A132705 For an integer n with prime factorization (p_1)*(p_2)*(p_3)* ... *(p_k), a(n) = (p_1+2)*(p_2+2)*(p_3+2)* ... *(p_k+2).

2, 3, 4, 5, 16, 7, 20, 9, 64, 25, 28, 13, 80, 15, 36, 35, 256, 19, 100, 21, 112, 45, 52, 25, 320, 49, 60, 125, 144, 31, 140, 33, 128, 65, 76, 63, 400, 49, 84, 75, 448, 43, 180, 45, 208, 175, 100, 49, 1280, 81, 196, 95, 240, 55, 500, 91, 576, 105, 124, 60

Offset: 0

Author

Jonathan Vos Post, Nov 16 2007

Keywords

• easy
• nonn

Comments

a(0)=2 and a(1)=3 by convention. For an integer n with prime factorization prime(i_1)*prime(i_2)*prime(i_3)* ... *prime(i_k), a(n) = A052147(i_1)*A052147(i_2)*A052147(i_3)* ... *A052147(i_k). This sequence is to p+2 as A064478 is to p+1 for primes p.

If a(1) were 1 rather than 3, the sequence would be completely multiplicative with a(p) = p + 2. - Charles R Greathouse IV, Sep 02 2009

Crossrefs

Cf. A000040, A003958, A003959, A064476, A064479, A052147, A064478.

Programs

Python

from math import prod
from sympy import factorint
def A132705(n): return prod((p+2)**e for p,e in factorint(n).items()) if n!=1 else 3 # Chai Wah Wu, Mar 26 2025