A331789 T(b,n) is the smallest m such that for any N, at least one 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.

1, 1, 3, 1, 2, 7, 1, 3, 5, 15, 1, 2, 3, 8, 31, 1, 3, 5, 7, 17, 63, 1, 2, 5, 4, 15, 26, 127, 1, 3, 3, 7, 9, 15, 53, 255, 1, 2, 5, 6, 5, 14, 31, 80, 511, 1, 3, 5, 7, 9, 11, 29, 63, 161, 1023, 1, 2, 3, 4, 9, 6, 23, 24, 63, 242, 2047, 1, 3, 5, 7, 9, 11, 13, 35, 49, 127, 485, 4095

Offset: 2

Author

Jianing Song, Jan 25 2020

Keywords

• nonn
• base
• easy
• tabl

Comments

The main sequence is A331787; this is added because some people may search for this.

Examples

			Table begins
  b\n  1  2  3   4   5   6    7    8    9    10
   2   1  3  7  15  31  63  127  255  511  1023
   3   1  2  5   8  17  26   53   80  161   242
   4   1  3  3   7  15  15   31   63   63   127
   5   1  2  5   4   9  14   29   24   49    74
   6   1  3  5   7   5  11   23   35   47    35
   7   1  2  3   6   9   6   13   20   27    48
   8   1  3  5   7   9  11    7   15   31    47
   9   1  2  5   4   9  10   13    8   17    26
  10   1  3  3   7   9   9   13   15    9    19
		

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)

Crossrefs

Cf. A331787.

Cf. A000225 (row 2), A062318 (row 3 with an offset shift), A331788 (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)-1

Formula

If n = (b-1)*s + t, 1 <= t <= b-1, then T(b,n) = b^s*(2*t-gcd(t,b-1)+1) - 1. See A331787 for a proof of the formula in base b.

T(b,k) = A331787(b,k) + 1.

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))).