A064518 For an integer n with prime factorization p_1*p_2*p_3* ... *p_m let n* = (p_1+1)*(p_2+1)*(p_3+1)* ... *(p_m+1); sequence gives n* such that n* is divisible by n, ordered by increasing value of n.

1, 12, 36, 144, 432, 1296, 1728, 5184, 15552, 20736, 46656, 62208, 186624, 248832, 559872, 746496, 1679616, 2239488, 2985984, 6718464, 8957952, 20155392, 26873856, 60466176, 35831808, 80621568, 107495424, 241864704, 322486272

Offset: 1

Author

Vladeta Jovovic, Oct 07 2001

Keywords

• nonn

Comments

It is not difficult to show that these numbers have the form a(n) = 3^i*4^j with j <= i <= 2j.

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..1000 (terms 1..50 from Harry J. Smith)

Crossrefs

Every term is also a term of A064476.

Cf. A064514, A064515.

Programs

PARI

ns(n)= { local(f,p=1); f=factor(n); for(i=1, matsize(f)[1], p*=(1 + f[i, 1])^f[i, 2]); return(p) } { n=0; for (m=1, 10^9, s=ns(m); if (s%m == 0, write("b064518.txt", n++, " ", s); if (n==50, break)) ) } \\ Harry J. Smith, Sep 17 2009

Formula

a(n) = 3^A064514(n) * 4^A064515(n). - Chai Wah Wu, Mar 26 2025

Extensions

Title clarified by Sean A. Irvine, Jul 15 2023