A211189 Number of prime divisors formed by {2} and the consecutive Pythagorean primes for the all 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

1,2

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.

Links

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

Example

a(8) = 4 because the set formed by the union of the prime divisors of the all 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.

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:

Crossrefs

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

Sequence in context: A076291 A275015 A370264 * A194968 A194980 A323607

Adjacent sequences: A211186 A211187 A211188 * A211190 A211191 A211192

Keyword

nonn,obsc

Author

Michel Lagneau, Feb 03 2013

Status

approved