A129557 Numbers k > 0 such that k^2 is a centered pentagonal number (A005891).
1, 4, 34, 151, 1291, 5734, 49024, 217741, 1861621, 8268424, 70692574, 313982371, 2684456191, 11923061674, 101938642684, 452762361241, 3870983965801, 17193046665484, 146995452057754, 652883010927151, 5581956194228851, 24792361368566254
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Eric Weisstein's World of Mathematics, Centered Pentagonal Number.
- Index entries for linear recurrences with constant coefficients, signature (0,38,0,-1).
Crossrefs
Programs
-
Mathematica
Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[ Sqrt[f] ] ], {n,1,40000} ] CoefficientList[Series[(1-x)*(1+5*x+x^2)/((1+6*x-x^2)*(1-6*x-x^2)),{x,0,30}],x] (* Vincenzo Librandi, Apr 11 2012 *)
-
PARI
A129557()={ for(n=1,1000000000, f=(5*n^2+5*n+2)/2 ; if(issquare(f), print1(sqrtint(f), ", ") ; ); ) ; } \\ R. J. Mathar, Oct 11 2007
-
PARI
Vec(x*(1-x)*(1+5*x+x^2)/((1+6*x-x^2)*(1-6*x-x^2)) + O(x^100)) \\ Colin Barker, Jan 28 2015
Formula
For n >= 5, a(n) = 38*a(n-2) - a(n-4). - Max Alekseyev, May 08 2009
G.f.: x*(1-x)*(1 + 5*x + x^2)/((1 + 6*x - x^2)*(1 - 6*x - x^2)). - Colin Barker, Apr 11 2012
From Andrea Pinos, Oct 07 2022: (Start)
The ratios of successive terms converge to two different limits:
lower: D = lim_{n->oo} a(2n)/a(2n-1) = (7 + 2*sqrt(10))/3;
upper: E = lim_{n->oo} a(2n+1)/a(2n) = (13 + 4*sqrt(10))/3.
So lim_{n->oo} a(n+2)/a(n) = D*E = 19 + 6*sqrt(10).
Extensions
More terms from R. J. Mathar, Oct 11 2007
More terms from Max Alekseyev, May 08 2009
Comments