A235366 Smallest odd prime factor of 3^n - 1.

13, 5, 11, 7, 1093, 5, 13, 11, 23, 5, 797161, 547, 11, 5, 1871, 7, 1597, 5, 13, 23, 47, 5, 11, 398581, 13, 5, 59, 7, 683, 5, 13, 103, 11, 5, 13097927, 1597, 13, 5, 83, 7, 431, 5, 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

Offset: 3

Author

Jonathan Sondow, Jan 19 2014

Keywords

• nonn

Comments

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

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

Examples

			3^3 - 1 = 26 = 2 * 13, so a(3) = 13.
3^4 - 1 = 80 = 2^4 * 5, so a(4) = 5.
3^5 - 1 = 242 = 2 * 11^2, so a(5) = 11.
		

References

• L. E. Dickson, History of the Theory of Numbers, Vol. II, Chelsea, NY 1992; see p. 731.

Links

• Max Alekseyev, Table of n, a(n) for n = 3..796 (terms to a(660) from Charles R Greathouse IV)
• P. Franklin, Problem 2927, Amer. Math. Monthly, 30 (1923), p. 81.
• A. Herschfeld, The equation 2^x - 3^y = d, Bull. Amer. Math. Soc., 42 (1936), 231-234.
• H. Lenstra Harmonic Numbers, MSRI, 1998.
• J. J. O'Connor and E. F. Robertson, Levi ben Gerson, The MacTutor History of Mathematics archive, 2009.
• I. Peterson, Medieval Harmony, Math Trek, MAA, 2012.
• Wikipedia, Gersonides

Crossrefs

See A235365 for 3^n + 1.

Cf. also A003586 (products 2^m * 3^n), A006899, A061987, A108906.

Programs

Mathematica

Table[FactorInteger[3^n - 1][[2, 1]], {n, 3, 50}]

PARI

a(n)=factor(3^n>>valuation(3^n-1,2))[1,1] \\ Charles R Greathouse IV, Jan 20 2014

Formula

a(4n) = 5 as 3^(4n)-1 = (3^4)^n-1 = 81^n-1 = (80+1)^n-1 == 0 (mod 5).

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).