Magistra Rules the Waves

Monday, 30th Mar, 2015 by Krilling for Company in Base Behavior, Binary numbers, Digit-sums, Integer Sequences, Magistra Mundi, Mathematics and tagged base 10, base 13, base 16, base 17, base 2, base 25, base 33, base 49, base 9, digit sum, digit sums, integer patterns, Integer Sequences, integer spirals, integers, math, mathematics, maths, number patterns, primes, randomness, recreational math, recreational mathematics, recreational maths, regularity, spiral of integers, Square Numbers, Squares, Ulam Spirals | Leave a comment

One of my favourite integer sequences has the simple formula n(i) = n(i-1) + digitsum(n(i-1)). If it’s seeded with 1, its first few terms go like this:

n(1) = 1
n(2) = n(1) + digitsum(n(1)) = 1 + digitsum(1) = 2
n(3) = 2 + digitsum(2) = 4
n(4) = 4 + digitsum(4) = 8
n(5) = 8 + digitsum(8) = 16
n(6) = 16 + digitsum(16) = 16 + 1+6 = 16 + 7 = 23
n(7) = 23 + digitsum(23) = 23 + 2+3 = 23 + 5 = 28
n(8) = 28 + digitsum(28) = 28 + 2+8 = 28 + 10 = 38

As a sequence, it looks like this:

1, 2, 4, 8, 16, 23, 28, 38, 49, 62, 70, 77, 91, 101, 103, 107, 115, 122, 127, 137, 148, 161, 169, 185, 199, 218, 229, 242, 250, 257, 271, 281, 292, 305, 313, 320, 325, 335, 346, 359, 376, 392, 406, 416, 427, 440, 448, 464, 478, 497, 517, 530, 538, 554, 568, 587, 607, 620, 628, 644, 658, 677, 697, 719, 736, 752, 766, 785, 805, 818, 835, 851, 865, 884, 904, 917, 934, 950, 964, 983, 1003…

Given a number at random, is there a quick way to say whether it appears in the sequence seeded with 1? Not that I know, with one exception. If the number is divisible by 3, it doesn’t appear, at least in base 10. In base 2, that rule doesn’t apply:

n(1) = 1
n(2) = 1 + digitsum(1) = 10 = 1 + 1 = 2
n(3) = 10 + digitsum(10) = 10 + 1 = 11 = 2 + 1 = 3
n(4) = 11 + digitsum(11) = 11 + 1+1 = 101 = 3 + 2 = 5
n(5) = 101 + digitsum(101) = 101 + 1+0+1 = 111 = 5 + 2 = 7
n(6) = 111 + digitsum(111) = 111 + 11 = 1010 = 7 + 3 = 10
n(7) = 1010 + digitsum(1010) = 1010 + 10 = 1100 = 10 + 2 = 12
n(8) = 1100 + digitsum(1100) = 1100 + 10 = 1110 = 12 + 2 = 14

1, 2, 3, 5, 7, 10, 12, 14, 17, 19, 22, 25, 28, 31, 36, 38, 41, 44, 47, 52, 55, 60, 64, 65, 67, 70, 73, 76, 79, 84, 87, 92, 96, 98, 101, 105, 109, 114, 118, 123, 129, 131, 134, 137, 140, 143, 148, 151, 156, 160, 162, 165, 169, 173, 178, 182, 187, 193, 196, 199, 204, 208, 211, 216, 220, 225, 229, 234, 239, 246, 252, 258, 260, 262, 265, 268, 271, 276, 279, 284, 288, 290, 293, 297, 301, 306, 310, 315, 321, 324, 327, 332, 336, 339, 344, 348, 353, 357, 362, 367, 374…

What patterns are there in these sequences? It’s easier to check when they’re represented graphically, so I converted them into patterns à la the Ulam spiral, where n is represented as a dot on a spiral of integers. This is the spiral for base 10:

Base 10

And these are the spirals for bases 2 and 3:

Base 2

Base 3

These sequences look fairly random to me: there are no obvious patterns in the jumps from n(i) to n(i+1), i.e. in the values for digitsum(n(i)). Now try the spirals for bases 9 and 33:

Base 9

Base 33

Patterns have appeared: there is some regularity in the jumps. You can see these regularities more clearly if you represent digitsum(n(i)) as a graph, with n(i) on the x axis and digitsum(n(i)) on the y axis. If the graph starts with n(i) = 1 on the lower left and proceeds left-right, left-right up the screen, it looks like this in base 10:

Base 10 (click to enlarge)

Here are bases 2 and 3:

Base 2

Base 3

The jumps seem fairly random. Now try bases 9, 13, 16, 17, 25, 33 and 49:

Base 9

Base 13

Base 16

Base 17

Base 25

Base 33

Base 49

In some bases, the formula n(i) = n(i-1) + digitsum(n(i-1)) generates mild randomness. In others, it generates strong regularity, like waves rolling ashore under a steady wind. I don’t understand why, but regularity seems to occur in bases that are one more than a power of 2 and also in some bases that are primes or squares.

