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%I #71 Jun 11 2023 12:08:23
%S 1,81,7921,776161,76055841,7452696241,730288175761,71560788528321,
%T 7012226987599681,687126683996240401,67331402804643959601,
%U 6597790348171111800481,646516122717964312487521,63351982236012331511976561,6207847743006490523861215441
%N Indices of pentagonal numbers (A000326) that are also squares (A000290).
%C 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
%C 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
%C 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
%C Numbers k such that the k-th pentagonal number is equal to the sum of two consecutive triangular numbers. - _Colin Barker_, Dec 11 2014
%C Indices of pentagonal numbers (A000326) that are also centered octagonal numbers (A016754). - _Colin Barker_, Jan 11 2015
%D 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
%H Colin Barker, <a href="/A046172/b046172.txt">Table of n, a(n) for n = 1..503</a>
%H Leonhard Euler, <a href="https://scholarlycommons.pacific.edu/euler-works/29/">De solutione problematum diophanteorum per numeros integros</a>, section 21.
%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2018volume18/FG201808index.html">Integer Sequences and Circle Chains Inside a Circular Segment</a>, Forum Geometricorum, Vol. 18 (2018), 47-55.
%H W. Sierpiński, <a href="http://popups.ulg.ac.be/0037-9565/index.php?id=3612">Sur les nombres pentagonaux</a>, Bull. Soc. Roy. Sci. Liege 33 (1964) 513-517.
%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_03">Index entries for linear recurrences with constant coefficients</a>, signature (99,-99,1).
%F 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
%F a(n+1) = 49*a(n) - 8 + 10*sqrt(8*(3a(n)^2 - a(n)) with a(1) = 1. - _Richard Choulet_, Apr 28 2009
%F 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
%F From _Paul Weisenhorn_, May 15 2009: (Start)
%F x(n+2) = 98*x(n+1) - x(n) with x(1)=5, x(2)=485;
%F y(n+2) = 98*y(n+1) - y(n) with y(n)=A046173(n)*2;
%F m(n+2) = 98*m(n+1) - m(n) with m(n)=A046173(n);
%F a(n) = A072256(n)^2.
%F (End)
%F 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
%F From _Ant King_, Nov 07 2011: (Start)
%F a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3).
%F a(n) = ceiling((1/12)*(sqrt(3) + sqrt(2))^(4*n-2)).
%F (End)
%F a(n) = (1 + x^(2n+1))^2 / (12*x^(2*n+1)), with x = 5 + 2*sqrt(6). - _Federico Provvedi_, Apr 24 2023
%t LinearRecurrence[{99, -99, 1}, {1, 81, 7921}, 13] (* _Ant King_, Nov 07 2011 *)
%t 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 *)
%Y Cf. A036353, A046173.
%Y Cf. A000217, A000290, A000326, A251914, A248205.
%K nonn,easy
%O 1,2
%A _Eric W. Weisstein_
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