A129557 - OEIS

%I #154 Feb 16 2025 08:33:05

%S 1,4,34,151,1291,5734,49024,217741,1861621,8268424,70692574,313982371,

%T 2684456191,11923061674,101938642684,452762361241,3870983965801,

%U 17193046665484,146995452057754,652883010927151,5581956194228851,24792361368566254

%N Numbers k > 0 such that k^2 is a centered pentagonal number (A005891).

%C Corresponding numbers m such that centered pentagonal number A005891(m) = (5*m^2 + 5*m + 2)/2 is a perfect square are listed in A129556 = {0, 2, 21, 95, 816, 3626, 31005, ...}.

%C Also positive integers x in the solutions to 2*x^2 - 5*y^2 + 5*y - 2 = 0, the corresponding values of y being A254332. - _Colin Barker_, Jan 28 2015

%H Vincenzo Librandi, <a href="/A129557/b129557.txt">Table of n, a(n) for n = 1..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/CenteredPentagonalNumber.html">Centered Pentagonal Number</a>.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,38,0,-1).

%F a(n) = sqrt( (5*A129556(n)^2 + 5*A129556(n) + 2)/2 ).

%F For n >= 5, a(n) = 38*a(n-2) - a(n-4). - _Max Alekseyev_, May 08 2009

%F 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

%F From _Andrea Pinos_, Oct 07 2022: (Start)

%F The ratios of successive terms converge to two different limits:

%F   lower: D = lim_{n->oo} a(2n)/a(2n-1) = (7 + 2*sqrt(10))/3;

%F   upper: E = lim_{n->oo} a(2n+1)/a(2n) = (13 + 4*sqrt(10))/3.

%F So lim_{n->oo} a(n+2)/a(n) = D*E = 19 + 6*sqrt(10).

%F a(n) = (A005667(n) - (-1)^n*A005667(n-1))/4. (End)

%t Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[ Sqrt[f] ] ], {n,1,40000} ]

%t 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 *)

%o (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

%o (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

%Y Cf. A005891 (centered pentagonal numbers).

%Y Cf. A129556 (k such that A005891(k) is a perfect square).

%Y Cf. A000290, A005667, A254332, A254333.

%K nonn,easy,changed

%O 1,2

%A _Alexander Adamchuk_, Apr 20 2007

%E More terms from _R. J. Mathar_, Oct 11 2007

%E More terms from _Max Alekseyev_, May 08 2009