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%I #33 Apr 28 2022 10:27:46
%S 13,5,11,7,1093,5,13,11,23,5,797161,547,11,5,1871,7,1597,5,13,23,47,5,
%T 11,398581,13,5,59,7,683,5,13,103,11,5,13097927,1597,13,5,83,7,431,5,
%U 11,47,1223,5,491,11,13,5,107,7,11,5,13,59,14425532687,5,603901,683,13,5,11,7,221101,5,13,11
%N Smallest odd prime factor of 3^n - 1.
%C Levi Ben Gerson (1288-1344) proved that 3^n - 1 = 2^m has no solution in integers if n > 2, by showing that 3^n - l has an odd prime factor. His proof uses remainders after division of powers of 3 by 8 and powers of 2 by 8; see the Lenstra and Peterson links. For an elegant short proof, see the Franklin link. - Sondow
%C One way to prove it is by the use of congruences. The powers of 3, modulo 80, are 3, 9, 27, 1, 3, 9, 27, 1, 3, 9, 27, 1, ... The powers of 2 are 2, 4, 8, 16, 32, 64, 48, 16, 32, 64, 48, 16, ... - _Alonso del Arte_, Jan 20 2014
%D L. E. Dickson, History of the Theory of Numbers, Vol. II, Chelsea, NY 1992; see p. 731.
%H Max Alekseyev, <a href="/A235366/b235366.txt">Table of n, a(n) for n = 3..796</a> (terms to a(660) from Charles R Greathouse IV)
%H P. Franklin, <a href="http://www.jstor.org/stable/2298495">Problem 2927</a>, Amer. Math. Monthly, 30 (1923), p. 81.
%H A. Herschfeld, <a href="http://dx.doi.org/10.1090/S0002-9904-1936-06275-0">The equation 2^x - 3^y = d</a>, Bull. Amer. Math. Soc., 42 (1936), 231-234.
%H H. Lenstra <a href="http://www.msri.org/publications/ln/msri/1998/mandm/lenstra/1/index.html">Harmonic Numbers</a>, MSRI, 1998.
%H J. J. O'Connor and E. F. Robertson, <a href="http://www-history.mcs.st-and.ac.uk/Biographies/Levi.html">Levi ben Gerson</a>, The MacTutor History of Mathematics archive, 2009.
%H I. Peterson, <a href="http://archive.is/iRXz">Medieval Harmony</a>, Math Trek, MAA, 2012.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Gersonides">Gersonides</a>
%F a(4n) = 5 as 3^(4n)-1 = (3^4)^n-1 = 81^n-1 = (80+1)^n-1 == 0 (mod 5).
%F a(6+12n) = 7 as 3^(6+12n)-1 = (3^6)^(1+2n)-1 = 729^(1+2n)-1 = (728+1)^(1+2n)-1 == 1^(1+2n)-1 == 0 (mod 7), but 729^(1+2n)-1 = (730-1)^(1+2n)-1 == (-1)^(1+2n)-1 == -2 (mod 5).
%e 3^3 - 1 = 26 = 2 * 13, so a(3) = 13.
%e 3^4 - 1 = 80 = 2^4 * 5, so a(4) = 5.
%e 3^5 - 1 = 242 = 2 * 11^2, so a(5) = 11.
%t Table[FactorInteger[3^n - 1][[2, 1]], {n, 3, 50}]
%o (PARI) a(n)=factor(3^n>>valuation(3^n-1,2))[1,1] \\ _Charles R Greathouse IV_, Jan 20 2014
%Y See A235365 for 3^n + 1.
%Y Cf. also A003586 (products 2^m * 3^n), A006899, A061987, A108906.
%K nonn
%O 3,1
%A _Jonathan Sondow_, Jan 19 2014
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