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

%I #10 Aug 23 2020 19:11:25

%S 0,0,0,1,0,3,0,2,1,3,0,7,0,3,3,3,0,7,0,6,3,3,0,11,1,3,2,6,0,15,0,4,3,

%T 3,3,13,0,3,3,10,0,14,0,6,7,3,0,15,1,7,3,6,0,11,3,10,3,3,0,26,0,3,7,5,

%U 3,14,0,6,3,15,0,19,0,3,7,6,3,14,0,14,3,3,0,25,3,3,3,9,0,27

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

%F a(n) = Sum_{d1|n, d2|n, d1 is prime, d1 < d2} 1.

%F a(n) = A332085(n) - omega(n).

%e a(7) = 0; There are two divisors of 7, {1,7}. There are no ordered pairs of divisors of n, (d1,d2) where d1 is prime and d1 < d2. So a(7) = 0.

%e a(8) = 2; There are four divisors of 8, {1,2,4,8}. There are 2 ordered pairs of divisors of n, (d1,d2) where d1 is prime and d1 < d2. They are: (2,4) and (2,8). So a(8) = 2.

%e a(9) = 1; There are three divisors of 9, {1,3,9}. There is one ordered pair of divisors of n, (d1,d2) where d1 is prime and d1 < d2. It is (3,9). So a(9) = 1.

%e a(10) = 3; There are four divisors of 10, {1,2,5,10}. There are three ordered pairs of divisors of n, (d1,d2) where d1 is prime and d1 < d2. They are: (2,5), (2,10) and (5,10). So a(10) = 3.

%t Table[Sum[Sum[(PrimePi[i] - PrimePi[i - 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, A337322.

%K nonn

%O 1,6

%A _Wesley Ivan Hurt_, Aug 23 2020