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A263884 a(n) = (m(n)*n)! / (n!)^(m(n)+1), where m(n) is the largest prime power <= n.

%I A263884 #8 Dec 24 2015 10:13:22
%S A263884 1,3,280,2627625,5194672859376,1903991899429620,
%T A263884 1461034854396267778567973305958400,
%U A263884 450538787986875167583433232345723106006796340625,146413934927214422927834111686633731590253260933067148964500000000,3752368324673960479843764075706478869144868251518618794695144146928706880
%N A263884 a(n) = (m(n)*n)! / (n!)^(m(n)+1), where m(n) is the largest prime power <= n.
%C A263884 Morris and Fritze (2015) prove that a(n) is an integer.
%H A263884 Howard Carry Morris and Daniel Fritze, <a href="http://dx.doi.org/10.4169/math.mag.88.4.285">Problem 1948</a>, Math. Mag., 88 (2015), 288-289.
%F A263884 a(n) = A057599(n) for n a prime power.
%e A263884 The largest prime power <= 6 is m(6) = 5, so a(6) = (5*6)! / (6!)^(5+1) = 30! / (6!)^6 = 1903991899429620.
%Y A263884 Cf. A057599.
%K A263884 nonn
%O A263884 1,2
%A A263884 _Jonathan Sondow_, Dec 19 2015