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

1, 2, 41, 937, 21506, 493697, 11333521, 260177282, 5972743961, 137112933817, 3147624733826, 72258255944177, 1658792261982241, 38079963769647362, 874180374439907081, 20068068648348215497, 460691398537569049346, 10575834097715739919457, 242783492848924449098161, 5573444501427546589338242, 127946440039984647105681401, 2937194676418219336841333977

Offset: 0

Author

Max Alekseyev, Jan 21 2022

Keywords

• nonn
• easy

Comments

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.

Other properties for all n:

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

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

Links

• Sebastian et al., Are there infinitely many positive integer solutions to (3+3k+l)^2=m(kl-k^3-1)?, MathOverflow, 2022.
• Index entries for linear recurrences with constant coefficients, signature (24,-24,1).

Crossrefs

Cf. A350916.

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.

Formula

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

G.f.: ( -1+22*x-17*x^2 ) / ( (x-1)*(x^2-23*x+1) ). - R. J. Mathar, Feb 07 2022