A211189 - OEIS

%I #21 Jun 25 2024 01:30:43

%S 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,

%T 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,

%U 2,9,5,4,12,2,4,6,10,1,4,1,2,9,2,5,2,4

%N 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).

%C 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).

%C 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, ...

%C 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.

%H Michel Lagneau, <a href="/A211189/b211189.txt">Table of n, a(n) for n = 1..10000</a>

%e 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.

%p with(numtheory) :lst:={2}:lst1:={}:

%p      for k from 1 to 1000 do: q:=4*k+1:

%p           if type(q,prime)=true then

%p           lst:=lst union {q}:else fi:

%p      od:

%p   L:=subsop(lst):

%p         for n from 2 to 1000 do:p:=n^2+1:x:=factorset(p):lst1:=lst1 union x:

%p           if type(p,prime)=true then

%p           z:=lst1 minus {p}: n1:=nops(z): jj:=0: d0:=0:

%p             for j from 1 to n1 while(jj=0) do:

%p                d:=nops(z intersect L[1..j]): if d>d0 then

%p               d0:=d:

%p               else

%p               jj:=1:fi:

%p             od:

%p             lst1:={}: printf(`%d, `,d0):

%p            fi:

%p           od:

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

%K nonn,obsc

%O 1,2

%A _Michel Lagneau_, Feb 03 2013