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 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). 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 (list; graph; refs; listen; history; text; internal format)
 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: A115624 A076291 A275015 * A194968 A194980 A323607 Adjacent sequences:  A211186 A211187 A211188 * A211190 A211191 A211192 KEYWORD nonn,obsc AUTHOR Michel Lagneau, Feb 03 2013 STATUS approved

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Last modified June 23 17:11 EDT 2021. Contains 345402 sequences. (Running on oeis4.)