

A274029


Product of infinitary divisors of n.


2



1, 2, 3, 4, 5, 36, 7, 64, 9, 100, 11, 144, 13, 196, 225, 16, 17, 324, 19, 400, 441, 484, 23, 331776, 25, 676, 729, 784, 29, 810000, 31, 1024, 1089, 1156, 1225, 1296, 37, 1444, 1521, 2560000, 41, 3111696, 43, 1936, 2025, 2116, 47, 2304, 49, 2500, 2601, 2704, 53, 8503056, 3025
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OFFSET

1,2


COMMENTS

The sequence consists of primes and squares. However, not all squares are present. The first square that does not appear is 576.
The positions of records of the sequence form A273011.


LINKS

Amiram Eldar, Table of n, a(n) for n = 1..10000


FORMULA

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 idivisors (or infinitary divisors) of n.
Indeed, a(n) = Product_{idn} id = Product_{idn} n/id, thus a(n)^2 = Product_{idn} n = n^id(n), and the formula follows. But, according to our comment in A037445, if k is the number of distinct A050376factors q_j such that n = Product(q_j), then id(n) = 2^k. So a(n) = n^(2^(k1)).


MATHEMATICA

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 *)


CROSSREFS

Cf. A007955, A050376, A037445, A273011.
Sequence in context: A062509 A061537 A306329 * A056925 A060067 A043310
Adjacent sequences: A274026 A274027 A274028 * A274030 A274031 A274032


KEYWORD

nonn


AUTHOR

Vladimir Shevelev, Jun 07 2016


STATUS

approved



