔ 21 ➔ 21 ↔ 15 ➔ 15 ↔ 12 ➔ 12 ↔ 6 ➔ 6 ↔ 3 ➔ 3 ↔ -3

999 and 1130 are also David and Goliath numbers:

999 ↔ 1130 ➔ 994 ↔ 1103 ➔ 989 ↔ 1081 ➔ 979 ↔ 1055 ➔ 968 ↔ 1030 ➔ 964 ↔ 1007 ➔ 956 ↔ 988 ➔ 931 ↔ 968 ➔ 908 ↔ 955 ➔ 889 ↔ 938 ➔ 869 ↔ 913 ➔ 856 ↔ 890 ➔ 839 ↔ 871 ➔ 823 ↔ 851 ➔ 809 ↔ 838 ➔ 790 ↔ 821 ➔ 779 ↔ 805 ➔ 766 ↔ 782 ➔ 749 ↔ 763 ➔ 733 ↔ 743 ➔ 719 ↔ 730 ➔ 709 ↔ 713 ➔ 698 ↔ 697 ➔ 676 ↔ 674 ➔ 659 ↔ 655 ➔ 643 ↔ 635 ➔ 629 ↔ 622 ➔ 619 ↔ 605 ➔ 608 ↔ 589 ➔ 586 ↔ 575 ➔ 569 ↔ 556 ➔ 553 ↔ 536 ➔ 539 ↔ 523 ➔ 529 ↔ 506 ➔ 518 ↔ 490 ➔ 505 ↔ 476 ➔ 488 ↔ 466 ➔ 472 ↔ 446 ➔ 458 ↔ 433 ➔ 448 ↔ 416 ➔ 437 ↔ 400 ➔ 433 ↔ 386 ➔ 416 ↔ 376 ➔ 400 ↔ 365 ➔ 386 ↔ 361 ➔ 376 ↔ 344 ➔ 365 ↔ 328 ➔ 352 ↔ 314 ➔ 344 ↔ 304 ➔ 337 ↔ 293 ➔ 323 ↔ 280 ➔ 313 ↔ 272 ➔ 302 ↔ 265 ➔ 289 ↔ 260 ➔ 281 ↔ 241 ➔ 274 ↔ 230 ➔ 269 ↔ 217 ➔ 259 ↔ 200 ➔ 257 ↔ 184 ➔ 244 ↔ 170 ➔ 236 ↔ 160 ➔ 229 ↔ 149 ➔ 215 ↔ 136 ➔ 205 ↔ 128 ➔ 194 ↔ 121 ➔ 190 ↔ 107 ➔ 182 ↔ 97 ➔ 166 ↔ 86 ➔ 152 ↔ 73 ➔ 142 ↔ 65 ➔ 131 ↔ 58 ➔ 118 ↔ 53 ➔ 110 ↔ 43 ➔ 103 ↔ 41 ➔ 98 ↔ 37 ➔ 88 ↔ 20 ➔ 86 ↔ 4 ➔ 82 ↔ -10

You can look in the other direction and find bully numbers, or numbers that beat all numbers smaller than themselves. In base 10, the numbers 2 to 9 obviously do. So do these:

35, 36, 37, 38, 39, 47, 48, 49, 58, 59, 64, 65, 66, 67, 68, 69, 76, 77, 78, 79, 189

In other bases, bullies are sometimes common, sometimes rare. Sometimes they don’t exist at all for n > b. Here are bully numbers for bases 2 to 30:

base=2: 3, 5, 7, 13, 15, 21, 27, 29, 31, 37, 43, 45, 47, 54, 59
b=3: 4, 5, 7, 8, 14
b=4: 5, 6, 7, 9, 10, 11, 14, 15, 27, 63
b=5: 12, 13, 14, 18, 19, 23, 24
b=6: 15, 16, 17, 22, 23, 26, 27, 28, 29, 32, 33, 34, 35, 65, 71, 101
b=7: 17, 18, 19, 20, 24, 25, 26, 27, 32, 33, 34, 40, 41, 45, 46, 47, 48, 76
b=8: 37, 38, 39, 46, 47, 59, 60, 61, 62, 63, 95, 103, 111, 119
b=9: 42, 43, 44, 52, 53, 61, 62
b=10: 35, 36, 37, 38, 39, 47, 48, 49, 58, 59, 64, 65, 66, 67, 68, 69, 76, 77, 78, 79, 189
b=11: 38, 39, 40, 41, 42, 43, 49, 50, 51, 52, 53, 54, 62, 63, 64, 65, 73, 74, 75, 76, 85, 86, 87
b=12: 57, 58, 59
b=13: 58, 59, 60, 61, 62, 63, 64, 74, 75, 76, 77, 87, 88, 89, 90, 101, 102, 103, 115, 116, 127, 128, 129
b=14: none (except 2 to 13)
b=15: 116, 117, 118, 119, 130, 131, 132, 133, 134, 147, 148, 149
b=16: 122, 123, 124, 125, 126, 127, 140, 141, 142, 143, 156, 157, 158, 159, 173, 174, 175, 190, 191, 222, 223
b=17: 151, 152, 168, 169, 185, 186
b=18: 85, 86, 87, 88, 89, 191, 192, 193, 194, 195, 196, 197, 212, 213, 214, 215
b=19: 242, 243, 244, 245, 246
b=20: none
b=21: 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 162, 163, 164, 165, 166, 167, 183, 184, 185, 186, 187, 188, 206, 207, 208, 209, 227, 228, 229, 230, 248, 249, 250, 251, 270, 271, 272
b=22: 477, 478, 479, 480, 481, 482, 483
b=23: none
b=24: none
b=25: 271, 272, 273, 274, 296, 297, 298, 299, 322, 323, 324, 348, 349, 372, 373, 374
b=26: none
b=27: none
b=28: none
b=29: 431, 432, 433, 434, 459, 460, 461, 462, 463, 490, 491, 492, 546, 547, 548, 549, 550
b=30: none

The Hex Fractor

by in Fractals, Geometry, Mathematics and tagged , , , , , , , , , , , , , , , , , , , | Leave a comment

A regular hexagon can be divided into six equilateral triangles. An equilateral triangle can be divided into three more equilateral triangles and a regular hexagon. If you discard the three triangles and repeat, you create a fractal, like this:

hexring
Adjusting the sides of the internal hexagon creates new fractals:
hexring2
hexring1
Discarding a hexagon after each subdivision creates new shapes:

hexring4
hexring5
hexring6
And you can start with another regular polygon, divide it into triangles, then proceed with the hexagons:
hexring3

Pair on a D-String

by in √2, Base Behavior, Binary numbers, π, Mathematics, Philosophy, Philosophy of mathematics and tagged , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

What’s special about the binary number 10011 and the ternary number 1001120221? To answer the question, you have to see double. 10011 contains all possible pairs of numbers created from 0 and 1, just as 1001120221 contains all possible pairs created from 0, 1 and 2. And each pair appears exactly once. Now try the quaternary number 10011202130322331. That contains exactly one example of all possible pairs created from 0, 1, 2 and 3.

