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A350917 a(0) = 1, a(1) = 2, and a(n) = 23*a(n-1) - a(n-2) - 4 for n >= 2.

%I A350917 #11 Feb 07 2022 10:54:49
%S A350917 1,2,41,937,21506,493697,11333521,260177282,5972743961,137112933817,
%T A350917 3147624733826,72258255944177,1658792261982241,38079963769647362,
%U A350917 874180374439907081,20068068648348215497,460691398537569049346,10575834097715739919457,242783492848924449098161,5573444501427546589338242,127946440039984647105681401,2937194676418219336841333977
%N A350917 a(0) = 1, a(1) = 2, and a(n) = 23*a(n-1) - a(n-2) - 4 for n >= 2.
%C A350917 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 A350917 Other properties for all n:
%C A350917 (a(n)+1)*(a(n+2)+1) = (a(n+1)+1)*(a(n+1)+26);
%C A350917 ((105*a(n) - 20)^2 - 50^2) / 21 is an integer square.
%H A350917 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 A350917 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (24,-24,1).
%F A350917 a(n) = 17/42*A090731(n) - 15/2*A097778(n-1) + 4/21.
%F A350917 G.f.: ( -1+22*x-17*x^2 ) / ( (x-1)*(x^2-23*x+1) ). - _R. J. Mathar_, Feb 07 2022
%Y A350917 Cf. A350916.
%Y A350917 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 A350917 nonn,easy
%O A350917 0,2
%A A350917 _Max Alekseyev_, Jan 21 2022