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# Talk:A000058

..."If one instead works with Sylvester's sequence ${\displaystyle ak=a1+...ak-1+1}$, then it is known [O1985] that the number of primes less than n that can divide any one of the ak is O(n / lognlogloglogn) rather than O(n / logn) (the prime number theorem bound). If we then factor the first k elements of this sequence, we must get a prime of size at least klogklogloglogk or so...."