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Number of ways to write n as n = u*v*w with 1 <= u < v <= w.
6

%I #26 Dec 17 2021 20:37:03

%S 0,0,0,1,0,1,0,1,1,1,0,2,0,1,1,2,0,3,0,2,1,1,0,4,1,1,1,2,0,4,0,3,1,1,

%T 1,5,0,1,1,4,0,4,0,2,2,1,0,7,1,3,1,2,0,4,1,4,1,1,0,8,0,1,2,4,1,4,0,2,

%U 1,4,0,9,0,1,3,2,1,4,0,6,2,1,0,8,1,1,1,4,0,8,1,2,1,1,1,9,0,3,2,6,0,4,0,4,4,1,0,9,0,4,1,6,0,4,1,2,2,1,1,14

%N Number of ways to write n as n = u*v*w with 1 <= u < v <= w.

%H Antti Karttunen, <a href="/A088432/b088432.txt">Table of n, a(n) for n = 1..3003</a>

%F a(n) = 0 iff n=1 or n is prime: a(A008578(n)) = 0, a(A002808(n)) > 0.

%F a(n) = 1 iff n has 3 or 4 divisors (A323644) (see examples). - _Bernard Schott_, Dec 13 2021

%F a(n) = 2 if n = p^2*q, p<q primes (A096156) or n = p^4 (A030514) (see examples). - _Bernard Schott_, Dec 16 2021

%e n=12: (1,2,6), (1,3,4): therefore a(12)=2;

%e n=18: (1,2,9), (1,3,6), (2,3,3): therefore a(18)=3.

%e For n = p*q, p < q primes: n = 1 * p * q, so a(n) = 1.

%e For n = p^2, p prime: n = 1 * p * p, so a(n) = 1.

%e For n = p^3, p prime: n = 1 * p * p^2, so a(n) = 1.

%e For n = p*q^2, p < q < p^2: n = 1 * p * pq = 1* q * p^2, so a(n) = 2 (see n=12).

%e For n = p*q^2, p < p^2 < q: n = 1 * p * pq = 1 * p^2 * q, so a(n) = 2

%e For n = p^4, p prime: n = 1 * p * p^3 = 1 * p^2 * p^2, so a(n) = 2.

%t a[n_] := Module[{s = 0}, Do[Do[Do[If[u v w == n, s++], {w, v, n}], {v, u + 1, n - 1}], {u, Divisors[n]}]; s];

%t Array[a, 120] (* _Jean-François Alcover_, Dec 10 2021, after _Antti Karttunen_ *)

%o (PARI) A088432(n) = { my(s=0); fordiv(n, u, for(v=u+1, n-1, for(w=v, n, if(u*v*w==n, s++)))); (s); }; \\ _Antti Karttunen_, Aug 24 2017

%Y Cf. A034836, A088433, A088434, A122179, A122180, A323644.

%K nonn

%O 1,12

%A _Reinhard Zumkeller_, Oct 01 2003

%E Data section extended to 120 terms by _Antti Karttunen_, Aug 24 2017