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A215054 a(n) = 1/11*(binomial(n,11) - floor(n/11)). 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
12,2
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]
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.
MATHEMATICA
Table[(Binomial[n, 11]-Floor[n/11])/11, {n, 12, 50}] (* Harvey P. Dale, Aug 06 2012 *)
PROG
(Maxima) A215054(n):=1/11*(binomial(n, 11) - floor(n/11))$ makelist(A215054(n), n, 12, 30); /* Martin Ettl, Oct 25 2012 */
CROSSREFS
A002620 (p = 2), A014125 (p = 3), A178904, A215052 (p = 5), A215053(p = 7).
Partial sums of A032169.
Sequence in context: A221036 A338232 A000605 * A350643 A114014 A229515
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Aug 01 2012
STATUS
approved

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Last modified June 27 16:21 EDT 2024. Contains 373746 sequences. (Running on oeis4.)