%I
%S 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,
%T 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,
%U 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
%N Number of steps for the map of primes p > 4p + 3, starting at the nth prime == 3 (mod 4), needed to reach the final prime in the trajectory.
%C The maximum number of iterations is 8 (proof in A179767).
%C 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.
%e a(1) = 0 because the first prime == 3 (mod 4) is A002145(1) = 3 and 3*4 + 3 = 15 is composite.
%e 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.
%e 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.
%p with(numtheory):for i from 1 to 300 do : n:=ithprime(i):if irem(n,4) = 3 then
%p nn:=n: id:=0:k:=0:for it from 1 to 8 do: p:=4*nn+3: if type (p,prime)=true and
%p id=0 then k:=k+1:nn:=p:else id:=1:fi:od:printf(`%d, `,k):else fi:od:
%Y Cf. A002144, A002145, A179767.
%K nonn
%O 0,2
%A _Michel Lagneau_, Jan 10 2011
