OFFSET
1,2
COMMENTS
Conjecture: a(n) = 1 only for n = 1, 5, 16.
In constrast, a classical theorem of Bang asserts that if n > 1 is different from 6 then 2^n - 1 has a prime divisor which does not divide any 2^k - 1 with 0 < k < n.
REFERENCES
A. S. Bang, Taltheoretiske Undersgelser, Tidsskrift fur Mat. 4(1886), no. 5, 70--80, 130--137.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..215
Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.
EXAMPLE
a(4) = 3 since 2^4 - 4 = 2^2*3 with 3 dividing none of 2^1 - 1 = 1, 2^2 - 2 = 2 and 2^3 - 3 = 5.
MATHEMATICA
u[n_]:=2^n-n
f[n_]:=FactorInteger[u[n]]
p[n_]:=Table[Part[Part[f[n], k], 1], {k, 1, Length[f[n]]}]
Do[If[u[n]<2, Goto[cc]]; Do[Do[If[Mod[u[i], Part[p[n], k]]==0, Goto[aa]], {i, 1, n-1}]; Print[n, " ", Part[p[n], k]]; Goto[bb]; Label[aa]; Continue, {k, 1, Length[p[n]]}]; Label[cc]; Print[n, " ", 1]; Label[bb]; Continue, {n, 1, 45}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 10 2014
STATUS
approved