Digital Rodeo

What a difference a digit makes. Suppose you take all representations of n in bases b <= n. When n = 3, the bases are 2 and 3, so 3 = 11 and 10, respectively. Next, count the occurrences of the digit 1:

digitcount(3, digit=1, n=11, 10) = 3

Add this digit-count to 3:

3 + digitcount(3, digit=1, n=11, 10) = 3 + 3 = 6.

Now apply the same procedure to 6. The bases will be 2 to 6:

6 + digitcount(6, digit=1, n=110, 20, 12, 11, 10) = 6 + 6 = 12

The procedure, n = n + digitcount(n,digit=1,base=2..n), continues like this:

12 + digcount(12,dig=1,n=1100, 110, 30, 22, 20, 15, 14, 13, 12, 11, 10) = 12 + 11 = 23
23 + digcount(23,dig=1,n=10111, 212, 113, 43, 35, 32, 27, 25, 23, 21, 1B, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 23 + 21 = 44
44 + digcount(44,dig=1,n=101100, 1122, 230, 134, 112, 62, 54, 48, 44, 40, 38, 35, 32, 2E, 2C, 2A, 28, 26, 24, 22, 20, 1L, 1K, 1J, 1I, 1H, 1G, 1F, 1E, 1D, 1C, 1B, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 44 + 31 = 75

And the sequence develops like this:

3, 6, 12, 23, 44, 75, 124, 202, 319, 503, 780, 1196, 1824, 2766, 4191, 6338, 9546, 14383, 21656, 32562, 48930, 73494, 110361, 165714, 248733, 373303, 560214, 840602, 1261237, 1892269, 2838926, 4258966, 6389157, 9584585, 14377879…

Now try the same procedure using the digit 0: n = n + digcount(n,dig=0,base=2..n). The first step is this:

3 + digcount(3,digit=0,n=11, 10) = 3 + 1 = 4

Next come these:

4 + digcount(4,dig=0,n=100, 11, 10) = 4 + 3 = 7
7 + digcount(7,dig=0,n=111, 21, 13, 12, 11, 10) = 7 + 1 = 8
8 + digcount(8,dig=0,n=1000, 22, 20, 13, 12, 11, 10) = 8 + 5 = 13
13 + digcount(13,dig=0,n=1101, 111, 31, 23, 21, 16, 15, 14, 13, 12, 11, 10) = 13 + 2 = 15
15 + digcount(15,dig=0,n=1111, 120, 33, 30, 23, 21, 17, 16, 15, 14, 13, 12, 11, 10) = 15 + 3 = 18
18 + digcount(18,dig=0,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 18 + 9 = 27
27 + digcount(27,dig=0,n=11011, 1000, 123, 102, 43, 36, 33, 30, 27, 25, 23, 21, 1D, 1C, 1B, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 27 + 7 = 34
34 + digcount(34,dig=0,n=100010, 1021, 202, 114, 54, 46, 42, 37, 34, 31, 2A, 28, 26, 24, 22, 20, 1G, 1F, 1E, 1D, 1C, 1B, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 34 + 8 = 42
42 + digcount(42,dig=0,n=101010, 1120, 222, 132, 110, 60, 52, 46, 42, 39, 36, 33, 30, 2C, 2A, 28, 26, 24, 22, 20, 1K, 1J, 1I, 1H, 1G, 1F, 1E, 1D, 1C, 1B, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 42 + 9 = 51

The sequence develops like this:

3, 4, 7, 8, 13, 15, 18, 27, 34, 42, 51, 59, 62, 66, 80, 94, 99, 111, 117, 125, 132, 151, 158, 163, 173, 180, 204, 222, 232, 244, 258, 279, 292, 307, 317, 324, 351, 364, 382, 389, 400, 425, 437, 447, 454, 466, 475, 483, 494, 509, 517, 536, 553, 566, 576, 612, 637, 649, 669, 679, 693, 712, 728, 753, 768, 801, 822, 835, 849, 862, 869, 883, 895, 906, 923, 932, 943, 949, 957, 967, 975, 999, 1011…

If you compare it with the sequence for digit=1, it appears that digcount(n,dig=1,b=2..n) is always larger than digcount(n,dig=0,b=2..n). That is in fact the case, with one exception, when n = 2:

digcount(2,dig=1,n=10) = 1
digcount(2,dig=0,n=10) = 1

When n = 10 (in base ten), there are twice as many ones as zeros:

digcount(10,dig=1,n=1010, 101, 22, 20, 14, 13, 12, 11, 10) = 10
digcount(10,dig=0,n=1010, 101, 22, 20, 14, 13, 12, 11, 10) = 5

As n gets larger, the difference grows dramatically:

digcount(100,dig=1,base=2..n) = 64
digcount(100,dig=0,base=2..n) = 16

digcount(1000,dig=1,base=2..n) = 533
digcount(1000,dig=0,base=2..n) = 25

digcount(10000,dig=1,base=2..n) = 5067
digcount(10000,dig=0,base=2..n) = 49

digcount(100000,dig=1,base=2..n) = 50140
digcount(100000,dig=0,base=2..n) = 73

digcount(1000000,dig=1,base=2..n) = 500408
digcount(1000000,dig=0,base=2..n) = 102

digcount(10000000,dig=1,base=2..n) = 5001032
digcount(10000000,dig=0,base=2..n) = 134

digcount(100000000,dig=1,base=2..n) = 50003137
digcount(100000000,dig=0,base=2..n) = 160

In fact, digcount(n,dig=1,b=2..n) is greater than the digit-count for any other digit: 0, 2, 3, 4, 5… (with the exception n = 2, as shown above). But digit=0 sometimes beats digits >= 2. For example, when n = 18:

digcount(18,dig=0,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 9
digcount(18,dig=2,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 7
digcount(18,dig=3,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 4
digcount(18,dig=4,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 2
digcount(18,dig=5,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 1

