A046173 Indices of square numbers that are also pentagonal.

1, 99, 9701, 950599, 93149001, 9127651499, 894416697901, 87643708742799, 8588189040096401, 841554882220704499, 82463790268588944501, 8080609891439495856599, 791817305570802005002201, 77590015336047156994359099, 7603029685627050583442189501

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• easy

Comments

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

a(n)^2 is of the form (2*m-1)*(3*m-2), and the corresponding values of m are 1, 41, 3961, 388081, 38027921, 3726348121, 365144087881, ..., with closed form ((5-2*sqrt(6))^(2n-1)+(5+2*sqrt(6))^(2n-1)+14)/24 (for n>0). - Bruno Berselli, Dec 12 2013

The terms of this sequence satisfy the Diophantine equation m^2 = k * (3k-1)/2, which is equivalent to (6k-1)^2 - 6*(2*m)^2 = 1. Now, with x=6k-1 and y=2*m, we get the Pell-Fermat equation x^2 - 6*y^2 = 1. The solutions (x,y) of this equation are respectively in A046174 and A046175. The indices m=y/2 of the square numbers which are also pentagonal are the terms of this sequence, the indices k=(x+1)/6 of the pentagonal numbers which are also square are in A046172, and the pentagonal square numbers are in A036353. - Bernard Schott, Mar 10 2019

Also, this sequence is related to A302330 by (sqrt(2) + sqrt(3))^(4*n-2) = A302330(n-1)*5 + a(n)*sqrt(24). - Bruno Berselli, Oct 29 2019

Examples

			G.f. = x + 99*x^2 + 9701*x^3 + 950599*x^4 + 93149001*x^5 + ...
99 is a term because 99^2 = 9801 = (1/2) * 81 * (3*81 - 1), so 9801 is the 99th square number, also the 81st pentagonal number, and the second pentagonal square number after 1. - _Bernard Schott_, Mar 10 2019
		

References

• E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 35.

Links

• Colin Barker, Table of n, a(n) for n = 1..503
• M. A. Asiru, All square chiliagonal numbers, Int J Math Edu Sci Technol, 47:7(2016), 1123-1134.
• L. Euler, De solutione problematum diophanteorum per numeros integros, par. 21
• Tanya Khovanova, Recursive Sequences
• Eric Weisstein's World of Mathematics, Pentagonal Square Number
• Index entries for linear recurrences with constant coefficients, signature (98,-1).

Crossrefs

Cf. A036353 (pentagonal square numbers), A046172 (indices of pentagonal numbers that are also square).

Cf. A046174, A046175 (solutions of x^2 - 6*y^2 = 1).

Cf. A302330.

Programs

Mathematica

CoefficientList[Series[(1 + x)/(1 - 98* x + x^2), {x, 0, 30}], x] (* T. D. Noe, Aug 01 2011 *)
LinearRecurrence[{98, -1}, {1, 99}, 30] (* Harvey P. Dale, Jul 31 2017 *)

PARI

{a(n) = subst( poltchebi(n) - poltchebi(n-1), 'x, 49) / 48}; /* Michael Somos, Sep 05 2006 */

PARI

Vec(x*(x+1)/(x^2-98*x+1) + O(x^30)) \\ Colin Barker, Jun 23 2015

Formula

a(n) = 98*a(n-1) - a(n-2); g.f.: (1+x)/(1-98*x+x^2). - Warut Roonguthai, Jan 05 2001

a(1-n) = -a(n) for all n in Z. - Michael Somos, Sep 05 2006

Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(f(a(n-1),5),5). - Marcos Carreira, Dec 27 2006

a(n) = ((12+5*sqrt(6))/24)*(5+2*sqrt(6))^(2*n)+((12-5*sqrt(6))/24)*(5-2*sqrt(6))^(2*n) for n>=0. - Richard Choulet, Apr 29 2009

a(n+1) = 49*a(n) + 10*sqrt(24*a(n)^2+1) for n > =0 with a(0)=1. - Richard Choulet, Apr 29 2009

a(n) = b such that (-1)^n*Integral_{x=-Pi/2..Pi/2} (cos(2*n-1)*x)/(5-sin(x)) dx = c + b*(log(2)-log(3)). - Francesco Daddi, Aug 01 2011

a(n) = floor((1/24) * sqrt(6) * (sqrt(2) + sqrt(3))^(4n-2)). - Ant King, Nov 07 2011

a(n) = A138288(n)*A054320(n). - Gerry Martens, May 13 2024