OFFSET
1,1
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).
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
PROG
(PARI) a(n) = if(n==1, 11, 4*n^3-4*n^2+2*n-1); \\ Jinyuan Wang, Apr 07 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Pierre CAMI, Apr 26 2005
EXTENSIONS
More terms from Colin Barker, Mar 06 2013
STATUS
approved