A331789 -id:A331789 - cp's OEIS frontend

A331788 a(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.

1, 3, 3, 7, 9, 9, 13, 15, 9, 19, 39, 39, 79, 99, 99, 139, 159, 99, 199, 399, 399, 799, 999, 999, 1399, 1599, 999, 1999, 3999, 3999, 7999, 9999, 9999, 13999, 15999, 9999, 19999, 39999, 39999, 79999, 99999, 99999, 139999, 159999, 99999, 199999, 399999, 399999, 799999

Offset: 1

Jianing Song, Jan 25 2020

The main sequence is A331786; this is added because some people may search for this.

			See A331786.

Jianing Song, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,10,-10).

Cf. A007953 (S(N)), A051885, A331786.

Row 10 of A331789.

a(n) = my(s=(n-1)\9, t=(n-1)%9+1); 10^s*(2*t-gcd(t,9)+1)-1

If n = 9*s + t, 1 <= t <= 9, then a(n) = 10^s*(2*t-gcd(t,9)+1) - 1. See A331787 for a proof of the formula in base b.
a(n) = A331786(n) + 1.
Conjectures from Colin Barker, Jan 26 2020: (Start)
G.f.: x*(1 + 2*x + 4*x^3 + 2*x^4 + 4*x^6 + 2*x^7 - 6*x^8) / ((1 - x)*(1 - 10*x^9)).
a(n) = a(n-1) + 10*a(n-9) - 10*a(n-10) for n>10.
(End) [This conjecture is correct.]
a(n) = O(10^(n/9)).

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

Jianing Song, Jan 25 2020

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.

			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

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.

Cf. A331789.

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

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

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))).

cp's OEIS Frontend

A331788 a(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.

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