A099676 Partial sums of repdigits of A002283.

9, 108, 1107, 11106, 111105, 1111104, 11111103, 111111102, 1111111101, 11111111100, 111111111099, 1111111111098, 11111111111097, 111111111111096, 1111111111111095, 11111111111111094, 111111111111111093, 1111111111111111092, 11111111111111111091

Offset: 1

Author

Labos Elemer, Nov 17 2004

Keywords

• base
• nonn
• easy

Comments

a(n) is the maximal positive integer k such that the sequence 1, 2, 3, 4, ..., k-1, k has a total of n*k digits. - Bui Quang Tuan, Mar 12 2015

Examples

			9 + 99 + 999 + 9999 + 99999 = a(5) = 111105.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (12,-21,10).

Crossrefs

Cf. A057932, A002275-A002283, A099669-A099674.

Programs

Magma

[(10/9)*(10^n-1)-n: n in [1..20]]; // Vincenzo Librandi, Mar 14 2014

Maple

a:=n->sum((10^(n-j)-1^(n-j)), j=0..n): seq(a(n), n=1..17); # Zerinvary Lajos, Jan 15 2007

Mathematica

<Vincenzo Librandi, Mar 14 2014 *)
LinearRecurrence[{12,-21,10},{9,108,1107},20] (* Harvey P. Dale, Apr 18 2015 *)

PARI

Vec(-9*x/((x-1)^2*(10*x-1)) + O(x^100)) \\ Colin Barker, Mar 12 2014

Sage

[gaussian_binomial(n,1,10)-n for n in range(2,19)] # Zerinvary Lajos, May 29 2009

Formula

a(n) = (10/9)*(10^n-1) - n. - R. Piyo (nagoya314(AT)yahoo.com), Dec 10 2004

From Colin Barker, Mar 12 2014: (Start)

a(n) = 12*a(n-1)-21*a(n-2)+10*a(n-3).

G.f.: -9*x / ((x-1)^2*(10*x-1)). (End)

E.g.f.: exp(x)*(10*(exp(9*x) - 1) - 9*x)/9. - Stefano Spezia, Sep 13 2023