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%I #104 Oct 30 2019 03:41:12
%S 1,99,9701,950599,93149001,9127651499,894416697901,87643708742799,
%T 8588189040096401,841554882220704499,82463790268588944501,
%U 8080609891439495856599,791817305570802005002201,77590015336047156994359099,7603029685627050583442189501
%N Indices of square numbers that are also pentagonal.
%C As n increases, this sequence is approximately geometric with common ratio r = lim_{n->infinity} a(n)/a(n-1)) = (sqrt(2) + sqrt(3))^4 = 49 + 20 * sqrt(6). - _Ant King_, Nov 07 2011
%C 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
%C 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
%C 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
%D E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 35.
%H Colin Barker, <a href="/A046173/b046173.txt">Table of n, a(n) for n = 1..503</a>
%H M. A. Asiru, <a href="http://dx.doi.org/10.1080/0020739X.2016.1164346">All square chiliagonal numbers</a>, Int J Math Edu Sci Technol, 47:7(2016), 1123-1134.
%H L. Euler, <a href="http://eulerarchive.maa.org/pages/E029.html">De solutione problematum diophanteorum per numeros integros</a>, par. 21
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PentagonalSquareNumber.html">Pentagonal Square Number</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (98,-1).
%F a(n) = 98*a(n-1) - a(n-2); g.f.: (1+x)/(1-98*x+x^2). - _Warut Roonguthai_, Jan 05 2001
%F a(1-n) = -a(n) for all n in Z. - _Michael Somos_, Sep 05 2006
%F 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
%F 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
%F 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
%F 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
%F a(n) = floor((1/24) * sqrt(6) * (sqrt(2) + sqrt(3))^(4n-2)). - _Ant King_, Nov 07 2011
%e G.f. = x + 99*x^2 + 9701*x^3 + 950599*x^4 + 93149001*x^5 + ...
%e 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
%t CoefficientList[Series[(1 + x)/(1 - 98* x + x^2), {x, 0, 30}], x] (* _T. D. Noe_, Aug 01 2011 *)
%t LinearRecurrence[{98, -1}, {1, 99}, 30] (* _Harvey P. Dale_, Jul 31 2017 *)
%o (PARI) {a(n) = subst( poltchebi(n) - poltchebi(n-1), 'x, 49) / 48}; /* _Michael Somos_, Sep 05 2006 */
%o (PARI) Vec(x*(x+1)/(x^2-98*x+1) + O(x^30)) \\ _Colin Barker_, Jun 23 2015
%Y Cf. A036353 (pentagonal square numbers), A046172 (indices of pentagonal numbers that are also square).
%Y Cf. A046174, A046175 (solutions of x^2 - 6*y^2 = 1).
%Y Cf. A302330.
%K nonn,easy
%O 1,2
%A _Eric W. Weisstein_
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