%I #21 Sep 16 2019 08:51:29
%S 1,2,3,4,5,36,7,64,9,100,11,144,13,196,225,16,17,324,19,400,441,484,
%T 23,331776,25,676,729,784,29,810000,31,1024,1089,1156,1225,1296,37,
%U 1444,1521,2560000,41,3111696,43,1936,2025,2116,47,2304,49,2500,2601,2704,53,8503056,3025
%N Product of infinitary divisors of n.
%C The sequence consists of primes and squares. However, not all squares are present. The first square that does not appear is 576.
%C The positions of records of the sequence form A273011.
%H Amiram Eldar, <a href="/A274029/b274029.txt">Table of n, a(n) for n = 1..10000</a>
%F As in A007955(n) = n^(d(n)/2), where d(n) is the number of divisors of n, a(n) = n^(id(n)/2), where id(n) is the number of i-divisors (or infinitary divisors) of n.
%F Indeed, a(n) = Product_{id|n} id = Product_{id|n} n/id, thus a(n)^2 = Product_{id|n} n = n^id(n), and the formula follows. But, according to our comment in A037445, if k is the number of distinct A050376-factors q_j such that n = Product(q_j), then id(n) = 2^k. So a(n) = n^(2^(k-1)).
%t f[x_] := If[x == 1, 1, Sort@ Flatten@ Outer[Times, Sequence @@ (FactorInteger[x] /. {p_, m_Integer} :> p^Select[Range[0, m], BitOr[m, #] == m &])]] ; Array[Times @@ f@ # &, 55] (* _Michael De Vlieger_, Jun 07 2016, after Paul Abbott at A077609 *)
%Y Cf. A007955, A050376, A037445, A273011.
%K nonn
%O 1,2
%A _Vladimir Shevelev_, Jun 07 2016
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