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Number of ordered pairs of divisors of n, (d1,d2), such that d2 is prime and d1 < d2.
3

%I #9 Aug 23 2020 15:35:51

%S 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 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 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 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) = A337228(n) - omega(n).

%e 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 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 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 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 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 Cf. A001221 (omega), A332085, A337228, A337320.

%K nonn

%O 1,6

%A _Wesley Ivan Hurt_, Aug 23 2020