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A175113 a(n) = ((2*n + 1)^6 + 1)/2. 3
1, 365, 7813, 58825, 265721, 885781, 2413405, 5695313, 12068785, 23522941, 42883061, 74017945, 122070313, 193710245, 297411661, 443751841, 645733985, 919132813, 1282863205, 1759371881, 2375052121, 3160681525, 4151882813 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

Convolution of the finite sequence 1, 358, 5279, 11764, 5279, 358, 1 with A000579. Partial sums of A175114.

Subsequence of A001844 because a(n)=(A050492(n+1)-1)^2+A050492(n+1)^2. - Bruno Berselli, Dec 28 2010

a(n) is also the first integer in a sum of (2*n + 1)^6 consecutive integers that equals (2*n + 1)^12. - Patrick J. McNab, Dec 26 2016

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

R. J. Mathar, Point counts of D_k and some A_k and E_k integer lattices inside hypercubes, arXiv:1002.3844, Variable V_6^(g)(n).

Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

FORMULA

a(n)= 7*a(n-1) -21*a(n-2) +35*a(n-3) -35*a(n-4) +21*a(n-5) -7*a(n-6) +a(n-7).

G.f.: (1+358*x+5279*x^2+11764*x^3+5279*x^4+358*x^5+x^6)/(1-x)^7.

a(n) = (2*n^2+2*n+1)*(16*n^4+32*n^3+20*n^2+4*n+1). - Bruno Berselli, Dec 27 2010

MATHEMATICA

CoefficientList[Series[(1 + 358*x + 5279*x^2 + 11764*x^3 + 5279*x^4 + 358*x^5 + x^6)/(1 - x)^7, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 20 2012 *)

PROG

(MAGMA) I:=[1, 365, 7813, 58825, 265721, 885781, 2413405]; [n le 7 select I[n] else 7*Self(n-1) - 21*Self(n-2) + 35*Self(n-3) - 35*Self(n-4) + 21*Self(n-5) - 7*Self(n-6) + Self(n-7): n in [1..40]]; // Vincenzo Librandi, Dec 20 2012

CROSSREFS

Sequence in context: A275182 A275130 A274727 * A222992 A276253 A233116

Adjacent sequences:  A175110 A175111 A175112 * A175114 A175115 A175116

KEYWORD

easy,nonn

AUTHOR

R. J. Mathar, Feb 13 2010

STATUS

approved

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Last modified May 25 11:27 EDT 2020. Contains 334592 sequences. (Running on oeis4.)