But there’s something more: in each case, the number is the smallest possible number with that property. As the bases get higher, that gets less obvious. In quinary, or base 5, the smallest number containing all possible pairs is 10011202130314042232433441. The digits look increasingly random. And what about base 10? There are 100 possible pairs of numbers created from the digits 0 to 9, starting with 00, 01, 02… and ending with …97, 98, 99. To accommodate 100 pairs, the all-pair number in base 10 has to be 101 digits long. It’s a string of digits, so let’s call it a d-string:

1, 0, 0, 1, 1, 2, 0, 2, 1, 3, 0, 3, 1, 4, 0, 4, 1, 5, 0, 5, 1, 6, 0, 6, 1, 7, 0, 7, 1, 8, 0, 8, 1, 9, 0, 9, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 4, 4, 5, 4, 6, 4, 7, 4, 8, 4, 9, 5, 5, 6, 5, 7, 5, 8, 5, 9, 6, 6, 7, 6, 8, 6, 9, 7, 7, 8, 7, 9, 8, 8, 9, 9, 1

Again, the digits look increasingly random. They aren’t: they’re strictly determined. The d-string is in harmony. As the digits are generated from the left, they impose restrictions on the digits that appear later. It might appear that you could shift larger digits to the right and make the number smaller, but if you do that you no longer meet the conditions and the d-string collapses into dischord.

Now examine d-strings containing all possible triplets created from the digits of bases 2, 3 and 4:

1, 0, 0, 0, 1, 0, 1, 1, 1, 0 in base 2 = 558 in base 10

1, 0, 0, 0, 1, 0, 1, 1, 0, 2, 0, 0, 2, 1, 1, 1, 2, 0, 1, 2, 1, 2, 2, 0, 2, 2, 2, 1, 0 in base 3 = 23203495920756 in base 10

1, 0, 0, 0, 1, 0, 1, 1, 0, 2, 0, 0, 2, 1, 0, 3, 0, 0, 3, 1, 1, 1, 2, 0, 1, 2, 1, 1, 3, 0, 1, 3, 1, 2, 2, 0, 2, 2, 1, 2, 3, 0, 2, 3, 1, 3, 2, 0, 3, 2, 1, 3, 3, 0, 3, 3, 2, 2, 2, 3, 2, 3, 3, 3, 1, 0 in base 4 = 1366872334420014346556556812432766057460 in base 10

Note that there are 8 possible triplets in base 2, so the all-triplet number has to be 10 digits long. In base 10, there are 1000 possible triplets, so the all-triplet number has to be 1002 digits long. Here it is:

1, 0, 0, 0, 1, 0, 1, 1, 0, 2, 0, 0, 2, 1, 0, 3, 0, 0, 3, 1, 0, 4, 0, 0, 4, 1, 0, 5, 0, 0, 5, 1, 0, 6, 0, 0, 6, 1, 0, 7, 0, 0, 7, 1, 0, 8, 0, 0, 8, 1, 0, 9, 0, 0, 9, 1, 1, 1, 2, 0, 1, 2, 1, 1, 3, 0, 1, 3, 1, 1, 4, 0, 1, 4, 1, 1, 5, 0, 1, 5, 1, 1, 6, 0, 1, 6, 1, 1, 7, 0, 1, 7, 1, 1, 8, 0, 1, 8, 1, 1, 9, 0, 1, 9, 1, 2, 2, 0, 2, 2, 1, 2, 3, 0, 2, 3, 1, 2, 4, 0, 2, 4, 1, 2, 5, 0, 2, 5, 1, 2, 6, 0, 2, 6, 1, 2, 7, 0, 2, 7, 1, 2, 8, 0, 2, 8, 1, 2, 9, 0, 2, 9, 1, 3, 2, 0, 3, 2, 1, 3, 3, 0, 3, 3, 1, 3, 4, 0, 3, 4, 1, 3, 5, 0, 3, 5, 1, 3, 6, 0, 3, 6, 1, 3, 7, 0, 3, 7, 1, 3, 8, 0, 3, 8, 1, 3, 9, 0, 3, 9, 1, 4, 2, 0, 4, 2, 1, 4, 3, 0, 4, 3, 1, 4, 4, 0, 4, 4, 1, 4, 5, 0, 4, 5, 1, 4, 6, 0, 4, 6, 1, 4, 7, 0, 4, 7, 1, 4, 8, 0, 4, 8, 1, 4, 9, 0, 4, 9, 1, 5, 2, 0, 5, 2, 1, 5, 3, 0, 5, 3, 1, 5, 4, 0, 5, 4, 1, 5, 5, 0, 5, 5, 1, 5, 6, 0, 5, 6, 1, 5, 7, 0, 5, 7, 1, 5, 8, 0, 5, 8, 1, 5, 9, 0, 5, 9, 1, 6, 2, 0, 6, 2, 1, 6, 3, 0, 6, 3, 1, 6, 4, 0, 6, 4, 1, 6, 5, 0, 6, 5, 1, 6, 6, 0, 6, 6, 1, 6, 7, 0, 6, 7, 1, 6, 8, 0, 6, 8, 1, 6, 9, 0, 6, 9, 1, 7, 2, 0, 7, 2, 1, 7, 3, 0, 7, 3, 1, 7, 4, 0, 7, 4, 1, 7, 5, 0, 7, 5, 1, 7, 6, 0, 7, 6, 1, 7, 7, 0, 7, 7, 1, 7, 8, 0, 7, 8, 1, 7, 9, 0, 7, 9, 1, 8, 2, 0, 8, 2, 1, 8, 3, 0, 8, 3, 1, 8, 4, 0, 8, 4, 1, 8, 5, 0, 8, 5, 1, 8, 6, 0, 8, 6, 1, 8, 7, 0, 8, 7, 1, 8, 8, 0, 8, 8, 1, 8, 9, 0, 8, 9, 1, 9, 2, 0, 9, 2, 1, 9, 3, 0, 9, 3, 1, 9, 4, 0, 9, 4, 1, 9, 5, 0, 9, 5, 1, 9, 6, 0, 9, 6, 1, 9, 7, 0, 9, 7, 1, 9, 8, 0, 9, 8, 1, 9, 9, 0, 9, 9, 2, 2, 2, 3, 2, 2, 4, 2, 2, 5, 2, 2, 6, 2, 2, 7, 2, 2, 8, 2, 2, 9, 2, 3, 3, 2, 3, 4, 2, 3, 5, 2, 3, 6, 2, 3, 7, 2, 3, 8, 2, 3, 9, 2, 4, 3, 2, 4, 4, 2, 4, 5, 2, 4, 6, 2, 4, 7, 2, 4, 8, 2, 4, 9, 2, 5, 3, 2, 5, 4, 2, 5, 5, 2, 5, 6, 2, 5, 7, 2, 5, 8, 2, 5, 9, 2, 6, 3, 2, 6, 4, 2, 6, 5, 2, 6, 6, 2, 6, 7, 2, 6, 8, 2, 6, 9, 2, 7, 3, 2, 7, 4, 2, 7, 5, 2, 7, 6, 2, 7, 7, 2, 7, 8, 2, 7, 9, 2, 8, 3, 2, 8, 4, 2, 8, 5, 2, 8, 6, 2, 8, 7, 2, 8, 8, 2, 8, 9, 2, 9, 3, 2, 9, 4, 2, 9, 5, 2, 9, 6, 2, 9, 7, 2, 9, 8, 2, 9, 9, 3, 3, 3, 4, 3, 3, 5, 3, 3, 6, 3, 3, 7, 3, 3, 8, 3, 3, 9, 3, 4, 4, 3, 4, 5, 3, 4, 6, 3, 4, 7, 3, 4, 8, 3, 4, 9, 3, 5, 4, 3, 5, 5, 3, 5, 6, 3, 5, 7, 3, 5, 8, 3, 5, 9, 3, 6, 4, 3, 6, 5, 3, 6, 6, 3, 6, 7, 3, 6, 8, 3, 6, 9, 3, 7, 4, 3, 7, 5, 3, 7, 6, 3, 7, 7, 3, 7, 8, 3, 7, 9, 3, 8, 4, 3, 8, 5, 3, 8, 6, 3, 8, 7, 3, 8, 8, 3, 8, 9, 3, 9, 4, 3, 9, 5, 3, 9, 6, 3, 9, 7, 3, 9, 8, 3, 9, 9, 4, 4, 4, 5, 4, 4, 6, 4, 4, 7, 4, 4, 8, 4, 4, 9, 4, 5, 5, 4, 5, 6, 4, 5, 7, 4, 5, 8, 4, 5, 9, 4, 6, 5, 4, 6, 6, 4, 6, 7, 4, 6, 8, 4, 6, 9, 4, 7, 5, 4, 7, 6, 4, 7, 7, 4, 7, 8, 4, 7, 9, 4, 8, 5, 4, 8, 6, 4, 8, 7, 4, 8, 8, 4, 8, 9, 4, 9, 5, 4, 9, 6, 4, 9, 7, 4, 9, 8, 4, 9, 9, 5, 5, 5, 6, 5, 5, 7, 5, 5, 8, 5, 5, 9, 5, 6, 6, 5, 6, 7, 5, 6, 8, 5, 6, 9, 5, 7, 6, 5, 7, 7, 5, 7, 8, 5, 7, 9, 5, 8, 6, 5, 8, 7, 5, 8, 8, 5, 8, 9, 5, 9, 6, 5, 9, 7, 5, 9, 8, 5, 9, 9, 6, 6, 6, 7, 6, 6, 8, 6, 6, 9, 6, 7, 7, 6, 7, 8, 6, 7, 9, 6, 8, 7, 6, 8, 8, 6, 8, 9, 6, 9, 7, 6, 9, 8, 6, 9, 9, 7, 7, 7, 8, 7, 7, 9, 7, 8, 8, 7, 8, 9, 7, 9, 8, 7, 9, 9, 8, 8, 8, 9, 8, 9, 9, 9, 1, 0

