|
%I #31 May 11 2023 13:15:16
%S 1,17,75,203,429,781,1287,1975,2873,4009,5411,7107,9125,11493,14239,
%T 17391,20977,25025,29563,34619,40221,46397,53175,60583,68649,77401,
%U 86867,97075,108053,119829,132431,145887,160225,175473,191659,208811
%N a(n) = (2*n - 1)*(7*n^2 - 7*n + 3)/3.
%C Interpret A176271 as 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
%H Harry J. Smith, <a href="/A063494/b063494.txt">Table of n, a(n) for n = 1..1000</a>
%H T. P. Martin, <a href="http://dx.doi.org/10.1016/0370-1573(95)00083-6">Shells of atoms</a>, Phys. Rep., 273 (1996), 199-241, eq. (10).
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F G.f.: x*(1+x)*(1+12*x+x^2)/(1-x)^4. - _Colin Barker_, Mar 02 2012
%F E.g.f.: (-3 + 6*x + 21*x^2 + 14*x^3)*exp(x)/3 + 1. - _G. C. Greubel_, Dec 01 2017
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - _Wesley Ivan Hurt_, May 11 2023
%t 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 *)
%o (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
%o (Magma) [(2*n - 1)*(7*n^2 - 7*n + 3)/3: n in [1..30]]; // _G. C. Greubel_, Dec 01 2017
%o (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
%Y 1/12*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.
%K nonn,easy
%O 1,2
%A _N. J. A. Sloane_, Aug 01 2001
|
