A129557 -id:A129557 - cp's OEIS frontend

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

Alexander Adamchuk, Apr 20 2007

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

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).

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

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

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 *)

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

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

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

A254332 Indices of centered pentagonal numbers (A005891) which are also squares (A000290).

Original entry on oeis.org

1, 3, 22, 96, 817, 3627, 31006, 137712, 1177393, 5229411, 44709910, 198579888, 1697799169, 7540806315, 64471658494, 286352060064, 2448225223585, 10873837476099, 92968086837718, 412919472031680, 3530339074609681, 15680066099727723, 134059916748330142

Offset: 1

Colin Barker, Jan 28 2015

Also positive integers y in the solutions to 2*x^2 - 5*y^2 + 5*y - 2 = 0, the corresponding values of x being A129557.

			3 is in the sequence because the 3rd centered pentagonal number is 16, which is also the 4th square number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. A000290, A005891, A129557, A254333.

LinearRecurrence[{1,38,-38,-1,1},{1,3,22,96,817},30] (* Harvey P. Dale, Mar 27 2017 *)

Vec(x*(2*x^3+19*x^2-2*x-1) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)) + O(x^100))

a(n) = a(n-1) + 38*a(n-2) - 38*a(n-3) - a(n-4) + a(n-5).
G.f.: x*(2*x^3 + 19*x^2 - 2*x - 1) / ((x-1)*(x^2 - 6*x - 1)*(x^2 + 6*x - 1)).
a(n) = (1/40)*(20 - b^n*(19 + 3*b) + (3 + b)*c^n - (b^n*(3 + b) + (1 - 3*b)*c^n)*(-1)^n) with b = sqrt(10) - 3 and c = sqrt(10) + 3. - Alan Michael Gómez Calderón, Jul 02 2024

A254333 Squares (A000290) which are also centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 16, 1156, 22801, 1666681, 32878756, 2403352576, 47411143081, 3465632747641, 68366835443776, 4997440018745476, 98584929298781641, 7206305041398228481, 142159399682007682276, 10391486872256226723856, 204993755756525779060081, 14984516863488437537571601

Offset: 1

Colin Barker, Jan 28 2015

			16 is in the sequence because it is the 4th square number and the 3rd centered pentagonal number.

Colin Barker, Table of n, a(n) for n = 1..634
Index entries for linear recurrences with constant coefficients, signature (1,1442,-1442,-1,1).

Cf. A000290, A005891, A129557, A254332.

LinearRecurrence[{1,1442,-1442,-1,1},{1,16,1156,22801,1666681},20] (* Harvey P. Dale, Jul 26 2015 *)

Vec(-x*(x^4+15*x^3-302*x^2+15*x+1) / ((x-1)*(x^2-38*x+1)*(x^2+38*x+1)) + O(x^100))

a(n) = a(n-1)+1442*a(n-2)-1442*a(n-3)-a(n-4)+a(n-5).

G.f.: -x*(x^4+15*x^3-302*x^2+15*x+1) / ((x-1)*(x^2-38*x+1)*(x^2+38*x+1)).

cp's OEIS Frontend

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

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A254332 Indices of centered pentagonal numbers (A005891) which are also squares (A000290).

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A254333 Squares (A000290) which are also centered pentagonal numbers (A005891).

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Mathematica

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