%I #23 Jun 06 2019 08:55:35
%S 1,3,6,5,8,8,8,14,13,9,14,17,8,18,23,18,14,17,13,33,23,10,19,36,15,22,
%T 32,22,19,26,17,39,24,18,50,45,8,22,39,38,22,27,13,50,45,16,27,52,24,
%U 39,38,27,20,50,45,72,24,12,31,58,15,28,69,45,49,39,12,52,40,33,33,66,12,33,64
%N Sum divides product: number of solutions (r,s,t), r>=s>=t>0, to the equation rst = n(r+s+t).
%C Number of solutions (r,s) in positive integers to the equation rs = n(r+s) is tau(n^2), cf. A048691. Number of solutions (r,s), r>=s>0, to the equation rs = n(r+s) is (tau(n^2)+1)/2, cf. A018892.
%C Conjecturally, includes all positive integers except 2, 4, 7 and 11 - _David W. Wilson_
%H Charles R Greathouse IV, <a href="/A063520/b063520.txt">Table of n, a(n) for n = 1..10000</a>
%H M. J. Pelling, <a href="https://www.jstor.org/stable/2589479">Problem 10745</a>, Amer. Math. Monthly, vol. 106 (1999), p. 587.
%H M. J. Pelling and F. W. Roush, <a href="https://www.jstor.org/stable/2695283">The Sum Divides the Product: Problem 10745</a>, Amer. Math. Monthly, vol. 108, (no. 7, Aug. 2001), pp. 668-669. [Gives upper bound]
%e There are 8 such solutions to rst = 5(r+s+t): (5, 4, 3), (7, 5, 2), (10, 4, 2), (11, 10, 1), (15, 8, 1), (20, 7, 1), (25, 3, 2), (35, 6, 1).
%t (* Assuming s <= 2n and t <= n*(n+2) *) redu[n_] := Reap[ Do[ red = Reduce[0 < r <= s <= t && r*s*t == n*(r+s+t), r, Integers]; If[red =!= False, Sow[{r, s, t} /. ToRules[red] ] ], {s, 1, 2*n}, {t, s, n*(n+2)}] ][[2, 1]]; a[n_] := redu[n] // Length; a[1] = 1; Table[ Print[n, " ", an = a[n]]; an, {n, 1, 75}] (* _Jean-François Alcover_, Feb 22 2013 *)
%o (PARI) a(n)=sum(t=1,sqrtint(3*n),sum(s=t,sqrtint(n^2+t)+n,my(N=n*(s+t), D=s*t-n);D&&denominator(N/D)==1&&N/D>=s)) \\ _Charles R Greathouse IV_, Feb 22 2013
%Y Cf. A018892, A004194, A063525.
%K nonn
%O 1,2
%A _Jud McCranie_ and _Vladeta Jovovic_, Aug 01 2001
%E More terms from _David W. Wilson_, Aug 01 2001