Consider the quadruplet number in base 10. There are 10000 possible quadruplets, so the all-quadruplet number is 10003 digits long. And so on. In general, the “all n-tuplet” number in base b contains b^n n-tuplets and is (b^n + n-1) digits long. If b = 10 and n = 4, the d-string starts like this:

1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2, 0, 0, 0, 2, 1, 0, 0, 3, 0, 0, 0, 3, 1, 0, 0, 4, 0, 0, 0, 4, 1, 0, 0, 5, 0, 0, 0, 5, 1, 0, 0, 6, 0, 0, 0, 6, 1, 0, 0, 7, 0, 0, 0, 7, 1, 0, 0, 8, 0, 0, 0, 8, 1, 0, 0, 9, 0, 0, 0, 9, 1, 0, 1, 0, 1, 1, 1, 0, 1, 2, 0, 0, 1, 2, 1, 0, 1, 3, 0, 0, 1, 3, 1, 0, 1, 4, 0, 0, 1, 4, 1, 0, 1, 5, 0, 0, 1, 5, 1, 0, 1, 6, 0, 0, 1, 6, 1, 0, 1, 7, 0, 0, 1, 7, 1, 0, 1, 8, 0, 0, 1, 8, 1, 0, 1, 9, 0, 0, 1, 9, 1, 0, 2, 0, 1, 0, 2, 1, 1, 0, 2, 2, 0, 0, 2, 2, 1, 0, 2, 3, 0, 0, 2, 3, 1, 0, 2, 4, 0, 0, 2, 4, 1, 0, 2, 5, 0, 0, 2, 5, 1, 0, 2, 6…

What about when n = 100? Now the d-string is ungraspably huge – too big to fit in the known universe. But it starts with 1 followed by a hundred 0s and every digit after that is entirely determined. Perhaps there’s a simple way to calculate any given digit, given its position in the d-string. Either way, what is the ontological status of the d-string for n=100? Does it exist in some Platonic realm of number, independent of physical reality?

Some would say that it does, just like √2 or π or e. I disagree. I don’t believe in a Platonic realm. If the universe or multiverse ceased to exist, numbers and mathematics in general would also cease to exist. But this isn’t to say that mathematics depends on physical reality. It doesn’t. Nor does physical reality depend on mathematics. Rather, physical reality necessarily embodies mathematics, which might be defined as “entity in interrelation”. Humans have invented small-m mathematics, a symbolic way of expressing the physical embodiment of big-m mathematics.

But small-m mathematics is actually more powerful and far-ranging, because it increases the number, range and power of entities and their interaction. Where are √2 and π in physical reality? Nowhere. You could say that early mathematicians saw their shadows, cast from a Platonic realm, and deduced their existence in that realm, but that’s a metaphor. Do all events, like avalanches or thunderstorms, exist in some Platonic realm before they are realized? No, they arise as physical entities interact according to laws of physics. In a more abstract way, √2 and π arise as entities of another kind interact according to laws of logic: the concepts of a square and its diagonal, of a circle and its diameter.

The d-strings discussed above arise from the interaction of simpler concepts: the finite set of digits in a base and the ways in which they can be combined. Platonism is unnecessary: the arc and spray of a fountain are explained by the pressure of the water, the design of the pipes, the arrangement of the nozzles, not by reference to an eternal archetype of water and spray. In small-m mathematics, there are an infinite number of fountains, because small-m mathematics opens a door to a big-U universe, infinitely larger and richer than the small-u universe of physical reality.

Summus

by in Base Behavior, Binary numbers, Digit-sums, Mathematics, Palindromic numbers and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

I’m interested in digit-sums and in palindromic numbers. Looking at one, I found the other. It started like this: 9^2 = 81 and 9 = 8 + 1, so digitsum(9^1) = digitsum(9^2). I wondered how long such a sequence of powers could be (excluding powers of 10). I quickly found that the digit-sum of 468 is equal to the digit-sum of its square and cube:

digsum(468) = digsum(219024) = digsum(102503232)

But I couldn’t find any longer sequence, although plenty of other numbers are similar to 468:

digsum(585) = digsum(342225) = digsum(200201625)
digsum(4680) = digsum(21902400) = digsum(102503232000)
digsum(5850) = digsum(34222500) = digsum(200201625000)
digsum(5851) = digsum(34234201) = digsum(200304310051)
digsum(5868) = digsum(34433424) = digsum(202055332032)
digsum(28845) = digsum(832034025) = digsum(24000021451125) […]
digsum(589680) = digsum(347722502400) = digsum(205045005215232000)

What about other bases? First came this sequence:

digsum(2) = digsum(11) (base = 3) (highest power = 2)

Then these:

digsum(4) = digsum(22) = digsum(121) (b=7) (highest power = 3)
digsum(8) = digsum(44) = digsum(242) = digsum(1331) (b=15) (hp=4)
digsum([16]) = digsum(88) = digsum(484) = digsum(2662) = digsum(14641) (b=31) (hp=5)

