OFFSET
2,3
COMMENTS
Write n = (b-1)*s + t, 1 <= t <= b-1. The smallest N_0 such that none of S(N_0), S(N_0+1), ..., S(N_0+m-1) is divisible by n is given by N_0 = b^(u_0) - b^s*(t-gcd(t,b-1)+1) + 1, where u_0 is the smallest nonnegative solution to (b-1)*u == -gcd(t,b-1) (mod n). See my link below for more detailed information.
LINKS
Jianing Song, Table of n, a(n) for n = 2..7627 (Note: T(b,n) occurs at the ((n+b-2)*(n+b-1)/2-b+3)-th place)
FORMULA
If n = (b-1)*s + t, 1 <= t <= b-1, then T(b,n) = b^s*(2*t-gcd(t,b-1)+1) - 2. See my link for a proof of the formula.
T(b,n) = T(b,n-1) + b*T(b,n-b+1) - b*T(b,n-b) for b >= 2, n >= b+1.
T(b,n) = O(b^(n/(b-1))).
EXAMPLE
Table begins
b\n 1 2 3 4 5 6 7 8 9 10
2 0 2 6 14 30 62 126 254 510 1022
3 0 1 4 7 16 25 52 79 160 241
4 0 2 2 6 14 14 30 62 62 126
5 0 1 4 3 8 13 28 23 48 73
6 0 2 4 6 4 10 22 34 46 34
7 0 1 2 5 8 5 12 19 26 47
8 0 2 4 6 8 10 6 14 30 46
9 0 1 4 3 8 9 12 7 16 25
10 0 2 2 6 8 8 12 14 8 18
PROG
(PARI) T(b, n) = my(s=(n-1)\(b-1), t=(n-1)%(b-1)+1); b^s*(2*t-gcd(t, b-1)+1)-2
CROSSREFS
KEYWORD
AUTHOR
Jianing Song, Jan 25 2020
STATUS
approved