A007117 a(0) = a(1) = 0; for n >= 2, a(n)*2^(n+2) + 1 is the smallest prime factor of the n-th Fermat number F(n) = 2^(2^n) + 1.
0, 0, 1, 8, 1024, 5, 1071, 116503103764643, 1209889024954, 1184, 11131, 39, 7, 82731770, 1784180997819127957596374417642156545110881094717, 9264, 3150, 59251857, 13, 33629
Offset: 0
Examples
From _Jianing Song_, Mar 02 2021: (Start) F(2) = 2^(2^2) + 1 = 1*2^4 + 1; F(3) = 2^(2^3) + 1 = 5*2^5 + 1; F(4) = 2^(2^4) + 1 = 1024*2^6 + 1; F(5) = 2^(2^5) + 1 = (5*2^7 + 1) * (52347*2^7 + 1); F(6) = 2^(2^6) + 1 = (1071*2^8 + 1) * (262814145745*2^8 + 1). (End)
References
- P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 71.
- H. Riesel, ``Prime numbers and computer methods for factorization,'' Progress in Mathematics, Vol. 57, Birkhauser, Boston, 1985, Chap. 4, see p. 377.
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Wilfrid Keller, Prime factors k.2^n + 1 of Fermat numbers F_m
- R. G. Wilson, V, Letter to N. J. A. Sloane, Aug. 1993
Crossrefs
Cf. A093179.
Programs
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PARI
a(n) = if(n<2, 0, my(lim=2^(2^n-(n+2))); for(k=1, lim, my(p=k*2^(n+2)+1); if(Mod(2,p)^(2^n)==-1, return(k)))) \\ Jianing Song, Mar 02 2021
Formula
a(n) = (A093179(n) - 1)/2^(n+2) for n >= 2. - Jianing Song, Mar 02 2021
Extensions
a(14)-a(19) added by Max Alekseyev, May 04 2010
Comments