|
%I #17 Sep 09 2017 22:30:05
%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 the all 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 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.
%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
|
