%I #12 Nov 27 2022 12:12:41
%S 1,7,43,253,1477,8611,50191,292537,1705033,9937663,57920947,337588021,
%T 1967607181,11468055067,66840723223,389576284273,2270616982417,
%U 13234125610231,77134136678971,449570694463597,2620290030102613,15272169486152083,89012726886809887,518804191834707241
%N Numbers k such that 8*k^2 + 8*k - 7 is a square.
%C a(n) is the n-th almost cobalancing number of second type (see Tekcan and Erdem).
%H Ahmet Tekcan and Alper Erdem, <a href="https://arxiv.org/abs/2211.08907">General Terms of All Almost Balancing Numbers of First and Second Type</a>, arXiv:2211.08907 [math.NT], 2022.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).
%F a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) for n > 3.
%F a(n) = (3*(3 - 2*sqrt(2))^n*(2 + sqrt(2)) + 3*(2 - sqrt(2))*(3 + 2*sqrt(2))^n - 4)/8.
%F O.g.f.: x*(1 + x^2)/((1 - x)*(1 - 6*x + x^2)).
%F E.g.f.: (3*(2 + sqrt(2))*(cosh(3*x - 2*sqrt(2)*x) + sinh(3*x - 2*sqrt(2)*x)) + 3*(2 - sqrt(2))*(cosh(3*x + 2*sqrt(2)*x) + sinh(3*x + 2*sqrt(2)*x)) - 4*(cosh(x) + sinh(x)) - 8)/8.
%F a(n) = 3*A011900(n) - 2 = 6*A053142(n) + 1. - _Hugo Pfoertner_, Nov 26 2022
%e a(2) = 7 is a term since 8*7^2 + 8*7 - 7 = 441 = 21^2.
%t LinearRecurrence[{7,-7,1},{1,7,43},24]
%Y Cf. A002315, A006452, A077443, A077446, A106328, A106329, A216134.
%K nonn,easy
%O 1,2
%A _Stefano Spezia_, Nov 26 2022