The pattern continues (a number between square brackets represents a single digit in the base):

digsum([32]) = digsum([16][16]) = digsum(8[16]8) = digsum(4[12][12]4) = digsum(28[12]82) = digsum(15[10][10]51) (b=63) (hp=6)
digsum([64]) = digsum([32][32]) = digsum([16][32][16]) = digsum(8[24][24]8) = digsum(4[16][24][16]4) = digsum(2[10][20][20][10]2) = digsum(16[15][20][15]61) (b=127) (hp=7)
digsum([128]) = digsum([64][64]) = digsum([32][64][32]) = digsum([16][48][48][16]) = digsum(8[32][48][32]8) = digsum(4[20][40][40][20]4) = digsum(2[12][30][40][30][12]2) = digsum(17[21][35][35][21]71) (b=255) (hp=8)
digsum([256]) = digsum([128][128]) = digsum([64][128][64]) = digsum([32][96][96][32]) = digsum([16][64][96][64][16]) = digsum(8[40][80][80][40]8) = digsum(4[24][60][80][60][24]4) = digsum(2[14][42][70][70][42][14]2) = digsum(18[28][56][70][56][28]81) (b=511) (hp=9)

After this, I looked at sequences in which n(i) = n(i-1) + digitsum(n(i-1)). How long could digitsum(n(i)) be greater than or equal to digitsum(n(i-1))? In base 10, I found these sequences:

1 (digitsum=1) → 2 → 4 → 8 → 16 (sum=7) (count=4) (base=10)
9 → 18 (sum=9) → 27 (s=9) → 36 (s=9) → 45 (s=9) → 54 (s=9) → 63 (s=9) → 72 (s=9) → 81 (s=9) → 90 (s=9) → 99 (s=18) → 117 (s=9) (c=11) (b=10)
801 (s=9) → 810 (s=9) → 819 (s=18) → 837 (s=18) → 855 (s=18) → 873 (s=18) → 891 (s=18) → 909 (s=18) → 927 (s=18) → 945 (s=18) → 963 (s=18) → 981 (s=18) → 999 (s=27) → 1026 (s=9) (c=13)

Base 2 does better:

1 → 10 (s=1) → 11 (s=2) → 101 (s=2) → 111 (s=3) → 1010 (s=2) (c=5) (b=2)
16 = 10000 (s=1) → 10001 (s=2) → 10011 (s=3) → 10110 (s=3) → 11001 (s=3) → 11100 (s=3) → 11111 (s=5) → 100100 (s=2) (c=7) (b=2)
962 = 1111000010 (s=5) → 1111000111 (s=7) → 1111001110 (s=7) → 1111010101 (s=7) → 1111011100 (s=7) → 1111100011 (s=7) → 1111101010 (s=7) → 1111110001 (s=7) → 1111111000 (s=7) → 1111111111 (s=10) → 10000001001 (s=3) (c=10) (b=2)
524047 = 1111111111100001111 (s=15) → 1111111111100011110 (s=15) → 1111111111100101101 (s=15) → 1111111111100111100 (s=15) → 1111111111101001011 (s=15) → 1111111111101011010 (s=15) → 1111111111101101001(s=15) → 1111111111101111000 (s=15) → 1111111111110000111 (s=15) → 1111111111110010110 (s=15) → 1111111111110100101 (s=15) → 1111111111110110100 (s=15) → 1111111111111000011 (s=15) → 1111111111111010010 (s=15) → 1111111111111100001 (s=15) → 1111111111111110000 (s=15) → 1111111111111111111 (s=19) → 10000000000000010010 (s=3) (c=17) (b=2)

The best sequence I found in base 3 is shorter than in base 10, but there are more sequences:

1 → 2 → 11 (s=2) → 20 (s=2) → 22 (s=4) → 110 (s=2) (c=5) (b=3)
31 = 1011 (s=3) → 1021 (s=4) → 1102 (s=4) → 1120 (s=4) → 1201 (s=4) → 1212 (s=6) → 2002 (s=4) (c=6) (b=3)
54 = 2000 (s=2) → 2002 (s=4) → 2020 (s=4) → 2101 (s=4) → 2112 (s=6) → 2202 (s=6) → 2222 (s=8) → 10021(s=4) (c=7) (b=3)
432 = 121000 (s=4) → 121011 (s=6) → 121101 (s=6) → 121121 (s=8) → 121220 (s=8) → 122012 (s=8) → 122111 (s=8) → 122210 (s=8) → 200002 (s=4) (c=8) (b=3)
648 = 220000 (s=4) → 220011 (s=6) → 220101 (s=6) → 220121 (s=8) → 220220 (s=8) → 221012 (s=8) → 221111 (s=8) → 221210 (s=8) → 222002 (s=8) → 222101 (s=8) → 222200 (s=8) → 222222 (s=12) → 1000102 (s=4) (c=12) (b=3)

And what about sequences in which digitsum(n(i)) is always greater than digitsum(n(i-1))? Base 10 is disappointing:

1 → 2 → 4 → 8 → 16 (sum=7) (count=4) (base=10)
50 (s=5) → 55 (s=10) → 65 (s=11) → 76 (s=13) → 89 (s=17) → 106 (s=7) (c=5) (b=10)

Some other bases do better:

2 = 10 (s=1) → 11 (s=2) → 101 (s=2) (c=2) (b=2)
4 = 100 (s=1) → 101 (s=2) → 111 (s=3) → 1010 (s=2) (c=3) (b=2)
240 = 11110000 (s=4) → 11110100 (s=5) → 11111001 (s=6) → 11111111 (s=8) → 100000111 (s=4) (c=4) (b=2)

1 → 2 → 11 (s=2) (c=2) (b=3)
19 = 201 (s=3) → 211 (s=4) → 222 (s=6) → 1012 (s=4) (c=3) (b=3)
58999 = 2222221011 (s=15) → 2222221201 (s=16) → 2222222022 (s=18) → 2222222222 (s=20) → 10000000201 (s=4) (c=4) (b=3)

1 → 2 → 10 (s=1) (c=2) (b=4)
4 = 10 (s=1) → 11 (s=2) → 13 (s=4) → 23 (s=5) → 100 (s=1) (c=4) (b=4)
977 = 33101 (s=8) → 33121 (s=10) → 33203 (s=11) → 33232 (s=13) → 33323 (s=14) → 100021 (s=4) (c=5) (b=4)

1 → 2 → 4 → 13 (s=4) (c=3) (b=5)
105 = 410 (s=5) → 420 (s=6) → 431 (s=8) → 444 (s=12) → 1021 (s=4) (c=4) (b=5)

1 → 2 → 4 → 12 (s=3) (c=3) (b=6)
13 = 21 (s=3) → 24 (s=6) → 34 (s=7) → 45 (s=9) → 102 (s=3) (c=4) (b=6)
396 = 1500 (s=6) → 1510 (s=7) → 1521 (s=9) → 1534 (s=13) → 1555 (s=16) → 2023 (s=7) (c=5) (b=6)

1 → 2 → 4 → 11 (s=2) (c=3) (b=7)
121 = 232 (s=7) → 242 (s=8) → 253 (s=10) → 266 (s=14) → 316 (s=10) (c=4) (b=7)
205 = 412 (s=7) → 422 (s=8) → 433 (s=10) → 446 (s=14) → 466 (s=16) → 521 (s=8) (c=5) (b=7)

1 → 2 → 4 → 10 (s=1) (c=3) (b=8)
8 = 10 (s=1) → 11 (s=2) → 13 (s=4) → 17 (s=8) → 27 (s=9) → 40 (s=4) (c=5) (b=8)
323 = 503 (s=8) → 513 (s=9) → 524 (s=11) → 537 (s=15) → 556 (s=16) → 576 (s=18) → 620 (s=8) (c=6) (b=8)

1 → 2 → 4 → 8 → 17 (s=8) (c=4) (b=9)
6481 = 8801 (s=17) → 8820 (s=18) → 8840 (s=20) → 8862 (s=24) → 8888 (s=32) → 10034 (s=8) (c=5) (b=9)

