# [seqfan] Primes by rank

N. J. A. Sloane njas at research.att.com
Fri May 28 18:30:22 CEST 2010

```Dear Seq Fans, I've edited this entry to clarify the text.
My question is, what is the sequence   a(n) = rank of n-th prime  which
underlies this?  For which this sequence gives the records?
Neil

%I A177854
%S A177854 2,3,11,131,1571,43717,5032843,1047774137
%N A177854 Smallest prime of rank n.
%C A177854 The Brillhart-Lehmer-Selfridge algorithm provides a general method for proving the primality of P as long as one can factor P+1 or P-1. Therefore for any prime number, when P+1 or P-1 is completely factored, the primality of any factors of P+1 or P-1 can also be proved by the same algorithm. The longest recursive primality proving chain depth is called the rank of P.
%H A177854 L. Zhou, <a href="http://bitc.bme.emory.edu/~lzhou/blogs/?p=117">The rank of primes</a>
%e A177854 The "trivial" prime 2 has rank 0. 3 = 2+1 takes one step to reduce to 2, so 3 has rank 1.
%e A177854 P=131: P+1=132=2^2*3*11. P1=2 has rank 0; P1=3 has rank 1; P1=11: P1+1=12=2^2*3; is one step from 3 and has recursion depth = 2. So P=131 has total maximum recursion depth 2+1 = 3 and therefore has rank 3.
%t A177854 The following program runs through all prime numbers until it finds the first rank 7 prime. (It took about a week.) Fr[n_]:= Module[{nm, np, fm, fp, szm, szp, maxm, maxp, thism, thisp, res, jm, jp}, If[n == 2, res = 0, nm = n - 1; np = n + 1; fm = FactorInteger[nm]; fp = FactorInteger[np]; szm = Length[fm]; szp = Length[fp]; maxm = 0; Do[thism = Fr[fm[[jm]][]]; If[maxm < thism, maxm = thism], {jm, 1, szm}]; maxp = 0; Do[thisp = Fr[fp[[jp]][]]; If[maxp maxp, res = maxp]; res++ ]; res]; i=1;While[p = Prime[i]; s = Fr[p];[p, s] >>> "prime_rank.out";s<7,i++ ]
%K A177854 hard,nonn,more,nice,new
%O A177854 0,1
%A A177854 Lei Zhou (lzhou5(AT)emory.edu), May 14 2010
%E A177854 Partially edited by N. J. A. Sloane, May 15 2010, May 28 2010

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