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f=x=>x>0&&4*ff=x=>x>1&&4*f(x-1)+f(x-2)+2
Recursion to the rescue! This uses one of the formulas on the OEIS page:
f(n < 01) = 0, f(n) = 4*a(n+1)+a(n)+2
(but shifted by one because the challenge shifts it by one)
f=x=>x>0&&4*f(x-1)+f(x-2)+2
f(n < 0) = 0, f(n) = 4*a(n+1)+a(n)+2
f=x=>x>1&&4*f(x-1)+f(x-2)+2
f(n < 1) = 0, f(n) = 4*a(n+1)+a(n)+2
