A106232 Least k > 0 such that (4*n^2 + 2)*(k^2) + (4*n^2 + 2)*k + 1 = j^2.

4, 4, 12, 24, 40, 60, 84, 112, 144, 180, 220, 264, 312, 364, 420, 480, 544, 612, 684, 760, 840, 924, 1012, 1104, 1200, 1300, 1404, 1512, 1624, 1740, 1860, 1984, 2112, 2244, 2380, 2520, 2664, 2812, 2964, 3120, 3280, 3444, 3612, 3784, 3960, 4140, 4324, 4512

Offset: 1

Author

Pierre CAMI, Apr 26 2005

Keywords

• nonn
• easy

Comments

For k there is always a recurrence.

For n=1, k(1,1) = 0, k(2,1) = 4 then k(i,1) = 10*k(i-1,1) + 4 - k(i-3,n).

For n>1, k(1,n) = 0, k(2,n) = 2*n^2 - 2*n, k(3,n) = 2*n^2 + 2*n, k(4,n) = (8*n^2+2)*k(2,n) + 4*n^2 then k(i,n) = (8*n^2+2)*k(i-2,n) + 4*n^2 - k(i-4,n).

Links

• Stefano Spezia, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A046092, A106231.

Programs

Magma

[4] cat [2*n*(n+1): n in [1..50]]; // Vincenzo Librandi, Apr 06 2020

Mathematica

LinearRecurrence[{3,-3,1},{4,4,12,24},50] (* Harvey P. Dale, Mar 05 2023 *)

PARI

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

Formula

a(1) = 4, a(n) = 2*n^2 - 2*n for n > 1, j sequence = A106231.

a(n) = A046092(n-1), n > 1. - R. J. Mathar, Aug 28 2008

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

E.g.f.: 4*x + 2*x^2*exp(x). - Stefano Spezia, Jun 06 2021

Extensions

More terms from Colin Barker, Mar 06 2013