A331787 T(b,n) is the largest m such that there exists N such that none of S(N), S(N+1), ..., S(N+m-1) is divisible by n, where S(N) is the sum of digits of N in base b. Square array read by ascending antidiagonals.
0, 0, 2, 0, 1, 6, 0, 2, 4, 14, 0, 1, 2, 7, 30, 0, 2, 4, 6, 16, 62, 0, 1, 4, 3, 14, 25, 126, 0, 2, 2, 6, 8, 14, 52, 254, 0, 1, 4, 5, 4, 13, 30, 79, 510, 0, 2, 4, 6, 8, 10, 28, 62, 160, 1022, 0, 1, 2, 3, 8, 5, 22, 23, 62, 241, 2046, 0, 2, 4, 6, 8, 10, 12, 34, 48, 126, 484, 4094
Offset: 2
Examples
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
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)
- Jianing Song, Proof of the formula, and the examples N_0 such that none of S(N_0), S(N_0+1), ..., S(N_0+m-1) is divisible by n.
Programs
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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
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))).
Comments