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A215052 a(n) = (binomial(n,5) - floor(n/5)) / 5. 3
1, 4, 11, 25, 50, 92, 158, 257, 400, 600, 873, 1237, 1713, 2325, 3100, 4069, 5266, 6729, 8500, 10625, 13155, 16145, 19655, 23750, 28500, 33981, 40274, 47466, 55650, 64925, 75397, 87178, 100387, 115150, 131600, 149878, 170132, 192518, 217200 (list; graph; refs; listen; history; text; internal format)
OFFSET

6,2

COMMENTS

Apparently a duplicate of A036837. - R. J. Mathar, Aug 06 2012

Not the same as A011851.

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 = 5. Other cases are A002620 (p = 2), A014125 (p = 3), A215053 (p = 7) and A215054 (p = 11).

There is a connection between these sequences and A178904. For a fixed prime p the o.g.f. for the sequence (1/p)*{binomial(n,p) - floor(n/p)} is a rational function of the form x^(p+1)*R(p,x)/((1-x^p)*(1-x)^p). The polynomial R(p,x) = sum {k = 0..p-1} (1/p)*{1 - (-1)^k*binomial(p-1,k)}*x^(k-1). For prime p >= 3, -R(p,x) is equal to the p-th row polynomial of A178904.

LINKS

Table of n, a(n) for n=6..44.

M. P. Saikia and J. Vogrinc, A simple number theoretic result. (arxiv.1207.6707v1 [mathNT]), J. Assam Academy of Mathematics, Vol. 3, 90-96, 2010.

FORMULA

a(n) = (1/5)*{binomial(n,5) - floor(n/5)}.

O.g.f.: sum {n>=0} a(n)*x^n = x^6*(1-x+x^2)/((1-x^5)*(1-x)^5) = x^6*(1 + 4*x + 11*x^2 + 25*x^3 + ...).

CROSSREFS

Cf. A002620 (p = 2), A014125 (p = 3), A178904, A215053 (p = 7), A215054( p = 11).

Sequence in context: A333643 A006522 A036837 * A011851 A193912 A136395

Adjacent sequences:  A215049 A215050 A215051 * A215053 A215054 A215055

KEYWORD

nonn,easy

AUTHOR

Peter Bala, Aug 01 2012

STATUS

approved

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Last modified September 26 02:39 EDT 2020. Contains 337346 sequences. (Running on oeis4.)