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%I #16 Apr 22 2014 02:46:45
%S 1,4,11,25,50,92,158,257,400,600,873,1237,1713,2325,3100,4069,5266,
%T 6729,8500,10625,13155,16145,19655,23750,28500,33981,40274,47466,
%U 55650,64925,75397,87178,100387,115150,131600,149878,170132,192518,217200
%N a(n) = (binomial(n,5) - floor(n/5)) / 5.
%C Apparently a duplicate of A036837. - _R. J. Mathar_, Aug 06 2012
%C Not the same as A011851.
%C 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).
%C 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.
%H M. P. Saikia and J. Vogrinc, <a href="http://arxiv.org/abs/1207.6707">A simple number theoretic result</a>. (arxiv.1207.6707v1 [mathNT]), J. Assam Academy of Mathematics, Vol. 3, 90-96, 2010.
%F a(n) = (1/5)*{binomial(n,5) - floor(n/5)}.
%F 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 + ...).
%Y Cf. A002620 (p = 2), A014125 (p = 3), A178904, A215053 (p = 7), A215054( p = 11).
%K nonn,easy
%O 6,2
%A _Peter Bala_, Aug 01 2012
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