Elsewhere other-posted:

• Mathematica Magistra Mundi

Dig Sum Fib

Monday, 16th Mar, 2015 by Krilling for Company in Base Behavior, Digit-sums, Fibonacci Sequence, Mathematics and tagged adding digits, Arithmetic, digit sums, digits, Fibonacci Sequence, Integer Sequences, integers, math, mathematics, maths, numbers, recreational math, recreational mathematics, recreational maths | Leave a comment

The Fibonacci sequence is an infinitely rich sequence based on a very simple rule: add the previous two numbers. If the first two numbers are 1 and 1, the sequence begins like this:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025…

Plainly, the numbers increase for ever. The hundredth Fibonacci number is 354,224,848,179,261,915,075, for example, and the two-hundredth is 280,571,172,992,510,140,037,611,932,413,038,677,189,525. But there are variants on the Fibonacci sequence that don’t increase for ever. The standard rule is n(i) = n(i-2) + n(i-1). What if the rule becomes n(i) = digitsum(n(i-2)) + digitsum(n(i-1))? Now the sequence falls into a loop, like this:

1, 1, 2, 3, 5, 8, 13, 12, 7, 10, 8, 9, 17, 17, 16, 15, 13, 10, 5, 6, 11, 8, 10, 9, 10, 10, 2, 3… (length=28)

But that’s in base 10. Here are the previous bases:

1, 1, 2, 2, 2… (base=2) (length=5)
1, 1, 2, 3, 3, 2, 3… (b=3) (l=7)
1, 1, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3… (b=4) (l=12)
1, 1, 2, 3, 5, 4, 5, 5, 2, 3… (b=5) (l=10)
1, 1, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6, 2, 3… (b=6) (l=24)
1, 1, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7, 2, 3… (b=7) (l=28)
1, 1, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3… (b=8) (l=20)
1, 1, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9, 2, 3… (b=9) (l=16)

Apart from base 2, all the bases repeat with (2, 3), which is set up in each case by (base, base) = (10, 10) in that base, equivalent to (1, 1). All bases > 2 appear to repeat with (2, 3), but I don’t understand why. The length of the sequence varies widely. Here it is in bases 29, 30 and 31:

1, 1, 2, 3, 5, 8, 13, 21, 34, 27, 33, 32, 9, 13, 22, 35, 29, 8, 9, 17, 26, 43, 41, 28, 41, 41, 26, 39, 37, 20, 29, 21, 22, 43, 37, 24, 33, 29, 6, 7, 13, 20, 33, 25, 30, 27, 29, 28, 29, 29, 2, 3… (b=29) (l=52)

1, 1, 2, 3, 5, 8, 13, 21, 34, 26, 31, 28, 30, 29, 30, 30, 2, 3 (b=30) (l=18)

1, 1, 2, 3, 5, 8, 13, 21, 34, 25, 29, 54, 53, 47, 40, 27, 37, 34, 11, 15, 26, 41, 37, 18, 25, 43, 38, 21, 29, 50, 49, 39, 28, 37, 35, 12, 17, 29, 46, 45, 31, 16, 17, 33, 20, 23, 43, 36, 19, 25, 44, 39, 23, 32, 25, 27, 52, 49, 41, 30, 41, 41, 22, 33, 25, 28, 53, 51, 44, 35, 19, 24, 43, 37, 20, 27, 47, 44, 31, 15, 16, 31, 17, 18, 35, 23, 28, 51, 49, 40, 29, 39, 38, 17, 25, 42, 37, 19, 26, 45, 41, 26, 37, 33, 10, 13, 23, 36, 29, 35, 34, 9, 13, 22, 35, 27, 32, 29, 31, 30, 31, 31, 2, 3 (b=31) (l=124)

The sequence for base 77 is short like that for base 30:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 68, 81, 73, 78, 75, 77, 76, 77, 77, 2, 3 (b=77) (l=22)

