A394867 Number of distinct cycles under iteration of the sum of n-th powers of digits.

9, 2, 9, 6, 16, 7, 17, 7, 19, 8, 24, 5, 14, 13, 12, 7, 22, 10, 17, 8, 20, 3, 23, 15, 31, 8, 28, 15, 31, 6, 30, 10, 25

Offset

1,1

Comments

Let f_n(k) denote the sum of the n-th powers of the decimal digits of k. For any positive integer k, the sequence k, f_n(k), f_n(f_n(k)), ... eventually enters a cycle. a(n) is the number of distinct such cycles.

Links

Table of n, a(n) for n=1..33.

f90boy, Is there any efficient algorithm for the integer black hole problem (in Chinese).

Example

For n=1, f1(k) is the digit sum, and iteration of f1 always leads to a single-digit fixed point, giving 9 distinct cycles: {1}, {2}, ..., {9}. So a(1) = 9.
For n=12, starting from 98840282759: f_12(98840282759) = 9^12 + 8^12 + 8^12 + 4^12 + 0^12 + 2^12 + 8^12 + 2^12 + 7^12 + 5^12 + 9^12 = 785119716404. Continuing the iteration gives the cycle (98840282759, 785119716404, 381286065015, 142281334933, 351184701607). There are 5 distinct cycles in total, with representative elements 1, 2700216437, 16068940818, 4876998775, and 98840282759. Every positive integer eventually enters one of these 5 cycles, so a(12) = 5.

Crossrefs

Cf. A007953, A003132.

Sequence in context: A260646 A200234 A242743 * A197812 A146492 A266565

Adjacent sequences: A394864 A394865 A394866 * A394868 A394869 A394870

Keyword

nonn,base,more

Author

Zhao Hui Du, Apr 04 2026

Status

approved