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%I #11 Aug 02 2023 13:55:44
%S 1,1,1,2,1,6,1,16,3,10,1,1728,1,14,15,2048,1,5832,1,8000,21,22,1,
%T 4586471424,5,26,81,21952,1,24300000,1,67108864,33,34,35,
%U 101559956668416,1,38,39,163840000000,1,130691232,1,85184,91125,46,1,16543163447903718821855232,7,125000,51,140608,1,1338925209984
%N Recursive product of aliquot divisors of n.
%H Michael De Vlieger, <a href="/A363595/b363595.txt">Table of n, a(n) for n = 1..719</a>
%F a(n) >= A007956(n).
%F a(p) = 1 for prime p.
%F a(p^2) = p.
%F a(p^e) = A000295(e).
%F a(p*q) = p*q for primes p, q, p < q.
%F A007947(n) | a(n) for n with omega(n) > 2.
%e Define S(n) to be the set of proper divisors of n.
%e a(2) = 1, since 2 is prime, S(2) = {1} and the product of S(2) is 1.
%e a(4) = 2, since S(4) = {1, 2}; S(2) = 1, hence we have (1 X 2) X 1 = 2.
%e a(6) = 6, since S(6) = {1, 2, 3}; 2 and 3 are primes p and both have S(p) = 1,
%e hence we have (1 X 2 X 3) X 1 X 1 = 6.
%e a(8) = 16, since S(8) = {1, 2, 4}; a(2) = 1, a(4) = 2,
%e therefore (1 X 2 X 4) X 1 X 2 = 16.
%e a(9) = 3, since S(9) = {1, 3}, a(3) = 1,
%e therefore (1 X 3) X 1 = 3.
%e a(10) = 10, since S(10) = {1, 2, 5};
%e a(2) = a(5) = 1, a(4) = 2,
%e therefore (1 X 2 X 5) X 1 X 1 = 10.
%e a(12) = 1728, since S(12) = {1, 2, 3, 4, 6};
%e a(2) = a(3) = 1, a(4) = 2, a(6) = 6,
%e therefore (1 X 2 X 3 X 4 X 6) X 1 X 1 X 2 X 6
%e = 144 X 12 = 1728.
%t f[x_] := f[x] = Times @@ # * Times @@ Map[f, #] &@ Most@ Divisors[x]; Table[f[n], {n, 120}]
%o (PARI) ali(n) = setminus(divisors(n), Set(n));
%o 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
%Y Cf. A000295, A007955, A007956, analogous to A255242.
%K nonn,easy
%O 1,4
%A _Michael De Vlieger_, Jul 10 2023
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