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A332085
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Number of ordered pairs of divisors of n, (d1,d2), such that d1 is prime and d1 <= d2.
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3
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0, 1, 1, 2, 1, 5, 1, 3, 2, 5, 1, 9, 1, 5, 5, 4, 1, 9, 1, 8, 5, 5, 1, 13, 2, 5, 3, 8, 1, 18, 1, 5, 5, 5, 5, 15, 1, 5, 5, 12, 1, 17, 1, 8, 9, 5, 1, 17, 2, 9, 5, 8, 1, 13, 5, 12, 5, 5, 1, 29, 1, 5, 9, 6, 5, 17, 1, 8, 5, 18, 1, 21, 1, 5, 9, 8, 5, 17, 1, 16, 4, 5, 1, 28, 5, 5, 5, 11, 1, 30
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OFFSET
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1,4
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LINKS
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FORMULA
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a(n) = Sum_{d1|n, d2|n, d1 is prime, d1 <= d2} 1.
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EXAMPLE
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a(7) = 1; There are two divisors of 7: {1,7}. If we list the ordered pairs of divisors of n, (d1,d2) where d1 is prime and d1 <= d2, we get (7,7). So a(7) = 1.
a(8) = 3; There are 4 divisors of 8: {1,2,4,8}. If we list the ordered pairs of divisors of n, (d1,d2) where d1 is prime and d1 <= d2, we get (2,2), (2,4) and (2,8). So a(8) = 3.
a(9) = 2; There are three divisors of 9: {1,3,9}. If we list the ordered pairs of divisors of n, (d1,d2) where d1 is prime and d1 <= d2, we get (3,3) and (3,9). So a(9) = 2.
a(10) = 5; There are four divisors of 10: {1,2,5,10}. If we list the ordered pairs of divisors of n, (d1,d2) where d1 is prime and d1 <= d2, we get (2,2), (2,5), (2,10), (5,5) and (5,10). So a(10) = 5.
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MATHEMATICA
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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}], {k, n}], {n, 100}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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