A215052 a(n) = (binomial(n,5) - floor(n/5)) / 5.

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

Offset: 6

Author

Peter Bala, Aug 01 2012

Keywords

• nonn
• easy

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

• 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.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,2,-5,10,-10,5,-1).

Crossrefs

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

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 + ...).