A066446 Number of unordered divisor pairs of n.
0, 1, 1, 3, 1, 6, 1, 6, 3, 6, 1, 15, 1, 6, 6, 10, 1, 15, 1, 15, 6, 6, 1, 28, 3, 6, 6, 15, 1, 28, 1, 15, 6, 6, 6, 36, 1, 6, 6, 28, 1, 28, 1, 15, 15, 6, 1, 45, 3, 15, 6, 15, 1, 28, 6, 28, 6, 6, 1, 66, 1, 6, 15, 21, 6, 28, 1, 15, 6, 28, 1, 66, 1, 6, 15, 15, 6, 28, 1, 45, 10, 6, 1, 66, 6, 6, 6, 28
Offset: 1
Examples
The divisors of 6 are 1, 2, 3 & 6. In unordered pairs they are {1, 2}, {1, 3}, {1, 6}, {2, 3}, {2, 6}, & {3, 6}. Since there are six pairs, a(6) = 6. Also d(6) = 4. 4*3/2 = 6.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..65537 (terms 1..1000 from Harry J. Smith)
Programs
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Haskell
a066446 = a000217 . subtract 1 . a000005' -- Reinhard Zumkeller, Sep 08 2015
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Maple
with(numtheory): seq(tau(n)*(tau(n)-1)/2, n=1..60); # Ridouane Oudra, Apr 15 2023
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Mathematica
Table[ Binomial[ DivisorSigma[0, n], 2], {n, 1, 100}] -
PARI
{ for (n=1, 1000, a=binomial(numdiv(n), 2); write("b066446.txt", n, " ", a) ) } \\ Harry J. Smith, Feb 15 2010
Formula
a(p) = 1 iff p is a prime.
Combinations of d(n), the number of divisors of n (A000005), taken two at a time. If the canonical factorization of n into prime powers is Product p^e(p) then d(n) = Product (e(p) + 1). Therefore a(n) = C(d(n), 2) = d(n)*{ d(n)-1 }/2 which is a triangular number (A000217).
a(n) = Sum_{k|n, i|n, i < k} 1. - Wesley Ivan Hurt, Aug 20 2020
a(n) = Sum_{d|n} A063647(d). - Ridouane Oudra, Apr 15 2023