But the sequence for base 51 is this:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 39, 44, 83, 77, 60, 37, 47, 84, 81, 65, 46, 61, 57, 18, 25, 43, 68, 61, 29, 40, 69, 59, 28, 37, 65, 52, 17, 19, 36, 55, 41, 46, 87, 83, 70, 53, 23, 26, 49, 75, 74, 49, 73, 72, 45, 67, 62, 29, 41, 70, 61, 31, 42, 73, 65, 38, 53, 41, 44, 85, 79, 64, 43, 57, 50, 57, 57, 14, 21, 35, 56, 41, 47, 88, 85, 73, 58, 31, 39, 70, 59, 29, 38, 67, 55, 22, 27, 49, 76, 75, 51, 26, 27, 53, 30, 33, 63, 46, 59, 55, 14, 19, 33, 52, 35, 37, 72, 59, 31, 40, 71, 61, 32, 43, 75, 68, 43, 61, 54, 15, 19, 34, 53, 37, 40, 77, 67, 44, 61, 55, 16, 21, 37, 58, 45, 53, 48, 51, 49, 50, 99, 99, 98, 97, 95, 92, 87, 79, 66, 45, 61, 56, 17, 23, 40, 63, 53, 16, 19, 35, 54, 39, 43, 82, 75, 57, 32, 39, 71, 60, 31, 41, 72, 63, 35, 48, 83, 81, 64, 45, 59, 54, 13, 17, 30, 47, 77, 74, 51, 25, 26, 51, 27, 28, 55, 33, 38, 71, 59, 30, 39, 69, 58, 27, 35, 62, 47, 59, 56, 15, 21, 36, 57, 43, 50, 93, 93, 86, 79, 65, 44, 59, 53, 12, 15, 27, 42, 69, 61, 30, 41, 71, 62, 33, 45, 78, 73, 51, 24, 25, 49, 74, 73, 47, 70, 67, 37, 54, 41, 45, 86, 81, 67, 48, 65, 63, 28, 41, 69, 60, 29, 39, 68, 57, 25, 32, 57, 39, 46, 85, 81, 66, 47, 63, 60, 23, 33, 56, 39, 45, 84, 79, 63, 42, 55, 47, 52, 49, 51, 50, 51, 51, 2, 3… (b=51) (l=304)

Summer Set Sequence

Monday, 23rd Feb, 2015 by Krilling for Company in Integer Sequences, Mathematics and tagged Arithmetic, cumulative sums, Integer Sequences, integer sums, integers, math, mathematics, maths, number sequences, recreational math, recreational mathematics, recreational maths | Leave a comment

I wondered what would happen if you added to a set of numbers, (a, b, c), the first number that wasn’t equal to the sum of any subset of the numbers: a + b, a + c, c + b, a + b + c. If the set begins with 1, the first number not equal to any subset of (1) is 2. So the set becomes (1, 2). 3 = 1 + 2, so 3 is not added. But 4 is added, making the set (1, 2, 4). The sequence of additions goes like this:

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536…

It’s the powers of 2, because some subset of the powers of 2 < 2^p will equal any number from 1 to (2^p)-1, therefore the first addition will be 2^p = the cumulative sum + 1:

1 (cumulative sum=1), 2 (cs=3), 4 (cs=7), 8 (cs=15), 16 (cs=31), 32 (cs=63), 64 (cs=127), 128 (cs=255), 256 (cs=511), 512 (cs=1023), 1024 (cs=2047), 2048 (cs=4095), 4096 (cs=8191), 8192 (cs=16383), 16384 (cs=32767), 32768 (cs=65535)…

If you seed the sequence with the set (2), the first addition is 3, but after that the powers of 2 re-appear:

2, 3, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536…

It becomes more complicated if the sequence is seeded with the set (3):

3, 4, 5, 6, 16, 17, 49, 50, 148, 149, 445, 446, 1336, 1337, 4009, 4010, 12028, 12029, 36085, 36086…

You can predict the pattern by looking at the cumulative sums again:

3, 4, 5, 6 (cumulative sum=18), 16, 17 (cs=51), 49, 50 (cs=150), 148, 149 (cs=447), 445, 446 (cs=1338), 1336, 1337 (cs=4011), 4009, 4010 (cs=12030), 12028, 12029 (cs=36087), 36085, 36086 (cs=108258)…

The sequence begins with a block of four consecutive numbers, followed by separate blocks of two consecutive numbers. The first number in each 2-block is predicted by the cumulative sum of the last number in the previous block, according to the formula n = cumulative sum – seed + 1. When the seed is 3, n = cs-3+1.

