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%I #21 Sep 08 2022 08:45:28
%S 1,3,24,147,864,5043,29400,171363,998784,5821347,33929304,197754483,
%T 1152597600,6717831123,39154389144,228208503747,1330096633344,
%U 7752371296323,45184131144600,263352415571283,1534930362283104
%N Numbers k such that A125650(k) is a perfect square.
%C Corresponding numbers m such that m^2 = A125650(a(n)) are listed in A125652.
%C 3 divides a(n) for n>1. For n>1 a(n) = 3*A001108(n-1), where A001108(k) = {0, 1, 8, 49, 288, 1681, ...}, A001108(k)-th triangular number is a square. - _Alexander Adamchuk_, Jan 19 2007
%C Disregarding the term 1, numbers k such that A071910(k) is a nonzero square; i.e., numbers k such that A000096(k) = k*(k+3)/2 is a nonzero square. - _Rick L. Shepherd_, Jul 13 2012
%H Vincenzo Librandi, <a href="/A125651/b125651.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7, -7, 1).
%F For n>1, a(n+2) = 6*a(n+1) - a(n) + 6.
%F For n>1, a(n) = ((3+2*sqrt(2))^(n-1) + (3-2*sqrt(2))^(n-1))*3/4 - 3/2.
%F a(2k) = 3*A002315(n)^2; a(2k+1) = 6*A001542(n)^2.
%F a(n) = 3*A001108(n-1) for n>1. - _Alexander Adamchuk_, Jan 19 2007
%F For n>1, a(2)=3, a(3)=24, a(4)=147, a(n)=7*a(n-1)-7*a(n-2)+a(n-3) [From Harvey P. Dale, May 15 2011]
%F G.f.: (-1+x(4+(-10+x)x))/((-1+x)(1+(-6+x)x)) [From Harvey P. Dale, May 15 2011]
%e a(2)=3 because A125650(3)=9=3^2; a(3)=24 because A125650(24)=81=9^2.
%t Join[{1},LinearRecurrence[{7,-7,1},{3,24,147},35]] (* or *) CoefficientList[Series[(-1+x(4+(-10+x)x))/((-1+x)(1+(-6+x) x)),{x,0,35}],x] (* _Harvey P. Dale_, May 15 2011 *)
%o (Magma) I:=[1, 3, 24, 147]; [n le 4 select I[n] else 7*Self(n-1)-7*Self(n-2)+Self(n-3): n in [1..30]]; _Vincenzo Librandi_, May 21 2012
%Y Cf. A125650, A125652.
%Y Cf. A001108, A071910, A000096, A000217.
%K nonn,easy
%O 1,2
%A _Alexander Adamchuk_, Nov 29 2006
%E Edited by _Max Alekseyev_, Jan 11 2007
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