|
%I #30 Aug 06 2019 12:03:16
%S 1,2,4,5,10,6,14,10,12,18,29,13,33,22,40,19,38,18,58,21,36,58,75,26,
%T 60,66,40,64,195,53,87,36,158,67,130,37,133,94,90,42,95,42,105,112,
%U 112,140,247,51,122,94,119,120,311,54,126,90,184,223,264,61
%N Least positive integer m such that {2^k - k: k = 1,...,m} contains a complete system of residues modulo n.
%C By a result of the author (see arXiv:1312.1166), for any integers a and n > 0, the set {a^k - k: k = 1, ..., n^2} contains a complete system of residues modulo n. (We may also replace a^k - k by a^k + k.) Thus a(n) always exists and it does not exceed n^2.
%C Conjectures:
%C (i) a(n) < 2*(prime(n)-1) for all n > 0.
%C (ii) The Diophantine equation x^n - n = y^m with m, n, x, y > 1 only has two integral solutions: 2^5 - 5 = 3^3 and 2^7 - 7 = 11^2. Also, the Diophantine equation x^n + n = y^m with m, n, x, y > 1 only has two integral solutions: 5^2 + 2 = 3^3 and 5^3 + 3 = 2^7.
%H Chai Wah Wu, <a href="/A232616/b232616.txt">Table of n, a(n) for n = 1..10000</a> (n = 1..700 from Zhi-Wei Sun)
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1312.1166">On a^n + b*n modulo m</a>, preprint, arXiv:1312.1166 [math.NT], 2013-2014.
%e a(3) = 4 since {2 - 1, 2^2 - 2, 2^3 - 3} = {1, 2, 5} does not contain a complete system of residues mod 3, but {2 - 1, 2^2 - 2, 2^3 - 3, 2^4 - 4} = {1, 2, 5, 12} does.
%t L[m_,n_]:=Length[Union[Table[Mod[2^k-k,n],{k,1,m}]]]
%t Do[Do[If[L[m,n]==n,Print[n," ",m];Goto[aa]],{m,1,n^2}];
%t Print[n," ",0];Label[aa];Continue,{n,1,60}]
%Y Cf. A000079, A000325, A231201, A231725, A232398, A232548, A232862.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Nov 26 2013
|
