A106231 Least j > 1 such that j^2 = (4*n^2 + 2)*(k^2) + (4*n^2 + 2)*k + 1.

11, 19, 77, 199, 409, 731, 1189, 1807, 2609, 3619, 4861, 6359, 8137, 10219, 12629, 15391, 18529, 22067, 26029, 30439, 35321, 40699, 46597, 53039, 60049, 67651, 75869, 84727, 94249, 104459, 115381, 127039, 139457, 152659, 166669, 181511, 197209, 213787

Offset: 1

Author

Pierre CAMI, Apr 26 2005

Keywords

• nonn
• easy

Comments

For j there is always a recurrence.

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).

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).

Links

• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A106232.

Programs

PARI

a(n) = if(n==1, 11, 4*n^3-4*n^2+2*n-1); \\ Jinyuan Wang, Apr 07 2020

Formula

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

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

Extensions

More terms from Colin Barker, Mar 06 2013