%I #14 Aug 02 2019 13:45:40

%S 1,0,-2,0,14,0,-62,0,254,0,-5110,0,2828954,0,-114674,0,237036478,0,

%T -11499383114,0,183092554714,0,-3584085584926,0,3965530936622474,0,

%U -573989008898786,0,6375197353574922166,0,-9251189109760413581110,0,33111281730973040956798,0

%N a(n) = B2(n) * C(n) where B2(n) are generalized Bernoulli numbers and C(n) the Clausen numbers.

%C The Clausen numbers C(n) are T(n, 1) in A160014.

%H Peter Luschny, <a href="http://www.luschny.de/math/euler/StirlingFrobeniusNumbers.html">Stirling-Frobenius numbers</a>

%H Peter Luschny, <a href="http://www.luschny.de/math/euler/GeneralizedBernoulliNumbers.html">Generalized Bernoulli numbers</a>.

%F Let B(n,m) = sum_{k = 0..n} sum_{j = 0..k} sum_{v = 0..j} ((-1)^(n-v)/(j+1)) *binomial(n,k)*binomial(j,v)*(m*v)^k then a(n) = B(n,2)*A141056(n).

%F Let B2(n) = sum_{k=0..n} ((-1)^k*k!/(k+1)) S_{2}(n, k) where S_{2}(n, k) the Stirling-Frobenius subset numbers A039755(n, k) then a(n) = B2(n)*A141056(n).

%e The numerators of 1/1, 0/2, -2/6, 0/2, 14/30, 0/2, -62/42, 0/2, 254/30, 0/2, -5110/66, 0/2, 2828954/2730, ... (the denominators are the Clausen numbers).

%p B := (n, m) -> add(add(add(((-1)^(n-v)/(j+1))*binomial(n,k)*binomial(j, v)*(m*v)^k, v = 0..j), j = 0..k), k = 0..n);

%p C := proc(n) numtheory[divisors](n);map(i->i+1,%);select(isprime,%);mul(i,i=%) end:

%p A225480 := n -> B(n, 2)*C(n); seq(A225480(n), n = 0..33);

%t B[n_, m_] := Sum[((-1)^(n - v)/(j + 1))*Binomial[n, k]*Binomial[j, v]*If[k == 0, 1, (m*v)^k], {k, 0, n}, {j, 0, k}, {v, 0, j}];

%t c[n_] := Denominator[Sum[Boole[PrimeQ[d + 1]]/(d + 1), {d, Divisors[n]}]];

%t a[n_] := B[n, 2]*c[n];

%t Table[a[n], {n, 0, 33}] (* _Jean-François Alcover_, Aug 02 2019, from Maple *)

%o @CachedFunction

%o def EulerianNumber(n, k, m) : # The Eulerian numbers

%o if n == 0: return 1 if k == 0 else 0

%o return (m*(n-k)+m-1)*EulerianNumber(n-1,k-1,m)+(m*k+1)*EulerianNumber(n-1,k,m)

%o @CachedFunction

%o def B(n, m): # The generalized Bernoulli numbers

%o return add(add(EulerianNumber(n, j, m)*binomial(j, n - k)

%o for j in (0..n))*(-1)^k/(k+1) for k in (0..n))

%o def A225480(n):

%o if n == 0: return 1

%o C = mul(filter(lambda s: is_prime(s) , map(lambda i: i+1, divisors(n))))

%o return C*B(n, 2)

%o [A225480(n) for n in (0..33)]

%Y Cf. A141056, A160014.

%K sign,frac

%O 0,3

%A _Peter Luschny_, May 30 2013