1,2

InterpretA176271as an infinite square array read by antidiagonals, with rows 1,5,11,19,...; 3,9,17,27,... and so on. The sum of the terms in the n X n upper submatrix are s(n) = 1, 18, 93, 296,... = n^2*(7*n^2-1)/6, and a(n) = s(n) - s(n-1) are the first differences. -J. M. Bergot,Jun 27 2013

Harry J. Smith,Table of n, a(n) for n = 1..1000

T. P. Martin,Shells of atoms,Phys. Rep., 273 (1996), 199-241, eq. (10).

Index entries for linear recurrences with constant coefficients,signature (4,-6,4,-1).

G.f.: x*(1+x)*(1+12*x+x^2)/(1-x)^4. -Colin Barker,Mar 02 2012

E.g.f.: (-3 + 6*x + 21*x^2 + 14*x^3)*exp(x)/3 + 1. -G. C. Greubel,Dec 01 2017

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). -Wesley Ivan Hurt,May 11 2023

Table[(2*n - 1)*(7*n^2 - 7*n + 3)/3, {n, 1, 30}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {1, 17, 75, 203}, 30] (*G. C. Greubel,Dec 01 2017 *)

(PARI) { for (n=1, 1000, write( "b063494.txt", n, "", (2*n - 1)*(7*n^2 - 7*n + 3)/3) ) } \\Harry J. Smith,Aug 23 2009

(Magma) [(2*n - 1)*(7*n^2 - 7*n + 3)/3: n in [1..30]]; //G. C. Greubel,Dec 01 2017

(PARI) x='x+O('x^30); Vec(serlaplace((-3+6*x+21*x^2+14*x^3)*exp(x)/3 + 1)) \\G. C. Greubel,Dec 01 2017

1/12*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6,... givesA049480,A005894,A063488,A001845,A063489,A005898,A063490,A057813,A063491,A005902,A063492,A005917,A063493,A063494,A063495,A063496.

Sequence in context:A097223A296113A231779*A146594A202138A357740

Adjacent sequences:A063491A063492A063493*A063495A063496A063497

nonn,easy

N. J. A. Sloane,Aug 01 2001

approved