|
%I #9 Apr 07 2013 07:26:07
%S 1,1,5,28,273,6251,578162,107060591,29911744769,27309372325966,
%T 100510174785157275,579282314757603925315,5692451844585536053973346,
%U 272831740026972379247127727751,36494329378701187545939734030067963
%N a(n) = Sum_{k=0..[n/2]} A034807(n,k)^n, where A034807 is a triangle of Lucas polynomials.
%F a(n) = Sum_{k=0..[n/2]} [C(n-k,k) + C(n-k-1,k-1)]^n.
%F Ignoring the zeroth term, equals the logarithmic derivative of A171186.
%e The n-th term equals the sum of the n-th powers of the n-th row of triangle A034807:
%e a(0) = 2^0 = 1;
%e a(1) = 1^1 = 1;
%e a(2) = 1^2 + 2^2 = 5;
%e a(3) = 1^3 + 3^3 = 28;
%e a(4) = 1^4 + 4^4 + 2^4 = 273;
%e a(5) = 1^5 + 5^5 + 5^5 = 6251;
%e a(6) = 1^6 + 6^6 + 9^6 + 2^6 = 578162;
%e a(7) = 1^7 + 7^7 + 14^7 + 7^7 = 107060591; ...
%o (PARI) {a(n)=sum(k=0, n\2, (binomial(n-k, k)+binomial(n-k-1, k-1))^n)}
%Y Cf. A171186, A034807, A067961.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 13 2009
| 