A177854 Smallest prime of rank n.
2, 3, 11, 131, 1571, 43717, 5032843, 1047774137, 7128418089643
Offset: 0
Examples
The "trivial" prime 2 has rank 0. 3 = 2+1 takes one step to reduce to 2, so 3 has rank 1. P=131: P+1=132=2^2*3*11. P1[1]=2 has rank 0; P1[2]=3 has rank 1; P1[3]=11: P1[3]+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.
Links
- J. Brillhart, D. H. Lehmer and J. L. Selfridge, New primality criteria and factorizations of 2^m+-1, Math. Compl. 29 (1975) 620-647.
- Wikipedia, Lucas-Lehmer-Riesel test.
- Lei Zhou, The rank of primes.
Crossrefs
These are the primes where records occur in A169818.
Programs
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Mathematica
(* The following program runs through all prime numbers until it finds the first rank n prime. (It took about a week for n = 7.) *) 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]][[1]]]; If[maxm < thism, maxm = thism], {jm, 1, szm}]; maxp = 0; Do[thisp = Fr[fp[[jp]][[1]]]; If[maxp < thisp, maxp = thisp], {jp, 1, szp}]; res = Min[maxm, maxp] + 1 ]; res]; target=0; Do[p = Prime[i]; s = Fr[p]; If[s == target, Print[p]; target++], {i, Infinity}]
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PARI
rank(p)=if(p<8, return(p>2)); vecmin(apply(k->vecmax(apply(rank, factor(k)[,1])), [p-1,p+1]))+1 print1(2); r=0;forprime(p=3,, t=rank(p); if(t>r, r=t; print1(", "p))) \\ Charles R Greathouse IV, Oct 03 2016
Extensions
Partially edited by N. J. A. Sloane, May 15 2010, May 28 2010
Definition corrected by Robert Gerbicz, May 28 2010
a(8) from Florian Baur, Sep 05 2024
Comments