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A112037
Go through all of the primes p and for each one, factor p-1 into primes. List the primes in order of their first appearance in the p-1 factorizations.
7
2, 3, 5, 11, 7, 23, 13, 29, 41, 17, 53, 37, 83, 43, 89, 19, 113, 131, 67, 47, 73, 31, 79, 173, 179, 61, 191, 97, 233, 239, 251, 127, 139, 281, 71, 293, 101, 103, 107, 163, 59, 359, 193, 199, 137, 419, 431, 443, 151, 491, 509, 181, 109, 277, 593, 149, 307, 641, 653
OFFSET
2,1
COMMENTS
The length of this list of distinct prime factors of p-1 encountered through p=prime(n) is given by A055768. - Ray Chandler, Nov 30 2005
If "p-1" is changed to "p+1" we get A236388. - N. J. A. Sloane, Jan 24 2014
A permutation of the primes by Dirichlet's theorem on arithmetic progressions: for any pair (r,s) of integers such that gcd(r,s)=1 there are infinitely many primes in the sequence r + k*s; choose r=1 and s=p. - Joerg Arndt, Mar 20 2016
EXAMPLE
We start with the second prime, 3. 3-1 = 2, so 2 is the first term.
5-1 = 2*2, nothing new.
7-1 = 2*3 and 3 is new, so that is the second term.
11-1 = 2*5 and we get 5; etc.
MATHEMATICA
lst = {}; r[n_] := (len = Length@lst; lst = Flatten@ Join[lst, Select[First /@ FactorInteger[Prime@n - 1], ! MemberQ[lst, # ] &]]; If[l < Length@lst, 1, 0]); Do[ r[n], {n, 214}]; lst (* Robert G. Wilson v, Nov 30 2005 *)
DeleteDuplicates[Rest[Flatten[FactorInteger[#][[All, 1]]&/@ (Prime[ Range[ 250]]-1)]]] (* Harvey P. Dale, May 26 2019 *)
PROG
(PARI) g=1; forprime(p=2, 299, f=factorint(p-1)[, 1]; z=factorback(f); r=z/gcd(z, g); g*=r; if(r>1, print(r, " ", p))); \\ Jack Brennen, Nov 28 2005
(GAP) Set(Flat(List(Filtered([3..1500], IsPrime), i->Factors(i-1)))); # Muniru A Asiru, Dec 06 2018
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Michel Dauchez (mdzdm(AT)yahoo.fr), Nov 28 2005
EXTENSIONS
Better description from Jack Brennen, Nov 28 2005
Extended by Ray Chandler and Robert G. Wilson v, Nov 30 2005
STATUS
approved