A325204 Numbers k such that k*(k+1)*(k+2) has exactly 4 distinct prime factors.
5, 9, 10, 11, 12, 14, 15, 17, 18, 22, 23, 24, 25, 26, 27, 30, 31, 32, 36, 46, 47, 48, 52, 62, 71, 72, 79, 80, 81, 96, 106, 107, 126, 127, 162, 191, 192, 241, 242, 256, 382, 431, 486, 512, 576, 862, 1151, 1152, 2186, 2591, 2592, 2916, 4372, 8191, 8746, 131071, 131072, 139967, 472391, 524287, 786431, 995326, 995327
Offset: 1
Keywords
Examples
a(3)=10 is in the sequence because 10*11*12 has four distinct prime factors: 2, 3, 5, 11.
Links
- Ray Chandler, Table of n, a(n) for n = 1..178 (terms < 10^1000; first 114 terms from Robert Israel)
- Ray Chandler, Mathematica code used to compute b-file.
- Math StackExchange, Three consecutive numbers with exactly different four prime factors.
Programs
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Maple
select(t -> nops(numtheory:-factorset(t) union numtheory:-factorset(t+1) union numtheory:-factorset(t+2))=4, [$1..10^6]);
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PARI
select(k->4==omega(k*(k+1)*(k+2)), [1..10000]) \\ Andrew Howroyd, Sep 05 2019
Comments