
COMMENTS

Chain addition mod 10 with window n: take an ndigit 'seed'. Take the sum of its digits mod 10 and append to the seed. Repeat with the last n digits of the string, until the seed appears again.
This sequence shows the lengths of the longest sequences for different window sizes.
a(1)a(10) all occur for seed 1 (among others). If this is always true, the sequence continues: 406224, 12695306, 4272460934, 380859180, 122070312496, 518798826, 3433227539058.  Lars Blomberg, Feb 12 2013
Comment from Michel Lagneau, Jan 20 2017, edited by N. J. A. Sloane, Jan 24 2017: (Start)
If seed 1 is always as good as or better than any other, as seems likely, then this sequence has the following alternative description.
Consider the n initial terms of an infinite sequence S(k, n) of decimal digits given by 0, 0,..., 0, 1. The succeeding terms are given by the final digits in the sum of the n immediately preceding terms. The sequence lists the period of each sequence corresponding to n = 2, 3, ...
a(2) = period of A000045 mod 10 (Fibonacci numbers mod 10) = A001175(10).
a(3) = period of A000073 mod 10 (tribonacci numbers mod 10) = A046738(10).
a(4) = period of A000078 mod 10 (tetranacci numbers mod 10) = A106295(10).
a(5) = period of A001591 mod 10 (pentanacci numbers mod 10) = A106303(10).
a(6) = period of A001592 mod 10 (hexanacci numbers mod 10).
a(7) = period of A122189 mod 10 (heptanacci numbers mod 10).
a(8) = period of A079262 mod 10 (octanacci numbers mod 10).
a(4) = 1560 because the four initial terms 0, 0, 0, 1 => S(k, 4) = 0, 0, 0, 1, 1, 2, 4, 8, 5, 9, 6, 8, 8, 1, 3, 0, 2, 6, 1, 9, 8, ... (tetranacci numbers mod 10). This sequence is periodic with period 1560:
S(1560 + 1, 4) = S(1, 4) = 0,
S(1560 + 2, 4) = S(2, 4) = 0,
S(1560 + 3, 4) = S(3, 4) = 0,
S(1560 + 4, 4) = S(4, 4) = 1.
(End)
