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Talk:Least prime factor of n

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Posted by David W. Wilson on Sequence of the Day of 18 Oct 2010

I'm listing the elements without the initial 1. I'm a purist, and given that 1 has no prime factors, I feel a(1) is undefined and should not appear in the sequence. I'm sure I did not include it when I authored the sequence.

Why do I find this sequence SOTD-worthy? It is because of a question I posed back in 1997. I noticed that in this sequence, between every pair of 2's, there was an element > 2, easy enough to prove. Similarly, I could show that between every pair of 3's there is an element > 3. I found the same to be true of 5, 7, 11, and 13. I began to think it might true for every prime number, but I could not find a general proof.

So I posed my question to the seqfan list. Apparently it was interesting, because it sent John Conway, Johan de Jong, Derek Smith, and Manjul Bhargava to the blackboard to find a counterexample. After two hours, they found one:

n = 126972592296404970720882679404584182254788131, with a(n) = a(n+226) = 113 and all intervening values < 113.

Later, Fred Helenius found that the smallest prime that generates a counterexample is 71, for which

n = 7310131732015251470110369, with a(n) = a(n+142) = 71, and all intervening values < 71.

He also found the earliest known counterexample,

n = 2061519317176132799110061, with a(n) = a(n+146) = 73, and all intervening values < 73.

I was rather amazed that you had to look so far into the sequence.

1 is the empty product (defined as the multiplicative identity, i.e. 1) of primes, so 1 has no least prime factor (Cf. A020639) and no greatest prime factor (Cf. A006530). I proposed to remove a(1) from both sequences (and both b-files) if the removal is approved. — Daniel Forgues 20:18, 3 July 2011 (UTC)