A215053 -id:A215053 - cp's OEIS frontend

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

Peter Bala, Aug 01 2012

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.

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).

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

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 + ...).

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

Original entry on oeis.org

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

Peter Bala, Aug 01 2012

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).

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).

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

Partial sums of A032169.

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

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

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.

cp's OEIS Frontend

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

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