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

Author

Jianing Song, Jan 25 2020

Keywords

• nonn
• base
• easy

Comments

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

Examples

			See A331786.
		

Links

• 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).

Crossrefs

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

Row 10 of A331789.

Programs

PARI

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

Formula

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)).