OFFSET
0,4
COMMENTS
The number of factorizations of (n+1)! into k distinct factors can be arranged into the following triangle:
2! 1;
3! 1, 1;
4! 1, 3, 1;
5! 1, 7, 7, 1;
...
EXAMPLE
3! = 6 = 2*3.
a(3) = 2 because there are 2 factorizations of 3!.
4! = 24 = 2*12 = 3*8 = 4*6 = 2*3*4.
a(4) = 5 because there are 5 factorizations of 4!.
5! = 120 (1)
5! = 2*60 = 3*40 = 4*30 = 5*24 = 6*20 = 8*15 = 10*12 (7)
5! = 2*3*20 = 2*4*15 = 2*5*12 = 2*6*10 = 3*4*10 = 3*5*8 = 4*5*6 (7)
5! = 2*3*4*5 (1)
a(5) = 16 because there are 16 factorizations of 5!.
MAPLE
with(numtheory):
b:= proc(n, k) option remember;
`if`(n>k, 0, 1) +`if`(isprime(n), 0,
add(`if`(d>k, 0, b(n/d, d-1)), d=divisors(n) minus {1, n}))
end:
a:= n-> b(n!$2):
seq(a(n), n=0..12); # Alois P. Heinz, May 26 2013
MATHEMATICA
b[n_, k_] := b[n, k] = If[n>k, 0, 1] + If[PrimeQ[n], 0, Sum[If[d>k, 0, b[n/d, d-1]], {d, Divisors[n] ~Complement~ {1, n}}]];
a[n_] := b[n!, n!];
Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 0, 16}] (* Jean-François Alcover, Mar 21 2017, after Alois P. Heinz *)
PROG
(PARI) \\ See A318286 for count.
a(n)={if(n<=1, 1, count(factor(n!)[, 2]))} \\ Andrew Howroyd, Feb 01 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Jaume Oliver Lafont, Mar 03 2009
EXTENSIONS
a(8)-a(12) from Ray Chandler, Mar 07 2009
a(13)-a(17) from Alois P. Heinz, May 26 2013
a(18)-a(19) from Alois P. Heinz, Jan 10 2015
a(20)-a(26) from Andrew Howroyd, Feb 01 2020
STATUS
approved