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a(0) = 1, a(1) = 2, and a(n) = 23*a(n-1) - a(n-2) - 4 for n >= 2.
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%I #11 Feb 07 2022 10:54:49

%S 1,2,41,937,21506,493697,11333521,260177282,5972743961,137112933817,

%T 3147624733826,72258255944177,1658792261982241,38079963769647362,

%U 874180374439907081,20068068648348215497,460691398537569049346,10575834097715739919457,242783492848924449098161,5573444501427546589338242,127946440039984647105681401,2937194676418219336841333977

%N a(0) = 1, a(1) = 2, and a(n) = 23*a(n-1) - a(n-2) - 4 for n >= 2.

%C One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

%C Other properties for all n:

%C (a(n)+1)*(a(n+2)+1) = (a(n+1)+1)*(a(n+1)+26);

%C ((105*a(n) - 20)^2 - 50^2) / 21 is an integer square.

%H Sebastian et al., <a href="https://mathoverflow.net/q/414378">Are there infinitely many positive integer solutions to (3+3k+l)^2=m(kl-k^3-1)?</a>, MathOverflow, 2022.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (24,-24,1).

%F a(n) = 17/42*A090731(n) - 15/2*A097778(n-1) + 4/21.

%F G.f.: ( -1+22*x-17*x^2 ) / ( (x-1)*(x^2-23*x+1) ). - _R. J. Mathar_, Feb 07 2022

%Y Cf. A350916.

%Y Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350919, A350920, A350921, A350922, A350923, A350924, A350925, A350926.

%K nonn,easy

%O 0,2

%A _Max Alekseyev_, Jan 21 2022