%I #27 Apr 16 2023 02:17:14
%S 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,
%T 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,
%U 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
%N Number of unordered divisor pairs of n.
%H Antti Karttunen, <a href="/A066446/b066446.txt">Table of n, a(n) for n = 1..65537</a> (terms 1..1000 from Harry J. Smith)
%F a(p) = 1 iff p is a prime.
%F 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).
%F a(n) = A184389(n) - A000005(n) = A035116(n) - A184389(n). - _Reinhard Zumkeller_, Sep 08 2015
%F a(n) = A000217(A000005(n)-1). - _Antti Karttunen_, Sep 21 2018
%F a(n) = Sum_{k|n, i|n, i < k} 1. - _Wesley Ivan Hurt_, Aug 20 2020
%F a(n) = Sum_{d|n} A063647(d). - _Ridouane Oudra_, Apr 15 2023
%e 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.
%p with(numtheory): seq(tau(n)*(tau(n)-1)/2, n=1..60); # _Ridouane Oudra_, Apr 15 2023
%t Table[ Binomial[ DivisorSigma[0, n], 2], {n, 1, 100}]
%o (PARI) { for (n=1, 1000, a=binomial(numdiv(n), 2); write("b066446.txt", n, " ", a) ) } \\ _Harry J. Smith_, Feb 15 2010
%o (Haskell)
%o a066446 = a000217 . subtract 1 . a000005'
%o -- _Reinhard Zumkeller_, Sep 08 2015
%Y Cf. A000005, A000217, A129510.
%Y Cf. A035116, A184389, A063647.
%K easy,nonn
%O 1,4
%A _Robert G. Wilson v_, Dec 28 2001