%I #50 Oct 14 2019 11:36:46
%S 1,1,2,4,48,96,1152,2304,276480,552960,6635520,13271040,33443020800,
%T 66886041600,802632499200,1605264998400,385263599616000,
%U 770527199232000,194172854206464000,388345708412928000,512616335105064960000,1025232670210129920000
%N a(n) is the LCM of denominators of polynomials of degree n which are integer-valued on primes together with their first divided differences.
%C a(n) is also the n-th Bhargava's factorial n_P^{{1}} of the set P of primes with respect to the first divided difference.
%H Alois P. Heinz, <a href="/A212429/b212429.txt">Table of n, a(n) for n = 1..170</a>
%H M. Bhargava, <a href="https://doi.org/10.1090/S0894-0347-09-00638-9">On P-orderings, Integer-Valued Polynomials, and Ultrametric Analysis</a>, J. Amer. Math. Soc., 22 (2009), 963-993.
%H J. L. Chabert, <a href="http://www.lamfa.u-picardie.fr/chabert/AboutPolynomials.pdf">About polynomials whose divided differences are integer-valued on prime numbers</a>, ICM 2012 Proceedings, vol. I, pp. 1-7. Complete <a href="http://icm.uaeu.ac.ae/pdf/TomeI.pdf">proceedings</a> (warning: file size is 26MB).
%F a(n) = Prod_{p prime} p^w_p(n-1) where w_p(n) = Sum_{k>=0} floor(n / ((p-1)*p^k)) - t_{p,n} and p^(t_{p,n}-1) <= n/(p-1) < p^t_{p,n}.
%e a(5) = 48 because f(x) = (x-1)(x-2)(x-3)(x-5)(x-7)/48 satisfies f(p) and (f(p)-f(q))/(p-q) are integers for all primes p,q.
%p a:= proc(n) local i, p, wp, r;
%p r:=1;
%p for i do p:= ithprime(i);
%p wp:= p^(w(p,n-1));
%p if wp=1 then break fi;
%p r:= r*wp
%p od; r
%p end:
%p w:= proc(p, n) local d, k, r;
%p r:= 0;
%p for k from 0 do d:= floor(n/((p-1)*p^k));
%p if d=0 then break fi;
%p r:= r+d;
%p od;
%p r -t(n,p)
%p end:
%p t:= proc(n, p) local h, q;
%p q:= n/(p-1);
%p for h from 0 while q>= p^h do od; h
%p end:
%p seq (a(n), n=1..30); # _Alois P. Heinz_, Jun 25 2012
%t a[n_] := Module[{i, p, wp, r}, r = 1; For[i = 1, True, i++, p = Prime[i]; wp = p^w[p, n - 1]; If[wp == 1, Break[]]; r = r*wp]; r];
%t w[p_, n_] := Module[{d, k, r}, r = 0; For[k = 0, True, k++, d = Floor[n/((p - 1)*p^k)]; If[d == 0, Break[]]; r = r + d]; r - t[n, p]];
%t t[n_, p_] := Module[{h, q}, q = n/(p - 1); For[h = 0, q >= p^h , h++]; h];
%t a /@ Range[1, 30] (* _Jean-François Alcover_, Oct 14 2019, after _Alois P. Heinz_ *)
%Y Cf. A053657.
%K nonn
%O 1,3
%A _Jean-Luc Chabert_, Jun 21 2012