1 → 2 → 4 → 8 → 16 (s=7) (c=4) (b=10)
50 (s=5) → 55 (s=10) → 65 (s=11) → 76 (s=13) → 89 (s=17) → 106 (s=7) (c=5) (b=10)

1 → 2 → 4 → 8 → 15 (s=6) (c=4) (b=11)
1013 = 841 (s=13) → 853 (s=16) → 868 (s=22) → 888 (s=24) → 8[10][10] (s=28) → 925 (s=16) (c=5) (b=11)

1 → 2 → 4 → 8 → 14 (s=5) (c=4) (b=12)
25 = 21 (s=3) → 24 (s=6) → 2[10] (s=12) → 3[10] (s=13) → 4[11] (s=15) → 62 (s=8) (c=5) (b=12)
1191 = 833 (s=14) → 845 (s=17) → 85[10] (s=23) → 879 (s=24) → 899 (s=26) → 8[11][11] (s=30) → 925 (s=16) (c=6) (b=12)

1 → 2 → 4 → 8 → 13 (s=4) (c=4) (b=13)
781 = 481 (s=13) → 491 (s=14) → 4[10]2 (s=16) → 4[11]5 (s=20) → 4[12][12] (s=28) → 521 (s=8) (c=5) (b=13)
19621 = 8[12]14 (s=25) → 8[12]33 (s=26) → 8[12]53 (s=28) → 8[12]75 (s=32) → 8[12]9[11] (s=40) → 8[12][12][12] (s=44) → 9034 (s=16) (c=6) (b=13)

1 → 2 → 4 → 8 → 12 (s=3) (c=4) (b=14)
72 = 52 (s=7) → 59 (s=14) → 69 (s=15) → 7[10] (s=17) → 8[13] (s=21) → [10]6 (s=16) (c=5) (b=14)
1275 = 671 (s=14) → 681 (s=15) → 692 (s=17) → 6[10]5 (s=21) → 6[11][12] (s=29) → 6[13][13] (s=32) → 723 (s=12) (c=6) (b=14)
19026 = 6[13]10 (s=20) → 6[13]26 (s=27) → 6[13]45 (s=28) → 6[13]65 (s=30) → 6[13]87 (s=34) → 6[13][10][13] (s=42) → 6[13][13][13] (s=45) → 7032 (s=12) (c=7) (b=14)

1 → 2 → 4 → 8 → 11 (s=2) (c=4) (b=15)
603 = 2[10]3 (s=15) → 2[11]3 (s=16) → 2[12]4 (s=18) → 2[13]7 (s=22) → 2[14][14] (s=30) → 31[14] (s=18) (c=5) (b=15)
1023 = 483 (s=15) → 493 (s=16) → 4[10]4 (s=18) → 4[11]7 (s=22) → 4[12][14] (s=30) → 4[14][14] (s=32) → 521 (s=8) (c=6) (b=15)
1891 = 861 (s=15) → 871 (s=16) → 882 (s=18) → 895 (s=22) → 8[10][12] (s=30) → 8[12][12] (s=32) → 8[14][14] (s=36) → 925 (s=16) (c=7) (b=15)

1 → 2 → 4 → 8 → 10 (s=1) (c=4) (b=16)
16 = 10 (s=1) → 11 (s=2) → 13 (s=4) → 17 (s=8) → 1[15] (s=16) → 2[15] (s=17) → 40 (s=4) (c=6) (b=16)
1396 = 574 (s=16) → 584 (s=17) → 595 (s=19) → 5[10]8 (s=23) → 5[11][15] (s=31) → 5[13][14] (s=32) → 5[15][14] (s=34) → 620 (s=8) (c=7) (b=16)
2131 = 853 (s=16) → 863 (s=17) → 874 (s=19) → 887 (s=23) → 89[14] (s=31) → 8[11][13] (s=32) → 8[13][13] (s=34) → 8[15][15] (s=38) → 925 (s=16) (c=8) (b=16)

1 → 2 → 4 → 8 → [16] (s=16) → 1[15] (s=16) (c=5) (b=17)

1 → 2 → 4 → 8 → [16] (s=16) → 1[14] (s=15) (c=5) (b=18)
5330 = [16]82 (s=26) → [16]9[10] (s=35) → [16][11]9 (s=36) → [16][13]9 (s=38) → [16][15][11] (s=42) → [16][17][17] (s=50) → [17]2[13] (s=32) (c=6) (b=18)

1 → 2 → 4 → 8 → [16] (s=16) → 1[13] (s=14) (c=5) (b=19)
116339 = [16][18]52 (s=41) → [16][18]75 (s=46) → [16][18]9[13] (s=56) → [16][18][12][12] (s=58) → [16][18][15][13] (s=62) → [16][18][18][18] (s=70) → [17]03[12] (s=32) (c=6) (b=19)

1 → 2 → 4 → 8 → [16] (s=16) → 1[12] (s=13) (c=5) (b=20)
100 = 50 (s=5) → 55 (s=10) → 5[15] (s=20) → 6[15] (s=21) → 7[16] (s=23) → 8[19] (s=27) → [10]6 (s=16) (c=6) (b=20)
135665 = [16][19]35 (s=43) → [16][19]58 (s=48) → [16][19]7[16] (s=58) → [16][19][10][14] (s=59) → [16][19][13][13] (s=61) → [16][19][16][14] (s=65) → [16][19][19][19] (s=73) → [17]03[12] (s=32) (c=7) (b=20)

Spijit

by in Base Behavior, Mathematics, Ulam Spirals and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

The only two digits found in all standard bases are 1 and 0. But they behave quite differently. Suppose you take the integers 1 to 100 and compare the number of 1s and 0s in the representation of each integer, n, in bases 2 to n-1. For example, 10 would look like this:

1010 in base 2
101 in base 3
22 in base 4
20 in base 5
14 in base 6
13 in base 7
12 in base 8
11 in base 9

So there are nine 1s and four 0s. If you check 1 to 100 using this all-base function, the count of 1s goes like this:

1, 1, 2, 3, 5, 5, 8, 5, 9, 9, 11, 10, 15, 12, 14, 13, 15, 12, 17, 14, 20, 19, 20, 15, 23, 19, 22, 22, 25, 24, 31, 21, 25, 24, 24, 27, 33, 27, 31, 29, 34, 29, 36, 30, 34, 35, 34, 30, 40, 33, 36, 35, 38, 34, 42, 37, 43, 40, 41, 37, 48, 39, 42, 42, 44, 43, 48, 43, 47, 46, 51, 42, 53, 44, 48, 50, 51, 50, 55, 48, 59, 55, 55, 54, 64, 57, 57, 55, 60, 57, 68, 60, 64, 63, 64, 59, 68, 58, 61, 63.

And the count of 0s goes like this:

0, 1, 0, 2, 1, 2, 0, 4, 4, 4, 2, 5, 1, 2, 2, 7, 4, 8, 4, 7, 4, 3, 1, 8, 4, 4, 6, 8, 4, 7, 1, 10, 8, 7, 7, 12, 5, 6, 5, 10, 4, 8, 2, 6, 7, 4, 2, 12, 6, 9, 7, 8, 4, 11, 6, 10, 5, 4, 2, 12, 2, 3, 5, 14, 11, 13, 7, 10, 8, 11, 5, 17, 7, 8, 10, 10, 8, 10, 4, 13, 12, 10, 8, 16, 8, 7, 7, 12, 6, 14, 6, 8, 5, 4, 4, 16, 6, 10, 11, 15.

