A129556 Numbers k such that the k-th centered pentagonal number A005891(k) = (5k^2 + 5k + 2)/2 is a square.

0, 2, 21, 95, 816, 3626, 31005, 137711, 1177392, 5229410, 44709909, 198579887, 1697799168, 7540806314, 64471658493, 286352060063, 2448225223584, 10873837476098, 92968086837717, 412919472031679, 3530339074609680, 15680066099727722, 134059916748330141

Offset

1,2

Comments

Corresponding numbers m > 0 such that m^2 is a centered pentagonal number are listed in A129557 = {1, 4, 34, 151, 1291, 5734, 49024, ...}.

From Andrea Pinos, Nov 02 2022: (Start)

By definition: 5*T(a(n)) = A129557(n)^2 - 1 where triangular number T(j) = j*(j+1)/2. This implies:

Every odd prime factor of a(n) and d(n)=a(n)+1 is present in b(n)=A129557(n)+1 or in c(n)=A129557(n)-1. (End)

From the law of cosines the non-Pythagorean triple {a(n), a(n)+1=A254332(n), A129557(n+1)} forms a near-isosceles triangle whose angle between the consecutive integer sides is equal to the central angle of the regular pentachoron polytope (4-simplex) (see A140244 and A140245). This implies that the terms {a(n)} are also those numbers k such that 1 + 5*A000217(k) is a square. - Federico Provvedi, Apr 04 2023

Links

Harvey P. Dale, 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 (1,38,-38,-1,1).

Formula

For n >= 5, a(n) = 38*a(n-2) - a(n-4) + 18. - Max Alekseyev, May 08 2009

G.f.: x^2*(x^3+2*x^2-19*x-2) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)). - Colin Barker, Feb 21 2013

a(n) = (A221874(n) - 1) / 2. - Bruno Berselli, Feb 21 2013

From Andrea Pinos, Oct 24 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). (End)

a(n) = (x^(2*(n+1)) + (-1)^n*(x^(2*n+1)+1) - x) / (2*x^n*(x^2 + 1)) - (1/2), with x=3+sqrt(10). - Federico Provvedi, Apr 04 2023

Maple

A005891 := proc(n) (5*n^2+5*n+2)/2 ; end: n := 0 : while true do if issqr(A005891(n)) then print(n) ; fi ; n := n+1 ; od : # R. J. Mathar, Jun 06 2007

Mathematica

Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[n] ], {n, 1, 40000} ]
LinearRecurrence[{1, 38, -38, -1, 1}, {0, 2, 21, 95, 816}, 30] (* Harvey P. Dale, Nov 09 2017 *)
Table[(((x^(n+2))+(((-1)^n*(x^(2*n+1)+1)-x)/(x^n)))/(x^2+1)-1)/2/.x->3+Sqrt[10], {n, 0, 50}]//Round (* Federico Provvedi, Apr 04 2023 *)

Prog

(PARI) a(n)=([0, 1, 0, 0, 0; 0, 0, 1, 0, 0; 0, 0, 0, 1, 0; 0, 0, 0, 0, 1; 1, -1, -38, 38, 1]^(n-1)*[0; 2; 21; 95; 816])[1, 1] \\ Charles R Greathouse IV, Feb 11 2019

Crossrefs

Cf. A005891 (centered pentagonal numbers), A129557 (numbers k>0 such that k^2 is a centered pentagonal number), A221874.

Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; A233450 for k=3; A001652 for k=4; this sequence for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013

Sequence in context: A034520 A111128 A213827 * A077209 A369754 A068045

Adjacent sequences: A129553 A129554 A129555 * A129557 A129558 A129559

Keyword

nonn,easy

Author

Alexander Adamchuk, Apr 20 2007

Extensions

More terms from R. J. Mathar, Jun 06 2007

Further terms from Max Alekseyev, May 08 2009

a(22)-a(23) from Colin Barker, Feb 21 2013

Status

approved