A337362 Number of pairs of divisors of n, (d1,d2), with d1 <= d2 such that d1 and d2 are nonconsecutive integers.
1, 2, 3, 5, 3, 8, 3, 9, 6, 9, 3, 18, 3, 9, 10, 14, 3, 19, 3, 19, 10, 9, 3, 33, 6, 9, 10, 20, 3, 33, 3, 20, 10, 9, 10, 42, 3, 9, 10, 34, 3, 33, 3, 20, 21, 9, 3, 52, 6, 20, 10, 20, 3, 34, 10, 34, 10, 9, 3, 73, 3, 9, 21, 27, 10, 34, 3, 20, 10, 35, 3, 74, 3, 9, 21, 20, 10, 34, 3, 53, 15
Offset: 1
Keywords
Examples
a(6) = 8; The divisors of 6 are {1,2,3,6}. There are 8 divisor pairs, (d1,d2), with d1 <= d2 that do not contain consecutive integers. They are (1,1), (1,3), (1,6), (2,2), (2,6), (3,3), (3,6) and (6,6). So a(6) = 8.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..20000
Programs
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Mathematica
Table[Sum[Sum[(1 - KroneckerDelta[i + 1, k]) (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k}], {k, n}], {n, 100}]
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PARI
a(n) = sumdiv(n, d1, sumdiv(n, d2, (d1<=d2) && (d1 + 1 != d2))); \\ Michel Marcus, Aug 25 2020
Formula
a(n) = Sum_{d1|n, d2|n, d1<=d2} (1 - [d1 + 1 = d2]), where [] is the Iverson bracket.
Comments