OFFSET
1,1
COMMENTS
For j^2 = (n^2 + 1)*(k^2) + (n^2 + 1)*k + 1, there is a sequence j(i,n) with a recurrence.
For n=1, j(1,1) = 1, j(2,1) = 5, j(i,1) = 6*j(i-1,1) - j(i-2,1).
For n=2, j(1,2) = 1, j(2,2) = 19, j(i,2) = 18*j(i-1,2) - j(i-2,2).
For n>2, j(1,n) = 1, j(2,n) = n^3 - 2*n^2 + n - 1, j(3,n) = n^3 + 2*n^2 + n + 1, j(4,n) = (4*n^2 + 2)*j(2,n) + 1 then j(i,n) = (4*n^2+2)*j(i-2,n) - j(i-4,n).
LINKS
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
For n > 2, a(n) = n^3 - 2*n^2 + n - 1.
MATHEMATICA
LinearRecurrence[{4, -6, 4, -1}, {5, 19, 11, 35, 79, 149}, 43] (* Georg Fischer, Oct 25 2020 *)
PROG
(PARI) a(n) = if(n<3, 14*n-9, n^3-2*n^2+n-1); \\ Jinyuan Wang, Apr 07 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Pierre CAMI, Apr 26 2005
EXTENSIONS
More terms from Jinyuan Wang, Apr 07 2020
STATUS
approved