OFFSET
0,2
COMMENTS
Binomial transform of A081909 4th binomial transform of (1,0,1,0,0,0,......). Case k=4 where a(n,k) = k^n*(n^2 - n + 2*k^2)/(2*k^2) with g.f. (1 - 2*k*x + (k^2+1)*x^2)/(1-k*x)^3.
LINKS
Vincenzo Librandi and Harvey P. Dale, Table of n, a(n) for n = 0..1000 [First 168 terms from Vincenzo Librandi]
Index entries for linear recurrences with constant coefficients, signature (12,-48,64).
FORMULA
a(n) = 4^n*(n^2 - n + 32)/32.
G.f.: (1-8*x+17*x^2)/(1-4*x)^3.
a(n) = 12*a(n-1) - 48*a(n-2) + 64*a(n-3), a(0)=1, a(1)=4, a(2)=17. - Harvey P. Dale, Jan 18 2014
E.g.f.: (1 + x^2/2)*exp(4*x). - Elmo R. Oliveira, Nov 12 2025
MAPLE
MATHEMATICA
Table[(4^n (n^2 - n + 32))/32, {n, 0, 30}] (* or *) LinearRecurrence[{12, -48, 64}, {1, 4, 17}, 30] (* Harvey P. Dale, Jan 18 2014 *)
CoefficientList[Series[(1 - 8 x + 17 x^2)/(1 - 4 x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 14 2014 *)
PROG
(Magma) [4^n*(n^2-n+32)/32: n in [0..40]]; // Vincenzo Librandi, Apr 27 2011
(PARI) a(n) = 4^n*(n^2-n+32)/32; \\ Joerg Arndt, Mar 12 2014
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Mar 31 2003
STATUS
approved