A337315 Number of divisor pairs, (d1,d2), of A094519(n) such that (d1+d2) | A094519(n) and d1 < d2.
1, 3, 2, 1, 5, 4, 6, 2, 3, 7, 3, 1, 12, 2, 1, 11, 2, 3, 9, 9, 9, 2, 2, 9, 1, 2, 2, 21, 7, 7, 2, 4, 16, 7, 7, 4, 4, 17, 2, 26, 1, 2, 11, 5, 4, 6, 14, 17, 4, 2, 3, 6, 4, 31, 2, 20, 2, 2, 13, 14, 1, 6, 9, 2, 1, 21, 5, 21, 5, 1, 12, 2, 5, 12, 11, 25, 2, 5, 6, 2, 2, 47, 2, 2, 6, 11, 3, 13
Offset: 1
Keywords
Examples
a(2) = 3; A094519(2) = 12 has divisors {1,2,3,4,6,12}. There are 3 divisor pairs, (d1,d2) such that (d1+d2) | 12 and where d1 < d2: (1,2), (1,3) and (2,4); so a(2) = 3.
a(3) = 2; A094519(3) = 18 has divisors {1,2,3,6,9,18}. There are 2 divisor pairs, (d1,d2) such that (d1+d2) | 18 and where d1 < d2: (1,2) and (3,6); so a(3) = 2.
a(4) = 1; A094519(4) = 20 has divisors {1,2,4,5,10,20}. There is 1 divisor pair, (d1,d2) such that (d1+d2) | 20 and where d1 < d2: (1,4). So a(4) = 1.
a(5) = 5; A094519(5) = 24 has divisors {1,2,3,4,6,8,12,24}. There are 5 divisor pairs, (d1,d2) such that (d1+d2) | 24 and where d1 < d2: (1,2), (1,3), (2,4), (2,6) and (4,8). So a(5) = 5.
Crossrefs
Cf. A094519.
Comments