OFFSET
1,1
COMMENTS
Here m = (11*(2*n-1)+3*(-1)^n)/4 for n>0.
More generally, (t*((11*(2*n-1)+k*(-1)^n)/4)^2 +1)/11 = ( 22*t*n*(n-1) +t*k*(2*n-1)*(-1)^n+(t*(k^2+121)+16)/22 )/8 for any natural number t == 2, 6, 7, 8, 10 (mod 11) and k = -5, -1, -9, 3, 7, respectively.
LINKS
B. Berselli, Table of n, a(n) for n = 1..10000.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).
FORMULA
a(n) = (330*n*(n-1)+45*(2*n-1)*(-1)^n+89)/4.
G.f.: x*(11+210*x+218*x^2+210*x^3+11*x^4)/((1+x)^2*(1-x)^3).
a(n) = a(-n+1) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5).
MATHEMATICA
LinearRecurrence[{1, 2, -2, -1, 1}, {11, 221, 461, 1091, 1571}, 40] (* Harvey P. Dale, Mar 03 2023 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Jul 11 2010 - Dec 10 2010
STATUS
approved