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A112037
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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.
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7
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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
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OFFSET
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2,1
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COMMENTS
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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
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
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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Michel Dauchez (mdzdm(AT)yahoo.fr), Nov 28 2005
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EXTENSIONS
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STATUS
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approved
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