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%I #34 Feb 01 2020 22:40:38
%S 1,1,2,7,21,98,392,2116,11830,70520,425240,2787810,19530213,144890639,
%T 1149978830,8558078111,76417516719,618437486332,6087770992601,
%U 54574732902278,525656554130914,5290117056157616,61626071051832409,555057889968635744,5809502058957961682
%N Number of distinct factorizations of n! with all factors > 1.
%F a(n) = A001055(n!).
%e a(3) = 2 because 3! = 6 = 2*3 has just 2 factorizations.
%e a(4) = 7 because 4! = 24 = 2*12 = 2*2*6 = 2*2*2*3 = 2*3*4 = 3*8 = 4*6 has 7 factorizations.
%p with(numtheory):
%p b:= proc(n, k) option remember;
%p `if`(n>k, 0, 1) +`if`(isprime(n), 0,
%p add(`if`(d>k, 0, b(n/d, d)), d=divisors(n) minus {1, n}))
%p end:
%p a:= n-> b(n!$2):
%p seq(a(n), n=1..12); # _Alois P. Heinz_, May 25 2013
%t c[1, r_] := c[1, r]=1; c[n_, r_] := c[n, r]=Module[{ds, i}, ds=Select[Divisors[n], 1<#<=r&]; Sum[c[n/ds[[i]], ds[[i]]], {i, 1, Length[ds]}]]; a[n_] := c[n!, n! ]; a/@Range[16] (* c[n, r] is the number of factorizations of n with factors <= r. - _Dean Hickerson_, Oct 29 2002 *)
%o (PARI) \\ See A318284 for count.
%o a(n)={if(n<=1, 1, count(factor(n!)[,2]))} \\ _Andrew Howroyd_, Feb 01 2020
%Y Cf. A001055, A157612, A162247, A318284.
%K nonn
%O 1,3
%A _Donald S. McDonald_, Oct 27 2002
%E Edited by _Robert G. Wilson v_, Oct 29 2002
%E 4 more terms from _Ryan Propper_, May 20 2007
%E a(20)-a(25) from _Andrew Howroyd_, Feb 01 2020
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