A163407 Sum of semiprime divisors of n with repetition.

0, 0, 0, 4, 0, 6, 0, 12, 9, 10, 0, 16, 0, 14, 15, 24, 0, 21, 0, 24, 21, 22, 0, 30, 25, 26, 27, 32, 0, 31, 0, 40, 33, 34, 35, 37, 0, 38, 39, 42, 0, 41, 0, 48, 39, 46, 0, 48, 49, 45, 51, 56, 0, 45, 55, 54, 57, 58, 0, 51, 0, 62, 51, 60, 65, 61, 0, 72, 69, 59, 0, 57, 0, 74, 55, 80, 77, 71, 0

Offset: 1

Author

Franklin T. Adams-Watters, Jul 26 2009

Keywords

• nonn

Comments

We regard each prime divisor of n as distinct, and count each product of an unordered, distinct pair of them as a semiprime divisor.

Examples

			For n = 12, the prime divisors with repetition are 2,2,3. Distinguishing the 2s as 2 and 2', we have semiprime divisors 2*2', 2*3, and 2'*3, totaling 4+6+6 = 16.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A076290, A001358, A001222, A067666.

Programs

PARI

a(n)=local(fn,p,e,s,ss);fn=factor(n);for(i=1,matsize(fn)[1],p=fn[i,1];e=fn[i,2];s+=p*e;ss+=p^2*e);(s^2-ss)\2

Formula

If s is the sum of the prime divisors of n with repetition, and ss is the sum of their squares, a(n) = (s^2 - ss) / 2.