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a(n) = Sum_{k=0,n-1} 5^k*B(k)*binomial(n,k) where B(k) is the k-th Bernoulli number.
9

%I #22 Sep 28 2016 09:54:34

%S 0,1,-4,6,16,-74,-264,1946,9056,-88434,-512024,6154786,42716496,

%T -607884394,-4920817384,80834386026,747784582336,-13923204233954,

%U -144898927180344,3015393801263666,34867899296006576,-801997872697905114,-10201104981227536904,256982712667627683706

%N a(n) = Sum_{k=0,n-1} 5^k*B(k)*binomial(n,k) where B(k) is the k-th Bernoulli number.

%H Seiichi Manyama, <a href="/A083009/b083009.txt">Table of n, a(n) for n = 0..467</a>

%F E.g.f.: 5x/(1+exp(x)+exp(2x)+exp(3x)+exp(4x)). - _Benoit Cloitre_, Oct 26 2012 (following I. Gessel).

%t Range[0, 15]! CoefficientList[ Series[ 5x/(1 + Exp[x] + Exp[ 2x] + Exp[ 3x] + Exp[ 4x]), {x, 0, 15}], x] (* _Robert G. Wilson v_, Oct 26 2012 *)

%t Table[Sum[5^k*BernoulliB[k] Binomial[n, k], {k, 0, n - 1}], {n, 0, 23}] (* _Michael De Vlieger_, Sep 28 2016 *)

%o (PARI) a(n)=sum(k=0,n-1,5^k*binomial(n,k)*bernfrac(k))

%Y Cf. A001469.

%Y Cf. A036968, A083007, A083008, A083010, A083011, A083012, A083013, A083014.

%K sign

%O 0,3

%A _Benoit Cloitre_, May 31 2003

%E Offset changed to 0 by _Seiichi Manyama_, Sep 28 2016