A363595 Recursive product of aliquot divisors of n.

1, 1, 1, 2, 1, 6, 1, 16, 3, 10, 1, 1728, 1, 14, 15, 2048, 1, 5832, 1, 8000, 21, 22, 1, 4586471424, 5, 26, 81, 21952, 1, 24300000, 1, 67108864, 33, 34, 35, 101559956668416, 1, 38, 39, 163840000000, 1, 130691232, 1, 85184, 91125, 46, 1, 16543163447903718821855232, 7, 125000, 51, 140608, 1, 1338925209984

Offset: 1

Author

Michael De Vlieger, Jul 10 2023

Keywords

• nonn
• easy

Examples

			Define S(n) to be the set of proper divisors of n.
a(2) = 1, since 2 is prime, S(2) = {1} and the product of S(2) is 1.
a(4) = 2, since S(4) = {1, 2}; S(2) = 1, hence we have (1 X 2) X 1 = 2.
a(6) = 6, since S(6) = {1, 2, 3}; 2 and 3 are primes p and both have S(p) = 1,
  hence we have (1 X 2 X 3) X 1 X 1 = 6.
a(8) = 16, since S(8) = {1, 2, 4}; a(2) = 1, a(4) = 2,
  therefore (1 X 2 X 4) X 1 X 2 = 16.
a(9) = 3, since S(9) = {1, 3}, a(3) = 1,
  therefore (1 X 3) X 1 = 3.
a(10) = 10, since S(10) = {1, 2, 5};
  a(2) = a(5) = 1, a(4) = 2,
  therefore (1 X 2 X 5) X 1 X 1 = 10.
a(12) = 1728, since S(12) = {1, 2, 3, 4, 6};
  a(2) = a(3) = 1, a(4) = 2, a(6) = 6,
  therefore (1 X 2 X 3 X 4 X 6) X 1 X 1 X 2 X 6
  = 144 X 12 = 1728.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..719

Crossrefs

Cf. A000295, A007955, A007956, analogous to A255242.

Programs

Mathematica

f[x_] := f[x] = Times @@ # * Times @@ Map[f, #] &@ Most@ Divisors[x]; Table[f[n], {n, 120}]

PARI

ali(n) = setminus(divisors(n), Set(n));
a(n) = my(list = List(), v = [n]); while (#v, my(w = []); for (i=1, #v, my(s=ali(v[i])); for (j=1, #s, w = concat(w, s[j]); listput(list, s[j]));); v = w;); vecprod(Vec(list)); \\ Michel Marcus, Jul 15 2023

Formula

a(n) >= A007956(n).

a(p) = 1 for prime p.

a(p^2) = p.

a(p^e) = A000295(e).

a(p*q) = p*q for primes p, q, p < q.

A007947(n) | a(n) for n with omega(n) > 2.