cp's OEIS Frontend

A180411 Sum of the semiprime divisors (with repetition) of the n-th number with two or more distinct semiprime divisors.

%I A180411 #3 Mar 30 2012 18:40:58
%S A180411 16,21,24,30,32,31,37,42,41,48,39,48,45,56,45,54,51,51,61,72,59,57,55,
%T A180411 80,71,64,65,78,61,96,70,77,75,69,91,90,71,67,87,80,101,120,87,75,128,
%U A180411 77,101,93,72,114,121,87,81,91,152,81,126,111,113,107,90,78,168,103,93,129,123,176
%N A180411 Sum of the semiprime divisors (with repetition) of the n-th number with two or more distinct semiprime divisors.
%C A180411 This is to A164865 [Sum of the distinct semiprime divisors of the n-th number with two or more distinct semiprime divisors], as bigomega [A001222, Number of prime divisors of n (counted with multiplicity)] is to omega [A001221, Number of distinct primes dividing n].
%C A180411 The sum of semiprime divisors (with multiplicity) of all k such that A086971(k) > 1.
%C A180411 This is to A001414 [Integer log of n: sum of primes dividing n (with repetition)], as semiprimes A001358 are to primes A000040.
%F A180411 a(n) = A163407(A102467(n+1)).
%e A180411 a(1) = 16 because the first number (greater than 1) such that the sum of numbers of prime factors with and without repetitions does not equal the number of divisors, is a(2) = 12 = (2^2)*3 whose semiprime factors are (2^2 = 4) once and (2*3) with multiplicity two hence (4*1)*1 + (3*3)*2 = 4 + 12 = 16.
%e A180411 a(6) = 31 because 30 = 2*3*5 has multiplicity one semiprime factors (2*3), (2*5), (3*5), which sum to 6+10+15 = 31.
%Y A180411 Cf. A001221, A001222, A001358, A086971, A164865.
%K A180411 nonn
%O A180411 1,1
%A A180411 _Jonathan Vos Post_, Sep 02 2010
%E A180411 Formula, edits, and more terms from _Charles R Greathouse IV_, Sep 03 2010