OFFSET
1,2
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
REFERENCES
Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing (2012), pages 35-36.
LINKS
Colin Barker, Table of n, a(n) for n = 1..503
Muniru A. Asiru, All square chiliagonal numbers, Int J Math Edu Sci Technol, 47:7(2016), 1123-1134.
Leonhard 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).
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
E.g.f.: exp(49*x)*(12*cosh(20*sqrt(6)*x) + 5*sqrt(6)*sinh(20*sqrt(6)*x))/12. - Stefano Spezia, Sep 20 2025
EXAMPLE
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
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved