A374941 a(n) = (sum of proper prime factors of n) + Sum_{d = proper composite factor of n} a(d).
0, 0, 0, 2, 0, 5, 0, 4, 3, 7, 0, 12, 0, 9, 8, 8, 0, 13, 0, 16, 10, 13, 0, 28, 5, 15, 6, 20, 0, 30, 0, 16, 14, 19, 12, 40, 0, 21, 16, 36, 0, 36, 0, 28, 19, 25, 0, 64, 7, 19, 20, 32, 0, 32, 16, 44, 22, 31, 0, 90, 0, 33, 23, 32, 18, 48, 0, 40, 26, 42, 0, 112, 0, 39
Offset: 1
Keywords
Examples
For n=24, the tree of recurrences "a(d)" is
24
/ / / \ \ \
2 3 4 6 8 12
/ / \ / \ \ \ \ \
2 2 3 2 4 2 3 4 6
/ / / \
2 2 2 3
a(24) = 2 + 3 + a(4) + a(6) + a(8) + a(12)
= 5 + 2 + 2 + 3 + 2 + a(4) + 2 + 3 + a(4) + a(6)
= 14 + 2 + 5 + 2 + 2 + 3
= 28
Links
- Michael S. Branicky, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A371075 (fixed points).
Programs
-
Python
from sympy import divisors, isprime def a(n): return sum(di if isprime(di) else a(di) for di in divisors(n)[1:-1]) print([a(n) for n in range(1, 75)]) # Michael S. Branicky, Jul 24 2024
Formula
a(p) = 0, iff p = 1 or a prime number.
a(p^k) == 2^(k-2)*p for p prime, k > 1. - Michael S. Branicky, Aug 01 2024
Extensions
More terms from Michael S. Branicky, Jul 24 2024