If the seed is 4, the sequences goes like this:

4, 5, 6, 7, 8, 27, 28, 29, 111, 112, 113, 447, 448, 449, 1791, 1792, 1793, 7167, 7168, 7169…

Now the sequence begins with a block of five consecutive numbers, followed by separate blocks of three consecutive numbers. The formula is n = cs-4+1:

4, 5, 6, 7, 8 (cumulative sum=30), 27, 28, 29 (cs=114), 111, 112, 113 (cs=450), 447, 448, 449 (cs=1794), 1791, 1792, 1793 (cs=7170), 7167, 7168, 7169 (cs=28674)…

And here’s the sequence seeded with (5):

5, 6, 7, 8, 9, 10, 41, 42, 43, 44, 211, 212, 213, 214, 1061, 1062, 1063, 1064, 5311, 5312, 5313, 5314…

5, 6, 7, 8, 9, 10 (cs=45), 41, 42, 43, 44 (cs=215), 211, 212, 213, 214 (cs=1065), 1061, 1062, 1063, 1064 (cs=5315), 5311, 5312, 5313, 5314 (cs=26565)…

M.i.P. Trip

Thursday, 8th Jan, 2015 by Krilling for Company in Fractals, Geometry, Mathematics and tagged animated gif, animated gifs, divide-and-discard, fractal, fractals, fractals based on squares, geometry, hexagon, hexagonal fractal, math, mathematics, maths, much in little, multum in parvo, recreational math, recreational mathematics, recreational maths, square, Squares | Leave a comment

The Latin phrase multum in parvo means “much in little”. It’s a good way of describing the construction of fractals, where the application of very simple rules can produce great complexity and beauty. For example, what could be simpler than dividing a square into smaller squares and discarding some of the smaller squares?

Yet repeated applications of divide-and-discard can produce complexity out of even a 2×2 square. Divide a square into four squares, discard one of the squares, then repeat with the smaller squares, like this:

Increase the sides of the square by a little and you increase the number of fractals by a lot. A 3×3 square yields these fractals:

And the 4×4 and 5×5 fractals yield more:

The Rite of Sling

Saturday, 27th Dec, 2014 by Krilling for Company in Base Behavior, Digit-sums, Mathematics and tagged bases, bully numbers, David and Goliath numbers, David numbers, digits, digitsum, digitsums, duelling numbers, duels, fighting numbers, Goliath numbers, integers, math, mathematics, maths, number bases, recreational math, recreational mathematics, recreational maths, sum of digits, sums of digits | 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

Monday, 1st Dec, 2014 by Krilling for Company in Fractals, Geometry, Mathematics and tagged animated gif, animated gifs, equilateral triangle, fractal, fractals, geometry, hex-ring, hex-rings, hexagon, hexagonal fractal, math, mathematics, maths, polygon, Polygons, recreational math, recreational mathematics, recreational maths, regular polygons, triangle | 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:

Adjusting the sides of the internal hexagon creates new fractals:

Discarding a hexagon after each subdivision creates new shapes:

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

Pair on a D-String

Thursday, 20th Nov, 2014 by Krilling for Company in √2, Base Behavior, Binary numbers, π, Mathematics, Philosophy, Philosophy of mathematics and tagged archetype, Arithmetic, √2, base, bases, binary numbers, combinatorics, digit, digit-strings, digits, double digits, π, number, numbers, pairs of digits, philosophy, philosophy of mathematics, Platonic realm, Platonism, quadruplets, recreational math, recreational mathematics, recreational maths, strings of digits, ternary numbers, triplets | 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

Wednesday, 29th Oct, 2014 by Krilling for Company in Base Behavior, Binary numbers, Digit-sums, Mathematics, Palindromic numbers and tagged Arithmetic, base 10, base 2, base 3, base arithmetic, base ten, bases, binary, binary numbers, different bases, digit sums, digits, digitsum, digitsums, integer, integer sequence, Integer Sequences, integers, math, mathematics, maths, number palindromes, palindromes, palindromic numbers, recreational math, recreational mathematics, recreational maths, sum of digits, sums of digits, ternary | 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

Monday, 20th Oct, 2014 by Krilling for Company in Base Behavior, Mathematics, Ulam Spirals and tagged all-base function, animated gif, animated gifs, Arithmetic, base 2, bases, binary, binary numbers, count of 0s, count of 1s, count of ones, count of zeroes, count of zeros, different bases, digit spiral, digit spirals, digits, integer, integers, math, mathematics, maths, number spiral, one, ones, prime 1014719, recreational math, recreational mathematics, recreational maths, spijit, spijits, spiral, spiral digit, spiral digits, spirals, zero, Zeroes, zeros | 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:

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:

And here’s an animated gif of the n-spiral for n = 0..9:

N-route

Tuesday, 30th Sep, 2014 by Krilling for Company in Binary numbers, Digit-sums, Geometry, Mathematics and tagged Arithmetic, binary, binary numbers, digit sums, digits, integer, integer spiral, integers, math, mathematics, maths, number spiral, recreational math, recreational mathematics, recreational maths, rectangles, Squares, ternary, ternary numbers | 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:

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

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

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

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

(Please open in new window etc)

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

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:

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:

And what about cubes, hypercubes and higher?

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