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%I #23 Dec 10 2023 17:33:11
%S 1,1,2,4,8,16,96,192,1152,768,1536,3072,18432,36864,221184,147456,
%T 884736,1769472,10616832,21233664,637009920,424673280,2548039680,
%U 5096079360,152882380800,61152952320,366917713920,81537269760,163074539520,326149079040,1956894474240
%N 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.
%C The challenge is to characterize the sequence purely arithmetically, i.e., without reference to the Eulerian numbers or the Bernoulli polynomials.
%H András Bazsó and István Mező, <a href="http://dx.doi.org/10.1016/j.jnt.2015.01.019">On the coefficients of power sums of arithmetic progressions</a>, J. Number Th., 153 (2015), 117-123.
%H J.-L. Chabert and P.-J. Cahen, <a href="https://web.archive.org/web/20120114162802/http://www.latp.cahen.u-3mrs.fr/Recherche/Pubs/oldpb.pdf">Old problems and new questions around integer-valued polynomials and factorial sequences</a> In: J. W. Brewer, S. Glaz, W. J. Heinzer, B. M. Olberding (eds), Multiplicative Ideal Theory in Commutative Algebra. Springer, Boston, MA., 2006.
%H Bernd C. Kellner and Jonathan Sondow, <a href="https://arxiv.org/abs/1705.03857">Power-Sum Denominators</a>, arXiv:1705.03857 [math.NT] 2017, Amer. Math. Monthly.
%F a(n) = A053657(n+1)/(n!*A144845(n)).
%F a(n) = (n+1)*A163176(n+1)/A144845(n).
%F a(n) = A341108(n)/A318256(n).
%F a(n) = A341107(n)*A324369(n+1).
%F a(n) = A341108(n)/A324370(n+1).
%F a(n) = A341108(n)*A007947(n+1)/A144845(n).
%F a(n) = A341108(n)*A324369(n+1)/A195441(n).
%F prime(n) divides a(k) for k >= A036689(n).
%F 2^(n-1) divides exactly a(n) for n >= 2.
%p 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:
%p seq(denom(Epoly(n, x)) / (n!*denom(bernoulli(n, x))), n = 0..30);
%t A053657[n_] := Product[p^Sum[Floor[(n-1)/((p-1) p^k)], {k,0,n}],{p, Prime[Range[n]]}];
%t A144845[n_] := Denominator[Together[BernoulliB[n, x]]];
%t A163176[n_] := A053657[n] / n!;
%t Table[(n + 1) A163176[n + 1] / A144845[n], {n, 0, 30}]
%o (Sage)
%o def A341109(n): # uses[A341108, A318256]
%o return A341108(n)//A318256(n)
%o print([A341109(n) for n in (0..30)])
%Y Cf. A100655, A053657 (Minkowski), A341107, A341108, A318256, A144845, A163176, A201637 (Eulerian2), A036689, A324370, A007947, A324369, A195441.
%K nonn
%O 0,3
%A _Peter Luschny_, Feb 06 2021
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