The bigger the numbers get, the bigger the discrepancies get. Sometimes the discrepancy is dramatic. For example, suppose you represented the prime 1014719 in bases 2 to 1014718. How 0s would there be? And how many 1s? There are exactly nine zeroes:

1014719 = 11110111101110111111 in base 2 = 1220112221012 in base 3 = 40B27B in base 12 = 1509CE in base 15 = 10[670] in base 1007.

But there are 507723 ones. The same procedure applied to the next integer, 1014720, yields 126 zeroes and 507713 ones. However, there is a way to see that 1s and 0s in the all-base representation are behaving in a similar way. To do this, imagine listing the individual digits of n in bases 2 to n-1 (or just base 2, if n <= 3). When the digits aren’t individual they look like this:

1 = 1 in base 2
2 = 10 in base 2
3 = 11 in base 2
4 = 100 in base 2; 11 in base 3
5 = 101 in base 2; 12 in base 3; 11 in base 4
6 = 110 in base 2; 20 in base 3; 12 in base 4; 11 in base 5
7 = 111 in base 2; 21 in base 3; 13 in base 4; 12 in base 5; 11 in base 6
8 = 1000 in base 2; 22 in base 3; 20 in base 4; 13 in base 5; 12 in base 6; 11 in base 7
9 = 1001 in base 2; 100 in base 3; 21 in base 4; 14 in base 5; 13 in base 6; 12 in base 7; 11 in base 8
10 = 1010 in base 2; 101 in base 3; 22 in base 4; 20 in base 5; 14 in base 6; 13 in base 7; 12 in base 8; 11 in base 9

So the list would look like this:

1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 2, 1, 1, 1, 1, 0, 2, 0, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 1, 0, 0, 0, 2, 2, 2, 0, 1, 3, 1, 2, 1, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 4, 1, 3, 1, 2, 1, 1, 1, 0, 1, 0, 1, 0, 1, 2, 2, 2, 0, 1, 4, 1, 3, 1, 2, 1, 1

Suppose that these digits are compared against the squares of a counter-clockwise spiral on a rectangular grid. If the spiral digit is equal to 1, the square is filled in; if the spijit is not equal to 1, the square is left blank. The 1-spiral looks like this:
1spiral
Now try zero. If the spijit is equal to 0, the square is filled in; if not, the square is left blank. The 0-spiral looks like this:
0spiral
And here’s an animated gif of the n-spiral for n = 0..9:
animspiral

N-route

by in Binary numbers, Digit-sums, Geometry, Mathematics and tagged , , , , , , , , , , , , , , , , , , | Leave a comment

In maths, one thing leads to another. I wondered whether, in a spiral of integers, any number was equal to the digit-sum of the numbers on the route traced by moving to the origin first horizontally, then vertically. To illustrate the procedure, here is a 9×9 integer spiral containing 81 numbers:

| 65 | 64 | 63 | 62 | 61 | 60 | 59 | 58 | 57 |
| 66 | 37 | 36 | 35 | 34 | 33 | 32 | 31 | 56 |
| 67 | 38 | 17 | 16 | 15 | 14 | 13 | 30 | 55 |
| 68 | 39 | 18 | 05 | 04 | 03 | 12 | 29 | 54 |
| 69 | 40 | 19 | 06 | 01 | 02 | 11 | 28 | 53 |
| 70 | 41 | 20 | 07 | 08 | 09 | 10 | 27 | 52 |
| 71 | 42 | 21 | 22 | 23 | 24 | 25 | 26 | 51 |
| 72 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |
| 73 | 74 | 75 | 76 | 77 | 78 | 79 | 80 | 81 |

Take the number 21, which is three places across and up from the bottom left corner of the spiral. The route to the origin contains the numbers 21, 22, 23, 8 and 1, because first you move right two places, then up two places. And 21 is what I call a route number, because 21 = 3 + 4 + 5 + 8 + 1 = digitsum(21) + digitsum(22) + digitsum(23) + digitsum(8) + digitsum(1). Beside the trivial case of 1, there are two more route numbers in the spiral:

58 = 13 + 14 + 6 + 7 + 7 + 6 + 4 + 1 = digitsum(58) + digitsum(59) + digitsum(60) + digitsum(61) + digitsum(34) + digitsum(15) + digitsum(4) + digitsum(1).

74 = 11 + 12 + 13 + 14 + 10 + 5 + 8 + 1 = digitsum(74) + digitsum(75) + digitsum(76) + digitsum(77) + digitsum(46) + digitsum(23) + digitsum(8) + digitsum(1).

Then I wondered about other possible routes to the origin. Think of the origin as one corner of a rectangle and the number being tested as the diagonal corner. Suppose that you always move away from the starting corner, that is, you always move up or right (or up and left, and so on, depending on where the corners lie). In a x by y rectangle, how many routes are there between the diagonal corners under those conditions?

It’s an interesting question, but first I’ve looked at the simpler case of an n by n square. You can encode each route as a binary number, with 0 representing a vertical move and 1 representing a horizontal move. The problem then becomes equivalent to finding the number of distinct ways you can arrange equal numbers of 1s and 0s. If you use this method, you’ll discover that there are two routes across the 2×2 square, corresponding to the binary numbers 01 and 10:

2x2

Across the 3×3 square, there are six routes, corresponding to the binary numbers 0011, 0101, 0110, 1001, 1010 and 1100:

3x3

Across the 4×4 square, there are twenty routes:
4x4

(Please open in new window if it fails to animate)

(Please open in new window if it fails to animate)

Across the 5×5 square, there are 70 routes:

5x5

(Please open in new window etc)

(Please open in new window etc)

Across the 6×6 and 7×7 squares, there are 252 and 924 routes:

6x6

7x7

After that, the routes quickly increase in number. This is the list for n = 1 to 14:

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600… (see A000984 at the Online Encyclopedia of Integer Sequences)

After that you can vary the conditions. What if you can move not just vertically and horizontally, but diagonally, i.e. vertically and horizontally at the same time? Now you can encode the route with a ternary number, or number in base 3, with 0 representing a vertical move, 1 a horizontal move and 2 a diagonal move. As before, there is one route across a 1×1 square, but there are three across a 2×2, corresponding to the ternary numbers 01, 2 and 10:

3x3t

There are 13 routes across a 3×3 square, corresponding to the ternary numbers 0011, 201, 021, 22, 0101, 210, 1001, 120, 012, 102, 0110, 1010, 1100:

4x4t

And what about cubes, hypercubes and higher?

