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%I #36 Aug 23 2020 15:35:26
%S 0,2,2,2,2,5,2,2,2,5,2,5,2,5,5,2,2,5,2,6,5,5,2,5,2,5,2,6,2,9,2,2,5,5,
%T 5,5,2,5,5,6,2,10,2,6,5,5,2,5,2,5,5,6,2,5,5,6,5,5,2,10,2,5,5,2,5,10,2,
%U 6,5,9,2,5,2,5,5,6,5,10,2,6,2,5,2,11,5,5,5,7,2,9,5,6,5,5
%N Number of ordered pairs of divisors of n, (d1,d2), such that d2 is prime and d1 <= d2.
%F a(n) = Sum_{d1|n, d2|n, d2 is prime, d1 <= d2} 1.
%F a(n) = A337322(n) + omega(n).
%e a(39) = 5; There are 4 divisors of 39, {1,3,13,39}. There are five ordered pairs of divisors, (d1,d2), such that d2 is prime and d1 <= d2. They are: (1,3), (1,13), (3,3), (3,13) and (13,13). So a(39) = 5.
%e a(40) = 6; There are 8 divisors of 40, {1,2,4,5,8,10,20,40}. There are six ordered pairs of divisors, (d1,d2), such that d2 is prime and d1 <= d2. They are: (1,2), (1,5), (2,2), (2,5), (4,5) and (5,5). So a(40) = 6.
%e a(41) = 2; There are 2 divisors of 41, {1,41}. There are two ordered pairs of divisors, (d1,d2), such that d2 is prime and d1 <= d2. They are: (1,41) and (41,41). So a(41) = 2.
%e a(42) = 10; There are 8 divisors of 42, {1,2,3,6,7,14,21,42}. There are ten ordered pairs of divisors, (d1,d2), such that d2 is prime and d1 <= d2. They are: (1,2), (1,3), (1,7), (2,2), (2,3), (2,7), (3,3), (3,7), (6,7) and (7,7). So a(42) = 10.
%t Table[Sum[Sum[(PrimePi[k] - PrimePi[k - 1]) (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k}], {k, n}], {n, 100}]
%Y Cf. A001221 (omega), A332085, A337320, A337322.
%K nonn
%O 1,2
%A _Wesley Ivan Hurt_, Aug 22 2020
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