A328683 Positive integers that are equal to 99...99 (repdigit with n digits 9) times the sum of their digits.

81, 1782, 26973, 359964, 4499955, 53999946, 629999937, 7199999928, 80999999919, 899999999910, 9899999999901, 107999999999892, 1169999999999883, 12599999999999874, 134999999999999865, 1439999999999999856, 15299999999999999847, 161999999999999999838

Offset: 1

Author

Bernard Schott, Feb 25 2020

Keywords

• nonn
• base

Comments

The idea of this sequence comes from a problem during the annual Moscow Mathematical Olympiad (MMO) in 2001 (see reference).

Examples

			359964 = 36 * 9999 and the digital sum of 359964 = 36 , so 359964 = a(4).
		

References

• Roman Fedorov, Alexei Belov, Alexander Kovaldzhi, Ivan Yashchenko, Moscow Mathematical Olympiads, 2000-2005, Level B, Problem 5, 2001, MSRI, 2011, p. 8 and 70/71.

Links

• Colin Barker, Table of n, a(n) for n = 1..995
• Index to sequences related to Olympiads.
• Index entries for linear recurrences with constant coefficients, signature (22,-141,220,-100).

Crossrefs

Cf. A057147, A086580, A110807.

Programs

Maple

C:=seq(9*n*(10^n-1),n=1..20);

Mathematica

Table[9*n*(10^n - 1), {n, 1, 18}] (* Amiram Eldar, Feb 25 2020 *)
LinearRecurrence[{22,-141,220,-100},{81,1782,26973,359964},20] (* Harvey P. Dale, Feb 02 2025 *)

PARI

Vec(81*x*(1 - 10*x^2) / ((1 - x)^2*(1 - 10*x)^2) + O(x^20)) \\ Colin Barker, Feb 25 2020

Formula

a(n) = 9 * n * (10^n - 1).

From Colin Barker, Feb 25 2020: (Start)

G.f.: 81*x*(1 - 10*x^2) / ((1 - x)^2*(1 - 10*x)^2).

a(n) = 22*a(n-1) - 141*a(n-2) + 220*a(n-3) - 100*a(n-4) for n>4.

(End)

From Michel Marcus, Feb 25 2020: (Start)

a(n) = 9*A110807(n).

a(n) = n*A086580(n). (End)