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%I #21 Feb 24 2022 08:46:01
%S 1,2,10,10,650,211900,414688300,56812297100,6226684574457100,
%T 6142063931090228011000,60585938964731049213533111000,
%U 1643471023248326636197980531190858000,12662130715971848810220521992462621415290000,214329322370515670487612822767624011121300533960940000
%N Product, for k <= n, of the squarefree parts of the total number of arrangements of a set with k elements.
%C In their abstract, Luca and Shparlinski write: "In this note, we show that if we write floor(e*n!) = s(n)*u(n)^2, where s(n) is square-free then S(N) = prod(n<=N) has at least C log log N distinct prime factors for some absolute constant C > 0 and sufficiently large N. A similar result is obtained for the total number of distinct primes dividing the m-th power-free part of s(n) as n ranges from 1 to N, where m = 3 is a positive integer. As an application of such results, we give an upper bound on the number of n <= N such that floor(e*n!) is a square."
%H F. Luca and I. E. Shparlinski, <a href="https://doi.org/10.1017/S0017089507003734">On the squarefree parts of floor(e*n!)</a>, Glasgow Math. J., 49 (2007), 391-403.
%F a(n) = Product_{k <= n} A222637(k) = Product_{k <= n} A007913(A000522(k)).
%o (PARI) a(n) = prod(i=1, n, core(i! * polcoef(exp(x + x*O(x^i)) / (1 - x), i)))
%Y Cf. A000522, A007913, A222637.
%K nonn
%O 0,2
%A _Michel Marcus_, Feb 27 2013
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