A100255 Squares of pentagonal numbers: a(n) = (1/4)*n^2*(3*n-1)^2.
0, 1, 25, 144, 484, 1225, 2601, 4900, 8464, 13689, 21025, 30976, 44100, 61009, 82369, 108900, 141376, 180625, 227529, 283024, 348100, 423801, 511225, 611524, 725904, 855625, 1002001, 1166400, 1350244, 1555009, 1782225, 2033476
Offset: 0
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..10000
- Leonhard Euler, De mirabilibus proprietatibus numerorum pentagonalium, The Euler Archive, 1783, par. 29.
- Leonhard Euler, On the remarkable properties of the pentagonal numbers, arXiv:math/0505373 [math.HO], 2005.
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Crossrefs
Programs
-
Mathematica
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 25, 144, 484}, 32] (* Ilya Gutkovskiy, Apr 13 2016 *) Table[(1/4) n^2 (3 n - 1)^2, {n, 0, 31}] (* Michael De Vlieger, Apr 13 2016 *)
-
PARI
a(n) = (1/4)*n^2*(3*n-1)^2 \\ Altug Alkan, Apr 13 2016
Formula
a(n) = A000326(n)^2.
G.f.: x*(1+20*x+29*x^2+4*x^3)/(1-x)^5. - Colin Barker, Feb 14 2012
From Ilya Gutkovskiy, Apr 13 2016: (Start)
E.g.f.: x*(4 + 46*x + 48*x^2 + 9*x^3)*exp(x)/4.
a(n) = 5*a(n-1) - 10*(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
Sum_{n>=1} 1/a(n) = 2*Pi^2/3 + 4*sqrt(3)*Pi - 36*log(3) + 4*psi_1(2/3), where psi_1 is the trigamma function. - Amiram Eldar, Apr 04 2025
Comments