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

1,4

Links

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

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.

Example

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.

Mathematica

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

Prog

(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

Crossrefs

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

Sequence in context: A059344 A109193 A322672 * A225769 A280736 A279095

Adjacent sequences: A363592 A363593 A363594 * A363596 A363597 A363598

Keyword

nonn,easy

Author

Michael De Vlieger, Jul 10 2023

Status

approved