A337228
Number of ordered pairs of divisors of n, (d1,d2), such that d2 is prime and d1 <= d2.
Original entry on oeis.org
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, 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, 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
Offset: 1
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.
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.
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.
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.
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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}]
A337320
Number of ordered pairs of divisors of n, (d1,d2), such that d1 is prime and d1 < d2.
Original entry on oeis.org
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
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.
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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}]
A337322
Number of ordered pairs of divisors of n, (d1,d2), such that d2 is prime and d1 < d2.
Original entry on oeis.org
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, 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, 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
Offset: 1
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.
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.
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.
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.
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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}]
Showing 1-3 of 3 results.