A211189 Number of prime divisors formed by {2} and the consecutive Pythagorean primes for all the composites k^2 + 1 between the two primes A002496(n) and A002496(n+1).

0, 2, 1, 3, 2, 1, 3, 4, 1, 4, 2, 7, 1, 4, 7, 6, 4, 2, 6, 4, 2, 4, 1, 2, 2, 4, 4, 3, 2, 5, 4, 3, 2, 10, 1, 2, 7, 4, 2, 3, 5, 4, 2, 2, 4, 5, 3, 4, 6, 5, 4, 7, 4, 7, 1, 5, 3, 2, 7, 5, 3, 4, 2, 8, 1, 2, 4, 7, 2, 9, 5, 4, 12, 2, 4, 6, 10, 1, 4, 1, 2, 9, 2, 5, 2, 4

Offset: 1

Author

Michel Lagneau, Feb 03 2013

Keywords

• nonn
• obsc

Comments

a(1)=0; for n > 1, a(n) = number of consecutive elements of the form {2, A002144(1), A002144(2), ...} of each row in A211175(n).

The immediate objective of this sequence is to show that it is difficult to obtain a large range of consecutive Pythagorean primes from the decomposition of n^2 + 1, because the growth of a(n) is very slow, for example a(351) = 29, a(22215) = 34, ...

These considerations confirm the opinion of the truthfulness of the conjecture about an infinity of primes of the form n^2 + 1. This sequence gives the length of a variety of conjecturally infinite subsequences of consecutive primes starting with {2, 5, ...}. If the number of primes of the form n^2 + 1 were finite, there should exist a last prime p such that this sequence stops abruptly from p because the length of A002144(n) is infinite. In this case, we should observe a contradictory behavior of this sequence between the stability of the slow growth of a(n) and the discontinuity from the prime p. But this case is highly improbable.

Examples

			a(8) = 4 because the set formed by the union of the prime divisors of all the numbers k^2+1 between the primes A002496(8) = 401 and A002496(9) = 577 are {2, 5, 13, 17, 53, 97} and the subset {2} union {5, 13, 17} contains 4 consecutive elements, hence 4 is in the sequence.
		

Links

• Michel Lagneau, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002144, A002496, A002522, A134406, A181413, A206400, A211175, A211188.

Programs

Maple

with(numtheory) :lst:={2}:lst1:={}:
     for k from 1 to 1000 do: q:=4*k+1:
          if type(q,prime)=true then
          lst:=lst union {q}:else fi:
     od:
  L:=subsop(lst):
        for n from 2 to 1000 do:p:=n^2+1:x:=factorset(p):lst1:=lst1 union x:
          if type(p,prime)=true then
          z:=lst1 minus {p}: n1:=nops(z): jj:=0: d0:=0:
            for j from 1 to n1 while(jj=0) do:
               d:=nops(z intersect L[1..j]): if d>d0 then
              d0:=d:
              else
              jj:=1:fi:
            od:
            lst1:={}: printf(`%d, `,d0):
           fi:
          od: