A337320 Number of ordered pairs of divisors of n, (d1,d2), such that d1 is prime and d1 < d2.
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, 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, 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
Offset: 1
Keywords
Examples
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.
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.
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.
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.
Programs
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Mathematica
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}]
Formula
a(n) = Sum_{d1|n, d2|n, d1 is prime, d1 < d2} 1.
a(n) = A332085(n) - omega(n).