cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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

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Author

Jianing Song, Jan 25 2020

Keywords

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.

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
		

Crossrefs

Cf. A331789.
Cf. A000918 (row 2), A164123 (row 3), A331786 (row 10).

Programs

  • 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))).