But as n gets larger, digcount(0) will fall permanently behind all these digits. However, digcount(0) will always be greater than some digit d, for the obvious reason that some digits only appear when the base is high enough. For example, the hexadecimal digit A (with the decimal value 10) first appears when n = 21:

digcount(21,dig=A,n=10101, 210, 111, 41, 33, 30, 25, 23, 21, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 1 digcount(21,dig=0,n=10101, 210, 111, 41, 33, 30, 25, 23, 21, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 5

There is a general rule for the n at which digit d first appears, n = 2d + 1 (this doesn’t apply when d = 0 or d = 1):

d = 2, n = 5 = 2*2 + 1
digcount(5,dig=2,n=101, 12, 11, 10) = 1

d = 3, n = 7 = 2*3 + 1
digcount(7,dig=3,n=111, 21, 13, 12, 11, 10) = 1

d = 4, n = 9 = 2*4 + 1
digcount(9,dig=4,n=1001, 100, 21, 14, 13, 12, 11, 10) = 1

d = 5, n = 11 = 2*5 + 1
digcount(11,dig=5,n=1011, 102, 23, 21, 15, 14, 13, 12, 11, 10) = 1

It should be apparent, then, that the digit-count for a particular digit starts at 1 and gets gradually higher. The rate at which the digit-count increases is highest for 1 and lowest for 0, with digits 2, 3, 4, 5… in between:

All-Base Graph

Graph for digcount(n,dig=d,b=2..n)


You could think of the graph as a digital rodeo in which these digits compete with each other. 1 is the clear and permanent winner, 0 the gradual loser. Now recall the procedure introduced at the start: n = n + digcount(n,dig=d,b=2..n). When it’s applied to the digits 0 to 5, these are the sequences that appear:

n = n + digcount(n,dig=0,b=2..n)

2, 3, 4, 7, 8, 13, 15, 18, 27, 34, 42, 51, 59, 62, 66, 80, 94, 99, 111, 117, 125, 132, 151, 158, 163, 173, 180, 204, 222, 232, 244, 258, 279, 292, 307, 317, 324, 351, 364, 382, 389, 400, 425, 437, 447, 454, 466, 475, 483, 494, 509, 517, 536, 553, 566, 576, 612, 637, 649, 669, 679, 693, 712, 728, 753, 768, 801, 822, 835, 849, 862, 869, 883, 895, 906, 923, 932, 943, 949, 957, 967, 975, 999, 1011…

n = n + digcount(n,dig=1,b=2..n)

2, 3, 6, 12, 23, 44, 75, 124, 202, 319, 503, 780, 1196, 1824, 2766, 4191, 6338, 9546, 14383, 21656, 32562, 48930, 73494, 110361, 165714, 248733, 373303, 560214, 840602, 1261237, 1892269, 2838926, 4258966, 6389157, 9584585, 14377879…

n = n + digcount(n,dig=2,b=2..n)

5, 6, 8, 12, 16, 22, 31, 37, 48, 60, 76, 94, 115, 138, 173, 213, 257, 311, 374, 454, 542, 664, 790, 935, 1109, 1310, 1552, 1835, 2167, 2548, 2989, 3509, 4120, 4832, 5690, 6687, 7829, 9166, 10727, 12568, 14697, 17182, 20089, 23470, 27425, 32042, 37477, 43768, 51113, 59687, 69705, 81379, 94998, 110910, 129488, 151153, 176429, 205923, 240331, 280490, 327396, 382067, 445858…

n = n + digcount(n,dig=3,b=2..n)

7, 8, 9, 10, 11, 13, 16, 18, 22, 25, 29, 34, 38, 44, 50, 56, 63, 80, 90, 104, 113, 131, 151, 169, 188, 210, 236, 261, 289, 320, 350, 385, 424, 463, 520, 572, 626, 684, 747, 828, 917, 999, 1101, 1210, 1325, 1446, 1577, 1716, 1871, 2040, 2228, 2429, 2642, 2875, 3133, 3413, 3719, 4044, 4402, 4786, 5196, 5645, 6140, 6673, 7257, 7900, 8582, 9315, 10130, 10998, 11942, 12954, 14058…

n = n + digcount(n,dig=4,b=2..n)

9, 10, 11, 12, 13, 14, 16, 18, 20, 23, 25, 28, 34, 41, 44, 52, 61, 67, 74, 85, 92, 102, 113, 121, 134, 148, 170, 184, 208, 229, 253, 269, 287, 306, 324, 356, 386, 410, 439, 469, 501, 531, 565, 604, 662, 703, 742, 794, 845, 895, 953, 1007, 1062, 1127, 1188, 1262, 1336, 1421, 1503, 1585, 1676, 1777, 1876, 2001, 2104, 2249, 2375, 2502, 2636, 2789, 2938, 3102, 3267, 3444, 3644, 3868, 4099…

n = n + digcount(n,dig=5,b=2..n)

11, 12, 13, 14, 15, 16, 17, 19, 21, 23, 26, 28, 29, 33, 37, 41, 48, 50, 55, 60, 64, 67, 72, 75, 83, 91, 96, 102, 107, 118, 123, 129, 137, 151, 159, 171, 180, 192, 202, 211, 224, 233, 251, 268, 280, 296, 310, 324, 338, 355, 380, 401, 430, 455, 488, 511, 536, 562, 584, 607, 638, 664, 692, 718, 748, 778, 807, 838, 874, 911, 951, 993, 1039, 1081, 1124, 1166, 1216, 1264, 1313, 1370, 1432…

Spijit

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