A341109 a(n) = denominator(p(n, x)) / (n!*denominator(bernoulli(n, x))), where p(n, x) = Sum_{k=0..n} E2(n, k)*binomial(x + k, 2*n) / Product_{j=1..n} (j - x) and E2(n, k) are the second-order Eulerian numbers A201637.

1, 1, 2, 4, 8, 16, 96, 192, 1152, 768, 1536, 3072, 18432, 36864, 221184, 147456, 884736, 1769472, 10616832, 21233664, 637009920, 424673280, 2548039680, 5096079360, 152882380800, 61152952320, 366917713920, 81537269760, 163074539520, 326149079040, 1956894474240

Offset: 0

Author

Peter Luschny, Feb 06 2021

Keywords

• nonn

Comments

The challenge is to characterize the sequence purely arithmetically, i.e., without reference to the Eulerian numbers or the Bernoulli polynomials.

Links

• András Bazsó and István Mező, On the coefficients of power sums of arithmetic progressions, J. Number Th., 153 (2015), 117-123.
• J.-L. Chabert and P.-J. Cahen, Old problems and new questions around integer-valued polynomials and factorial sequences In: J. W. Brewer, S. Glaz, W. J. Heinzer, B. M. Olberding (eds), Multiplicative Ideal Theory in Commutative Algebra. Springer, Boston, MA., 2006.
• Bernd C. Kellner and Jonathan Sondow, Power-Sum Denominators, arXiv:1705.03857 [math.NT] 2017, Amer. Math. Monthly.

Crossrefs

Cf. A100655, A053657 (Minkowski), A341107, A341108, A318256, A144845, A163176, A201637 (Eulerian2), A036689, A324370, A007947, A324369, A195441.

Programs

Maple

Epoly := proc(n, x) add(combinat:-eulerian2(n, k)*binomial(x+k, 2*n), k = 0..n) / mul(j-x, j = 1..n): simplify(expand(%)) end:
seq(denom(Epoly(n, x)) / (n!*denom(bernoulli(n, x))), n = 0..30);

Mathematica

A053657[n_] := Product[p^Sum[Floor[(n-1)/((p-1) p^k)], {k,0,n}],{p, Prime[Range[n]]}];
A144845[n_] := Denominator[Together[BernoulliB[n, x]]];
A163176[n_] := A053657[n] / n!;
Table[(n + 1) A163176[n + 1] / A144845[n], {n, 0, 30}]

Sage

def A341109(n): # uses[A341108, A318256]
    return A341108(n)//A318256(n)
print([A341109(n) for n in (0..30)])

Formula

a(n) = A053657(n+1)/(n!*A144845(n)).

a(n) = (n+1)*A163176(n+1)/A144845(n).

a(n) = A341108(n)/A318256(n).

a(n) = A341107(n)*A324369(n+1).

a(n) = A341108(n)/A324370(n+1).

a(n) = A341108(n)*A007947(n+1)/A144845(n).

a(n) = A341108(n)*A324369(n+1)/A195441(n).

prime(n) divides a(k) for k >= A036689(n).

2^(n-1) divides exactly a(n) for n >= 2.