OFFSET
3,1
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
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
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).
EXAMPLE
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.
MATHEMATICA
Table[FactorInteger[3^n - 1][[2, 1]], {n, 3, 50}]
PROG
(PARI) a(n)=factor(3^n>>valuation(3^n-1, 2))[1, 1] \\ Charles R Greathouse IV, Jan 20 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Jan 19 2014
STATUS
approved