A046172 Indices of pentagonal numbers (A000326) that are also squares (A000290).

1, 81, 7921, 776161, 76055841, 7452696241, 730288175761, 71560788528321, 7012226987599681, 687126683996240401, 67331402804643959601, 6597790348171111800481, 646516122717964312487521, 63351982236012331511976561, 6207847743006490523861215441

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• easy

Comments

if P_x = y^2 is a pentagonal number that is also a square, the least both pentagonal and square number that is greater as P_x, is P_(49*x + 40*y - 8) = (60*x + 49*y - 10)^2 (in fact, P_(49*x + 40*y - 8) - (60*x + 49*y - 10)^2 = (3/2)*x^2 - (1/2)*x - y^2). - Richard Choulet, Apr 28 2009

a(n)*(3*a(n)-1)/2 = m^2 is equivalent to the Pell equation (6*a(n)-1)^2 - 6*(2*m)^2 = 1 or x(n)^2 - 6*y(n)^2 = 1. - Paul Weisenhorn, May 15 2009

As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> oo} a(n)/a(n-1) = (sqrt(2) + sqrt(3))^4 = 49 + 20*sqrt(6). - Ant King, Nov 07 2011

Numbers k such that the k-th pentagonal number is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 11 2014

Indices of pentagonal numbers (A000326) that are also centered octagonal numbers (A016754). - Colin Barker, Jan 11 2015

References

• Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7; http://dx.doi.org/10.1080/0020739X.2016.1164346

Links

• Colin Barker, Table of n, a(n) for n = 1..503
• Leonhard Euler, De solutione problematum diophanteorum per numeros integros, section 21.
• Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
• W. Sierpiński, Sur les nombres pentagonaux, Bull. Soc. Roy. Sci. Liege 33 (1964) 513-517.
• Eric Weisstein's World of Mathematics, Pentagonal Square Number.
• Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Crossrefs

Cf. A036353, A046173.

Cf. A000217, A000290, A000326, A251914, A248205.

Programs

Mathematica

LinearRecurrence[{99, -99, 1}, {1, 81, 7921}, 13] (* Ant King, Nov 07 2011 *)
Table[Round[(1 + x^(2*n+1))^2 / (12*x^(2*n+1)) /. x->5+2*Sqrt@6],{n,0,99}] (* Federico Provvedi, Apr 24 2023 *)

Formula

a(n) = 98*a(n-1) - a(n-2) - 16; g.f.: x*(1 - 18*x + x^2)/((1-x)*(1 - 98*x + x^2)). - Warut Roonguthai Jan 05 2001 - Corrected by Colin Barker, Jan 11 2015

a(n+1) = 49*a(n) - 8 + 10*sqrt(8*(3*a(n)^2 - a(n))) with a(1) = 1. - Richard Choulet, Apr 28 2009

a(n) = 1/6+((5 + 2*sqrt(6))^(2*n+1)/12) + ((5 - 2*sqrt(6))^(2*n+1)/12) for n >= 0. - Richard Choulet, Apr 29 2009

From Paul Weisenhorn, May 15 2009: (Start)

x(n+2) = 98*x(n+1) - x(n) with x(1)=5, x(2)=485;

y(n+2) = 98*y(n+1) - y(n) with y(n)=A046173(n)*2;

m(n+2) = 98*m(n+1) - m(n) with m(n)=A046173(n);

a(n) = A072256(n)^2.

(End)

a(n) = b(n)*b(n), b(n) = 10*b(n-1)- b(n-2), b(1)=1, b(2)=9, b(n)=((5 + sqrt(24))^n - (5 - sqrt(24))^n)/(2*sqrt(24)). - Sture Sjöstedt, Sep 21 2009

From Ant King, Nov 07 2011: (Start)

a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3).

a(n) = ceiling((1/12)*(sqrt(3) + sqrt(2))^(4*n-2)).

(End)

a(n) = (1 + x^(2n+1))^2 / (12*x^(2*n+1)), with x = 5 + 2*sqrt(6). - Federico Provvedi, Apr 24 2023