One of my favourite integer sequences uses the formula n(i) = n(i-1) + digsum(n(i-1)), where digsum(n) sums the digits of n. In base 10, it goes 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…

Another interesting sequence uses the formula n(i) = n(i-1) + digprod(n(i-1)), where digprod(n) multiplies the digits of n (excluding 0). In base 10, it goes like this:

1, 2, 4, 8, 16, 22, 26, 38, 62, 74, 102, 104, 108, 116, 122, 126, 138, 162, 174, 202, 206, 218, 234, 258, 338, 410, 414, 430, 442, 474, 586, 826, 922, 958, 1318, 1342, 1366, 1474, 1586, 1826, 1922, 1958, 2318, 2366, 2582, 2742, 2854, 3174, 3258, 3498, 4362, 4506, 4626, 4914, 5058, 5258, 5658, 6858, 8778, 11914, 11950, 11995…

You can apply these formulae in other bases and it’s trivially obvious that the sequences rise most slowly in base 2, because you’re never summing or multiplying anything but the digit 1. However, there is a sequence for which base 2 is by far the best performer. It has the formula n(i) = n(i-1) + blockmult(n(i-1)), where blockmult(n) counts the lengths of distinct blocks of the same digit, including 0, then multiplies those lengths together. For example:

blockmult(3,b=2) = blockmult(11) = 2

blockmult(28,b=2) = blockmult(11100) = 3 * 2 = 6

blockmult(51,b=2) = blockmult(110011) = 2 * 2 * 2 = 8

blockmult(140,b=2) = blockmult(10001100) = 1 * 3 * 2 * 2 = 12

blockmult(202867,b=2) = blockmult(110001100001110011) = 2 * 3 * 2 * 4 * 3 * 2 * 2 = 576

The full sequence begins like this (numbers are represented in base 10, but the formula is being applied to their representations in binary):

1, 2, 3, 5, 6, 8, 11, 13, 15, 19, 23, 26, 28, 34, 37, 39, 45, 47, 51, 59, 65, 70, 76, 84, 86, 88, 94, 98, 104, 110, 116, 122, 126, 132, 140, 152, 164, 168, 171, 173, 175, 179, 187, 193, 203, 211, 219, 227, 245, 249, 259, 271, 287, 302, 308, 316, 332, 340, 342, 344, 350, 354, 360, 366, 372, 378, 382, 388, 404, 412, 436, 444, 460, 484, 500, 510, 518, 530, 538, 546, 555, 561, 579, 595, 603, 611, 635, 651, 657, 663, 669, 675, 681…

In higher bases, it rises much more slowly. This is base 3:

1, 2, 3, 4, 6, 7, 8, 10, 11, 12, 14, 16, 17, 19, 20, 21, 22, 24, 26, 29, 31, 33, 34, 35, 37, 39, 42, 44, 48, 49, 51, 53, 56, 58, 60, 61, 62, 64, 65, 66, 68, 70, 71, 73, 75, 77, 79, 82, 85, 89, 93, 95, 97, 98, 100, 101, 102, 103, 105, 107, 110, 114, 116, 120, 124, 127, 129, 131, 133, 137, 139, 141, 142, 143, 145, 146, 147, 149, 151, 152, 154, 156, 158, 160, 163…

And this is base 10:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 90…

Note how, in bases 3 and 10, blockmult(n) often equals 1. In base 3, the sequence contains [141, 142, 143, 145]:

blockmult(141,b=3) = blockmult(12020) = 1 * 1 * 1 * 1 = 1

blockmult(142,b=3) = blockmult(12021) = 1 * 1 * 1 * 1 = 1

blockmult(143,b=3) = blockmult(12022) = 1 * 1 * 1 * 2 = 2

The formula also returns 1 much further along the sequence in base 3. For example, the 573809th number in the sequence, or n(573809), is 5775037 and blockmult(5775037) = blockmult(101212101212021) = 1^15 = 1. But in base 2, blockmult(n) = 1 is very rare. It happens three times at the beginning of the sequence:

1, 2, 3, 5, 6, 8, 11…

After that, I haven’t found any more examples of blockmult(n) = 1, although blockmult(n) = 2 occurs regularly. For example,

blockmult(n(100723)) = blockmult(44739241) = blockmult(10101010101010101010101__00__1) = 2

blockmult(n(100724)) = blockmult(44739243) = blockmult(101010101010101010101010__11__) = 2

blockmult(n(100725)) = blockmult(44739245) = blockmult(1010101010101010101010__11__01) = 2

Does the sequence in base 2 return another example of blockmult(n) = 1? The odds seem against it. For any given number of digits in base 2, there is only one number for which blockmult(n) = 1. For example: 1, 10, 101, 1010, 10101, 101010, 1010101… As the sequence increases, the percentage of these numbers becomes smaller and smaller. But the sequence is infinite, so who knows what happens in the end? Perhaps blockmult(n) = 1 occurs infinitely often.