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A007996
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Primes that divide at least one term of the sequence f given by f(1) = 2, f(n+1) = f(n)^2-f(n)+1 = A000058(n).
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13
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2, 3, 7, 13, 43, 73, 139, 181, 547, 607, 1033, 1171, 1459, 1861, 1987, 2029, 2287, 2437, 4219, 4519, 6469, 7603, 8221, 9829, 12763, 13147, 13291, 13999, 15373, 17881, 17977, 19597, 20161, 20479, 20641, 20857, 20929, 21661, 23689, 23773, 27031
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OFFSET
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1,1
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COMMENTS
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Or, let S_1 = [2] and let S_{n+1} = list formed by sorting the union of S_n together with all prime factors of 1 + Product_i S_n(i) into increasing order; sequence is limit as n -> infinity of S_n.
Because all terms of the sequence f(n) are coprime, a prime can divide at most one term. Odoni shows that primes p>3 in this sequence must satisfy p=1 (mod 6). - T. D. Noe, Sep 25 2010
See A180871(n) for the index of the first term of A000058 (this is one less than the index of the f-sequence) divisible by a(n). - M. F. Hasler, Apr 24 2014
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LINKS
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MAPLE
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n := 1; for p do if isprime(p) then x := 2 mod p; S := {}; while not member(x, S) do if x=0 then a[n] := p; n := n+1; break; fi; S := S union {x}; x := (x^2-x+1) mod p; od; fi; od;
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MATHEMATICA
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t={}; p=1; While[Length[t]<100, p=NextPrime[p]; s=Mod[2, p]; k=0; modSet={}; While[s>0 && !MemberQ[modSet, s], AppendTo[modSet, s]; k++; s=Mod[s^2-s+1, p]]; If[s==0, AppendTo[t, {p, k}]]]; Transpose[t][[1]] (* T. D. Noe, Sep 25 2010 *)
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PROG
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(PARI) is(n)=my(k=Mod(2, n)); for(i=1, n, k=(k-1)*k+1; if(k==0, return(isprime(n)))); n==2 \\ Charles R Greathouse IV, Sep 30 2015
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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Bennett Battaile (bennett.battaile(AT)autodesk.com)
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EXTENSIONS
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STATUS
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approved
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