|
|
A177854
|
|
Smallest prime of rank n.
|
|
2
|
|
|
|
OFFSET
|
0,1
|
|
COMMENTS
|
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 shortest recursive primality proving chain depth is called the rank of P (cf. A169818).
|
|
LINKS
|
|
|
EXAMPLE
|
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.
|
|
MATHEMATICA
|
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]][[1]]]; If[maxm < thism, maxm = thism], {jm, 1, szm}]; maxp = 0; Do[thisp = Fr[fp[[jp]][[1]]]; If[maxp maxp, res = maxp]; res++ ]; res]; i=1; While[p = Prime[i]; s = Fr[p]; [p, s] >>> "prime_rank.out"; s<7, i++ ]
|
|
PROG
|
(PARI) rank(p)=if(p<8, return(p>2)); vecmin(apply(k->vecmax(apply(rank, factor(k)[, 1])), [p-1, p+1]))+1
|
|
CROSSREFS
|
These are the primes where records occur in A169818.
|
|
KEYWORD
|
hard,nonn,more,nice,changed
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|