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Number of distinct products ijk with 0 <= i < j < k <= n.
4

%I #21 May 28 2021 15:35:37

%S 0,0,1,2,5,11,17,30,43,61,76,112,127,178,207,239,275,362,397,508,555,

%T 614,678,839,884,1005,1093,1199,1278,1530,1591,1882,1999,2134,2276,

%U 2433,2519,2922,3097,3279,3392,3885,4015,4564,4751,4939,5187,5841,5988,6423

%N Number of distinct products ijk with 0 <= i < j < k <= n.

%H Michael S. Branicky, <a href="/A027429/b027429.txt">Table of n, a(n) for n = 0..1000</a> (terms 0..200 from T. D. Noe)

%F a(n) = A027430(n) + 1. - _T. D. Noe_, Jan 16 2007

%e a(3) = 2 (0 and 6 being the only products) and a(4) = 5 (with products 0, 6, 8, 12 and 24).

%t nn=50; prod=Table[0, {1+nn^3}]; t=Table[Do[prod[[1+i*j*k]]=1, {i,0,n}, {j,i+1,n}, {k,j+1,n}]; Count[Take[prod,1+n^3],1], {n,0,nn}] (* _T. D. Noe_, Jan 16 2007 *)

%o (Haskell)

%o import Data.List (nub)

%o a027429 n = length $ nub [i*j*k | k<-[2..n], j<-[1..k-1], i<-[0..j-1]]

%o -- _Reinhard Zumkeller_, Jan 01 2012

%o (Python)

%o from itertools import combinations as C

%o def a(n): return len(set(i*j*k for i, j, k in C(range(n+1), 3)))

%o print([a(n) for n in range(50)]) # _Michael S. Branicky_, May 28 2021

%Y Cf. A027425, A027426, A027427, A027430.

%K nonn

%O 0,4

%A _N. J. A. Sloane_

%E Corrected by _T. D. Noe_, Jan 16 2007