Will Two Power?

by in Mathematics, Powers and tagged , , , , , , | Leave a comment

It’s such a simple thing: repeatedly doubling a number: 1, 2, 4, 8, 16, 32, 61, 128… And yet it yields such riches, reminiscent of DNA or a literary text:

2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^10 = 1024
2^20 = 1048576
2^30 = 1073741824
2^40 = 1099511627776
2^50 = 1125899906842624
2^60 = 1152921504606846976
2^70 = 1180591620717411303424
2^80 = 1208925819614629174706176
2^90 = 1237940039285380274899124224
2^100 = 1267650600228229401496703205376
2^200 = 1606938044258990275541962092341162602522202993782792835301376

Although, by Benford’s law*, 1 is the commonest leading digit, do all numbers eventually appear as the leading digits of some power of 2? I conjecture that they do. indeed, I conjecture that they do infinitely often. If the function first(n) returns the power of 2 whose leading digits are the same as the digits of n, then:

first(1) = 2^0 = 1
first(2) = 2^1 = 2
first(3) = 2^5 = 32
first(4) = 2^2 = 4
first(5) = 2^9 = 512
first(6) = 2^6 = 64
first(7) = 2^46 = 70368744177664
first(8) = 2^3 = 8
first(9) = 2^53 = 9007199254740992
first(10) = 2^10 = 1024

And I conjecture that this is true of all bases except bases that are powers of 2, like 2, 4, 8, 16 and so on. A related question is whether the leading digits of any 2^n are the same as the digits of n. Yes:

2^6 = 64
2^10 = 1024
2^1542 = 1.54259995… * 10^464
2^77075 = 7.70754024… * 10^23201
2^113939 = 1.13939932… * 10^34299
2^1122772 = 1.12277217… * 10^337988

That looks like a look of calculation, but there’s a simple way to cut it down: restrict the leading digits. Eventually they will lose accuracy, because the missing digits are generating carries. With four leading digits, this happens:

1: 0001
2: 0002
4: 0004
8: 0008
16: 0016
32: 0032
64: 0064
128: 0128
256: 0256
512: 0512
1024: 1024
2048: 2048
4096: 4096
8192: 8192
16384: 1638…
32768: 3276…
65536: 6552…

But working with only fifteen leading digits, you can find that 1122772 = the leading digits of 2^1122772, which has 337989 digits when calculated in full.

Previously pre-posted (please peruse):

• Talcum Power

*Not Zipf’s law, as I originally said.

Talcum Power

by in Base Behavior, Mathematics, Powers, Prime numbers and tagged , , , , , , , , , , , , , , , , , , , , , | Leave a comment

If primes are like diamonds, powers of 2 are like talc. Primes don’t crumble under division, because they can’t be divided by any number but themselves and one. Powers of 2 crumble more than any other numbers. The contrast is particularly strong when the primes are Mersenne primes, or equal to a power of 2 minus 1:

3 = 4-1 = 2^2 – 1.
4, 2, 1.

7 = 8-1 = 2^3 – 1.
8, 4, 2, 1.

31 = 32-1 = 2^5 – 1.
32, 16, 8, 4, 2, 1.

127 = 2^7 – 1.
128, 64, 32, 16, 8, 4, 2, 1.

8191 = 2^13 – 1.
8192, 4096, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1.

131071 = 2^17 – 1.
131072, 65536, 32768, 16384, 8192, 4096, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1.

524287 = 2^19 – 1.
524288, 262144, 131072, 65536, 32768, 16384, 8192, 4096, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1.

2147483647 = 2^31 – 1.
2147483648, 1073741824, 536870912, 268435456, 134217728, 67108864, 33554432, 16777216, 8388608, 4194304, 2097152, 1048576, 524288, 262144, 131072, 65536, 32768, 16384, 8192, 4096, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1.

Are Mersenne primes infinite? If they are, then there will be just as many Mersenne primes as powers of 2, even though very few powers of 2 create a Mersenne prime. That’s one of the paradoxes of infinity: an infinite part is equal to an infinite whole.

But are they infinite? No-one knows, though some of the greatest mathematicians in history have tried to find a proof or disproof of the conjecture. A simpler question about powers of 2 is this: Does every integer appear as part of a power of 2? I can’t find one that doesn’t:

0 is in 1024 = 2^10.
1 is in 16 = 2^4.
2 is in 32 = 2^5.
3 is in 32 = 2^5.
4 = 2^2.
5 is in 256 = 2^8.
6 is in 16 = 2^4.
7 is in 32768 = 2^15.
8 = 2^3.
9 is in 4096 = 2^12.
10 is in 1024 = 2^10.
11 is in 1099511627776 = 2^40.
12 is in 128 = 2^7.
13 is in 131072 = 2^17.
14 is in 262144 = 2^18.
15 is in 2097152 = 2^21.
16 = 2^4.
17 is in 134217728 = 2^27.
18 is in 1073741824 = 2^30.
19 is in 8192 = 2^13.
20 is in 2048 = 2^11.

666 is in 182687704666362864775460604089535377456991567872 = 2^157.
1066 is in 43556142965880123323311949751266331066368 = 2^135.
1492 is in 356811923176489970264571492362373784095686656 = 2^148.
2014 is in 3705346855594118253554271520278013051304639509300498049262642688253220148477952 = 2^261.

I’ve tested much higher than that, but testing is no good: where’s a proof? I don’t have one, though I conjecture that all integers do appear as part or whole of a power of 2. Nor do I have a proof for another conjecture: that all integers appear infinitely often as part or whole of powers of 2. Or indeed, of powers of 3, 4, 5 or any other number except powers of 10.

I conjecture that this would apply in all bases too: In any base b all n appear infinitely often as part or whole of powers of any number except those equal to a power of b.

1 is in 11 = 2^2 in base 3.
2 is in 22 = 2^3 in base 3.
10 is in 1012 = 2^5 in base 3.
11 = 2^2 in base 3.
12 is in 121 = 2^4 in base 3.
20 is in 11202 = 2^7 in base 3.
21 is in 121 = 2^4 in base 3.
22 = 2^3 in base 3.
100 is in 100111 = 2^8 in base 3.
101 is in 1012 = 2^5 in base 3.
102 is in 2210212 = 2^11 in base 3.
110 is in 1101221 = 2^10 in base 3.
111 is in 100111 = 2^8 in base 3.
112 is in 11202 = 2^7 in base 3.
120 is in 11202 = 2^7 in base 3.
121 = 2^4 in base 3.
122 is in 1101221 = 2^10 in base 3.
200 is in 200222 = 2^9 in base 3.
201 is in 12121201 = 2^12 in base 3.
202 is in 11202 = 2^7 in base 3.

1 is in 13 = 2^3 in base 5.
2 is in 112 = 2^5 in base 5.
3 is in 13 = 2^3 in base 5.
4 = 2^2 in base 5.
10 is in 1003 = 2^7 in base 5.
11 is in 112 = 2^5 in base 5.
12 is in 112 = 2^5 in base 5.
13 = 2^3 in base 5.
14 is in 31143 = 2^11 in base 5.
20 is in 2011 = 2^8 in base 5.
21 is in 4044121 = 2^16 in base 5.
22 is in 224 = 2^6 in base 5.
23 is in 112341 = 2^12 in base 5.
24 is in 224 = 2^6 in base 5.
30 is in 13044 = 2^10 in base 5.
31 = 2^4 in base 5.
32 is in 230232 = 2^13 in base 5.
33 is in 2022033 = 2^15 in base 5.
34 is in 112341 = 2^12 in base 5.
40 is in 4022 = 2^9 in base 5.

1 is in 12 = 2^3 in base 6.
2 is in 12 = 2^3 in base 6.
3 is in 332 = 2^7 in base 6.
4 = 2^2 in base 6.
5 is in 52 = 2^5 in base 6.
10 is in 1104 = 2^8 in base 6.
11 is in 1104 = 2^8 in base 6.
12 = 2^3 in base 6.
13 is in 13252 = 2^11 in base 6.
14 is in 144 = 2^6 in base 6.
15 is in 101532 = 2^13 in base 6.
20 is in 203504 = 2^14 in base 6.
21 is in 2212 = 2^9 in base 6.
22 is in 2212 = 2^9 in base 6.
23 is in 1223224 = 2^16 in base 6.
24 = 2^4 in base 6.
25 is in 13252 = 2^11 in base 6.
30 is in 30544 = 2^12 in base 6.
31 is in 15123132 = 2^19 in base 6.
32 is in 332 = 2^7 in base 6.

