A179769 Number of steps for the map of primes p -> 4p + 3, starting at the n-th prime == 3 (mod 4), needed to reach the final prime in the trajectory.

0, 2, 2, 1, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 3, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 2, 0, 0, 2, 0, 1, 0, 1, 0, 0, 0, 0, 3, 1, 1, 0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 3, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1

Offset

0,2

Comments

The maximum number of iterations is 8 (proof in A179767).

The similar problem with the map p -> 4p + 1 is less interesting because that sequence contains numbers <= 1: if we consider just the first two iterations 4p + 1 -> 16p + 5 -> 64p + 21, then (i) if p==0 (mod 3), 64p + 21 is composite, (ii) if p==1 (mod 3), 16p + 5 is composite and (iii) if p==2 (mod 3), 4p + 1 is composite. Thus the maximum number of iterations is 1.

Links

Table of n, a(n) for n=0..113.

Example

a(1) = 0 because the first prime == 3 (mod 4) is A002145(1) = 3 and 3*4 + 3 = 15 is composite.
a(2) = 2 because the second prime == 3 (mod 4) is A002145(2) = 7, and 4*7 + 3 = 31 -> 4*31 + 3  = 127 are primes.
a(1761) = 8 because the 1761st prime == 3 (mod 4) is A002145(1761) = 32611 and 32611 -> 130447 -> 521791 -> 2087167 -> 8348671 -> 33394687 -> 133578751 -> 534315007 -> 2137260031 are primes.

Maple

with(numtheory):for i from 1 to 300 do : n:=ithprime(i):if irem(n, 4) = 3 then
  nn:=n: id:=0:k:=0:for it from 1 to 8 do: p:=4*nn+3: if type (p, prime)=true and
  id=0 then k:=k+1:nn:=p:else id:=1:fi:od:printf(`%d, `, k):else fi:od:

Crossrefs

Cf. A002144, A002145, A179767.

Sequence in context: A368948 A368579 A287331 * A340594 A361685 A227193

Adjacent sequences: A179766 A179767 A179768 * A179770 A179771 A179772

Keyword

nonn

Author

Michel Lagneau, Jan 10 2011

Status

approved