A215054 a(n) = 1/11*(binomial(n,11) - floor(n/11)).

1, 7, 33, 124, 397, 1125, 2893, 6871, 15269, 32065, 64130, 122916, 226922, 405218, 702378, 1185263, 1952198, 3145208, 4966118, 7697483, 11729498, 17594247, 26008887, 37929627, 54618663, 77726559, 109392935, 152368731, 210163767, 287223815, 389141943

Offset: 12

Author

Peter Bala, Aug 01 2012

Keywords

• nonn
• easy

Comments

Let p be a prime. Saikia and Vogrinc have proved that 1/p*{binomial(n,p) - floor(n/p)} is an integer sequence. The present sequence is the case p = 11. Other cases are A002620 (p = 2), A014125 (p = 3), A215052 (p = 5) and A215053 (p = 7).

Links

• M. P. Saikia and J. Vogrinc, A simple number theoretic result arxiv.1207.6707v1 [mathNT]
• Index entries for linear recurrences with constant coefficients, signature (11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 2, -11, 55, -165, 330, -462, 462, -330, 165, -55, 11, -1).

Crossrefs

A002620 (p = 2), A014125 (p = 3), A178904, A215052 (p = 5), A215053(p = 7).

Partial sums of A032169.

Programs

Mathematica

Table[(Binomial[n,11]-Floor[n/11])/11,{n,12,50}] (* Harvey P. Dale, Aug 06 2012 *)

Maxima

A215054(n):=1/11*(binomial(n,11) - floor(n/11))$ makelist(A215054(n),n,12,30); /* Martin Ettl, Oct 25 2012 */

Formula

a(n) = 1/11*(binomial(n,11) - floor(n/11)).

O.g.f.: sum_{n>=0} a(n)*x^n = x^12*(1 - 4*x + 11*x^2 - 19*x^3 + 23*x^4 - 19*x^5 + 11*x^6 - 4*x^7 + x^8)/((1-x^11)*(1-x)^11) = x^12*(1 + 7*x + 33*x^2 + 124*x^3 + ...). The numerator polynomial 1 - 4*x + 11*x^2 - 19*x^3 + 23*x^4 - 19*x^5 + 11*x^6 - 4*x^7 + x^8 is the negative of the row generating polynomial for row 11 of A178904.