1 is in 11 = 2^3 in base 7.
2 is in 22 = 2^4 in base 7.
3 is in 1331 = 2^9 in base 7.
4 = 2^2 in base 7.
5 is in 514 = 2^8 in base 7.
6 is in 2662 = 2^10 in base 7.
10 is in 1054064 = 2^17 in base 7.
11 = 2^3 in base 7.
12 is in 121 = 2^6 in base 7.
13 is in 1331 = 2^9 in base 7.
14 is in 514 = 2^8 in base 7.
15 is in 35415440431 = 2^30 in base 7.
16 is in 164351 = 2^15 in base 7.
20 is in 362032 = 2^16 in base 7.
21 is in 121 = 2^6 in base 7.
22 = 2^4 in base 7.
23 is in 4312352 = 2^19 in base 7.
24 is in 242 = 2^7 in base 7.
25 is in 11625034 = 2^20 in base 7.
26 is in 2662 = 2^10 in base 7.

1 is in 17 = 2^4 in base 9.
2 is in 152 = 2^7 in base 9.
3 is in 35 = 2^5 in base 9.
4 = 2^2 in base 9.
5 is in 35 = 2^5 in base 9.
6 is in 628 = 2^9 in base 9.
7 is in 17 = 2^4 in base 9.
8 = 2^3 in base 9.
10 is in 108807 = 2^16 in base 9.
11 is in 34511011 = 2^24 in base 9.
12 is in 12212 = 2^13 in base 9.
13 is in 1357 = 2^10 in base 9.
14 is in 314 = 2^8 in base 9.
15 is in 152 = 2^7 in base 9.
16 is in 878162 = 2^19 in base 9.
17 = 2^4 in base 9.
18 is in 218715 = 2^17 in base 9.
20 is in 70122022 = 2^25 in base 9.
21 is in 12212 = 2^13 in base 9.
22 is in 12212 = 2^13 in base 9.

Prime Climb Time

by in Base Behavior, Binary numbers, Digit-sums, Mathematics, Prime numbers and tagged , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

The third prime is equal to the sum of the first and second primes: 2 + 3 = 5. After that, for obvious reasons, the prime-sum climbs much more rapidly than the primes themselves:

2, 3, 05, 07, 11, 13, 17, 19, 023, 029...
2, 5, 10, 17, 28, 41, 58, 77, 100, 129...

But what if you use digit-sum(p1..pn), i.e., the sum of the digits of the primes from the first to the nth? For example, the digit-sum(p1..p5) = 2 + 3 + 5 + 7 + 1+1 = 19, whereas the sum(p1..p5) = 2 + 3 + 5 + 7 + 11 = 28. Using the digit-sums of the primes, the comparison now looks like this:

2, 3, 05, 07, 11, 13, 17, 19, 23, 29...
2, 5, 10, 17, 19, 23, 31, 41, 46, 57...

The sum climbs more slowly, but still too fast. So what about a different base? In base-2, the digit-sum(p1..p3) = (1+0) + (1+1) + (1+0+1) = 1 + 2 + 2 = 5. The comparison looks like this:

2, 3, 05, 07, 11, 13, 17, 19, 23, 29...
1, 3, 05, 08, 11, 14, 16, 19, 23, 27...

For primes 3, 5, 11, 19, and 23, p = digit-sum(primes <= p) in base-2. But the cumulative digit-sum soon begins to climb too slowly:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271...

1, 3, 5, 8, 11, 14, 16, 19, 23, 27, 32, 35, 38, 42, 47, 51, 56, 61, 64, 68, 71, 76, 80, 84, 87, 091, 096, 101, 106, 110, 117, 120, 123, 127, 131, 136, 141, 145, 150, 155, 160, 165, 172, 175, 179, 184, 189, 196, 201, 206, 211, 218, 223, 230, 232, 236, 240, 245...

So what about base-3?

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59...
2, 3, 6, 9, 12, 15, 20, 23, 28, 31, 34, 37, 42, 47, 52, 59, 64...

In base-3, for p = 2, 3 and 37, p = digit-sum(primes <= p), while for p = 23, 31, 47 and 59, p = digit-sum(primes < p), like this:

2 = 2.
3 = 2 + (1+0).
37 = 2 + (1+0) + (1+2) + (2+1) + (1+0+2) + (1+1+1) + (1+2+2) + (2+0+1) + (2+1+2) + (1+0+0+2) + (1+0+1+1) + (1+1+0+1) = 2 + 1 + 3 + 3 + 3 + 3 + 5 + 3 + 5 + 3 + 3 + 3.

23 = 2 + (1+0) + (1+2) + (2+1) + (1+0+2) + (1+1+1) + (1+2+2) + (2+0+1) = 2 + 1 + 3 + 3 + 3 + 3 + 5 + 3.
31 = 2 + (1+0) + (1+2) + (2+1) + (1+0+2) + (1+1+1) + (1+2+2) + (2+0+1) + (2+1+2) + (1+0+0+2) = 2 + 1 + 3 + 3 + 3 + 3 + 5 + 3 + 5 + 3.
47 = 2 + (1+0) + (1+2) + (2+1) + (1+0+2) + (1+1+1) + (1+2+2) + (2+0+1) + (2+1+2) + (1+0+0+2) + (1+0+1+1) + (1+1+0+1) + (1+1+1+2) + (1+1+2+1) = 2 + 1 + 3 + 3 + 3 + 3 + 5 + 3 + 5 + 3 + 3 + 3 + 5 + 5.
59 = 2 + (1+0) + (1+2) + (2+1) + (1+0+2) + (1+1+1) + (1+2+2) + (2+0+1) + (2+1+2) + (1+0+0+2) + (1+0+1+1) + (1+1+0+1) + (1+1+1+2) + (1+1+2+1) + (1+2+0+2) + (1+2+2+2) = 2 + 1 + 3 + 3 + 3 + 3 + 5 + 3 + 5 + 3 + 3 + 3 + 5 + 5 + 5 + 7.

This carries on for a long time. For these primes, p = digit-sum(primes < p):

23, 31, 47, 59, 695689, 698471, 883517, 992609, 992737, 993037, 1314239, 1324361, 1324571, 1326511, 1327289, 1766291, 3174029

And for these primes, p = digit-sum(primes <= p):

3, 37, 695663, 695881, 1308731, 1308757, 1313153, 1314301, 1326097, 1766227, 3204779, 14328191

Now try the cumulative digit-sum in base-4:

2, 3, 5, 07, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59...
2, 5, 7, 11, 16, 20, 22, 26, 31, 36, 43, 47, 52, 59, 67, 72, 80...

The sum of digits climbs too fast. Base-3 is the Goldilocks base, climbing neither too slowly, like base-2, nor too fast, like all bases greater than 3.

Prime Time #2

by in Mathematics, Prime numbers and tagged , , , , , , , , , , , , , | Leave a comment

“2n2 + 29 is prime for all values of n for 1 to 28.” — The Penguin Dictionary of Curious and Interesting Numbers, David Wells (1986).

• 31, 37, 47, 61, 79, 101, 127, 157, 191, 229, 271, 317, 367, 421, 479, 541, 607, 677, 751, 829, 911, 997, 1087, 1181, 1279, 1381, 1487, 1597.

Elsewhere other-posted:

• Prime Time #1
• Poulet’s Propellor — Musings on Math and Mathculinity
• La Spirale è Mobile