cp's OEIS Frontend

A337322 Number of ordered pairs of divisors of n, (d1,d2), such that d2 is prime and d1 < d2.

%I A337322 #13 May 29 2025 03:03:41
%S A337322 0,1,1,1,1,3,1,1,1,3,1,3,1,3,3,1,1,3,1,4,3,3,1,3,1,3,1,4,1,6,1,1,3,3,
%T A337322 3,3,1,3,3,4,1,7,1,4,3,3,1,3,1,3,3,4,1,3,3,4,3,3,1,7,1,3,3,1,3,7,1,4,
%U A337322 3,6,1,3,1,3,3,4,3,7,1,4,1,3,1,8,3,3,3,5,1,6,3,4,3,3,3,3
%N A337322 Number of ordered pairs of divisors of n, (d1,d2), such that d2 is prime and d1 < d2.
%F A337322 a(n) = Sum_{d1|n, d2|n, d2 is prime, d1 < d2} 1.
%F A337322 a(n) = A337228(n) - omega(n).
%F A337322 a(n) = A248577(n) - A332085(n). - _Ridouane Oudra_, May 28 2025
%e A337322 a(39) = 3; There are 4 divisors of 39, {1,3,13,39}. There are three ordered pairs of divisors, (d1,d2), such that d2 is prime and d1 < d2. They are: (1,3), (1,13) and (3,13). So a(39) = 3.
%e A337322 a(40) = 4; There are 8 divisors of 40, {1,2,4,5,8,10,20,40}. There are four ordered pairs of divisors, (d1,d2), such that d2 is prime and d1 < d2. They are: (1,2), (1,5), (2,5) and (4,5). So a(40) = 4.
%e A337322 a(41) = 1; There are 2 divisors of 41, {1,41}. There is one ordered pair of divisors, (d1,d2), such that d2 is prime and d1 < d2. It is (1,41). So a(41) = 1.
%e A337322 a(42) = 7; There are 8 divisors of 42, {1,2,3,6,7,14,21,42}. There are seven ordered pairs of divisors, (d1,d2), such that d2 is prime and d1 < d2. They are: (1,2), (1,3), (1,7), (2,3), (2,7), (3,7) and (6,7). So a(42) = 7.
%t A337322 Table[Sum[Sum[(PrimePi[k] - PrimePi[k - 1]) (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k - 1}], {k, n}], {n, 100}]
%Y A337322 Cf. A001221 (omega), A332085, A337228, A337320, A248577, A332085.
%K A337322 nonn
%O A337322 1,6
%A A337322 _Wesley Ivan Hurt_, Aug 23 2020