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A105037 a(0) = 0, a(1) = 4, a(2) = 6, a(3) = 98, for n>3 a(n) = 22*a(n-2) - a(n-4) + 10. 2

%I

%S 0,4,6,98,142,2162,3128,47476,68684,1042320,1507930,22883574,33105786,

%T 502396318,726819372,11029835432,15956920408,242153983196,

%U 350325429614,5316357794890,7691202531110,116717717504394,168856130254816

%N a(0) = 0, a(1) = 4, a(2) = 6, a(3) = 98, for n>3 a(n) = 22*a(n-2) - a(n-4) + 10.

%C It appears this sequence gives all nonnegative m such that 120*m^2 + 120*m + 1 is a square.

%F G.f.:-2*x*(2*x^2+x+2)/((x-1)*(x^4-22*x^2+1)) [From Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009]

%F a(n)=-1/2-1/5*(11 + 2*sqrt(30))^(1/4*(-1)^n)*(-1)^n*(11 + 2*sqrt(30))^(1/2*n)*(11 + 2 *sqrt(30))^(-1/4)*sqrt(30) + 5/4*(11 + 2*sqrt(30))^(1/4*(-1)^n)*(11 + 2*sqrt(30))^(1/2 *n)*(11 + 2*sqrt(30))^(-1/4) + 5/24*(11 + 2*sqrt(30))^(1/4*(-1)^n)*(11 + 2*sqrt(30))^(1/2*n)*(11 + 2*sqrt(30))^(-1/4)*sqrt(30)-5/24*(11-2*sqrt(30))^(-1/4)*sqrt(30)*(11-2 *sqrt(30))^(1/4*(-1)^n)*(11-2*sqrt(30))^(1/2*n) + 5/4*(11-2*sqrt(30))^(-1/4)*(11-2 *sqrt(30))^(1/4*(-1)^n)*(11-2*sqrt(30))^(1/2*n)-(11 + 2*sqrt(30))^(1/4*(-1)^n)*( -1)^n*(11 + 2*sqrt(30))^(1/2*n)*(11 + 2*sqrt(30))^(-1/4) + 1/5*(-1)^n*(11-2 *sqrt(30))^(-1/4)*sqrt(30)*(11-2*sqrt(30))^(1/4*(-1)^n)*(11-2*sqrt(30))^(1/2*n) -(-1)^n*(11-2*sqrt(30))^(-1/4)*(11-2*sqrt(30))^(1/4*(-1)^n)*(11-2*sqrt(30))^(1/2*n), with n>=0 [From _Paolo P. Lava_, Aug 28 2009]

%Y Cf. A077421.

%K nonn

%O 0,2

%A _Gerald McGarvey_, Apr 03 2005

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Last modified July 11 01:42 EDT 2020. Contains 335600 sequences. (Running on oeis4.)