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A270592 List of pairs: smallest pair of successive primes p & q such that p+1 & q+1 have n distinct prime divisors in common. 3
2, 3, 3, 5, 23, 29, 18869, 18899, 1588761719, 1588761929 (list; graph; refs; listen; history; text; internal format)
All first 5 pairs are found and are proved to be a part of the sequence (using a computer program to find and prove them).
No other pairs were found yet.
The lower bound of the next pair (the 6th pair) is: 1425172824437699411. This is the lower end of the largest increasing primes gap confirmed (look up sequence A002386), that is because of that the gap is 1476 (look up sequence A005250), and the lowest difference possible for the 6th pair is 2310 or prime(5)# (look up sequence A002110), therefore the 2310 gap must be between larger primes than the pair of the maximal increasing prime gap, 1476.
The difference between the pairs is conjectured to be the primorials sequence: A002110.
p - The lower end of the sequence
q - The upper end of the sequence
prime - The primes sequence
x# - The primorial function for x
x|y - x is a divisor of y
q(n)-p(n)=prime(n)# (conjectured)
prime(n)#|(p(n)+1) (conjectured)
prime(n)#|(q(n)+1) (conjectured)
For n=2 the solution pair is p(2)=23, q(2)=29. p(2) & q(2) are successive primes, their consecutive numbers are p(2)+1=24, q(2)+1=30. Both 24 and 30 are multiples of both 2 and 3, which are two distinct primes. Since there are no smaller successive primes pair which is answering these criteria the pair (23, 29) is a part of the sequence, specifically the third pair in the sequence (n starts from 0).
(PARI) a(n) = {p = 2; f = factor(p+1)[, 1]~; forprime(q=3, , g = factor(q+1)[, 1]~; if (#setintersect(f, g) == n, return (p)); p = q; f = g; ); }
Sequence in context: A065854 A263769 A064776 * A096659 A154695 A154646
Alon Kellner, Mar 19 2016

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Last modified April 12 22:48 EDT 2024. Contains 371639 sequences. (Running on oeis4.)