%I #16 Apr 07 2020 08:48:44

%S 11,19,77,199,409,731,1189,1807,2609,3619,4861,6359,8137,10219,12629,

%T 15391,18529,22067,26029,30439,35321,40699,46597,53039,60049,67651,

%U 75869,84727,94249,104459,115381,127039,139457,152659,166669,181511,197209,213787

%N Least j > 1 such that j^2 = (4*n^2 + 2)*(k^2) + (4*n^2 + 2)*k + 1.

%C For j there is always a recurrence.

%C For n=1, j(1,1) = 1, j(2,1) = 10*j(1,1) + 1, then j(i,1) = 10*j(i-1,1) - j(i-3).

%C For n>1, j(1,n) = 1, j(2,n) = 4*n^3 - 4*n^2 + 2*n - 1, j(3,n) = 4*n^3 + 4*n^2 + 2*n+1, j(4,n) = (8*n^2+2)*j(2,n) + 1 then j(i,n) = (8*n^2+2)*j(i-2) - j(i-4,n).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(1) = 11, a(n) = 4*n^3 - 4*n^2 + 2*n - 1 for n > 1, k sequence = A106232.

%F G.f.: x*(10*x^4-39*x^3+67*x^2-25*x+11) / (x-1)^4. - _Colin Barker_, Mar 06 2013

%o (PARI) a(n) = if(n==1, 11, 4*n^3-4*n^2+2*n-1); \\ _Jinyuan Wang_, Apr 07 2020

%Y Cf. A106232.

%K nonn,easy

%O 1,1

%A _Pierre CAMI_, Apr 26 2005

%E More terms from _Colin Barker_, Mar 06 2013