A347498 Least k such that there exists an n-element subset S of {1,2,...,k} with the property that all products i * j are distinct for i <= j.
1, 2, 3, 5, 6, 7, 9, 11, 13, 15, 17, 19, 20, 23, 25, 28, 29, 31, 33, 37, 40, 41, 42, 43, 47, 51, 53, 55, 57, 59, 61, 67, 69, 71, 73, 75, 79, 83
Offset: 1
Examples
n | example set -----+------------------------------------------------------- 1 | {1} 2 | {1, 2} 3 | {1, 2, 3} 4 | {1, 2, 3, 5} 5 | {1, 3, 4, 5, 6} 6 | {1, 3, 4, 5, 6, 7} 7 | {1, 2, 5, 6, 7, 8, 9} 8 | {1, 2, 5, 6, 7, 8, 9, 11} 9 | {1, 2, 5, 6, 7, 8, 9, 11, 13} 10 | {1, 2, 5, 7, 8, 9, 11, 12, 13, 15} 11 | {1, 2, 5, 7, 8, 9, 11, 12, 13, 15, 17} 12 | {1, 2, 5, 7, 8, 9, 11, 12, 13, 15, 17, 19} 13 | {1, 5, 6, 7, 9, 11, 13, 14, 15, 16, 17, 19, 20} 14 | {1, 2, 5, 7, 11, 12, 13, 16, 17, 18, 19, 20, 21, 23} For n = 4, the set {1,2,3,4} does not have distinct products because 2*2 = 1*4. However, the set {1,2,3,5} does have distinct products because 1*1, 1*2, 1*3, 1*5, 2*2, 2*3, 2*5, 3*3, 3*5, and 5*5 are all distinct.
Programs
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Mathematica
Table[k=1;While[!Or@@(Length[s=Union[Sort/@Tuples[#,{2}]]]==Length@Union[Times@@@s]&/@Subsets[Range@k,{n}]),k++];k,{n,12}] (* Giorgos Kalogeropoulos, Sep 08 2021 *)
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Python
from itertools import combinations, combinations_with_replacement def a(n): k = n while True: for Srest in combinations(range(1, k), n-1): S = Srest + (k, ) allprods = set() for i, j in combinations_with_replacement(S, 2): if i*j in allprods: break else: allprods.add(i*j) else: return k k += 1 print([a(n) for n in range(1, 15)]) # Michael S. Branicky, Sep 08 2021
Formula
a(n) = min {k >= 1; A338006(k) = n}. - Pontus von Brömssen, Sep 09 2021
Extensions
a(15)-a(20) from Michael S. Branicky, Sep 08 2021
a(21)-a(38) (based on the terms in A338006) from Pontus von Brömssen, Sep 09 2021
Comments