A274029 Product of infinitary divisors of n.

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

Offset: 1

Author

Vladimir Shevelev, Jun 07 2016

Keywords

• nonn

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

Crossrefs

Cf. A007955, A050376, A037445, A273011.

Programs

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 *)

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 i-divisors (or infinitary divisors) of n.

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)).