

A232616


Least positive integer m such that {2^k  k: k = 1,...,m} contains a complete system of residues modulo n.


6



1, 2, 4, 5, 10, 6, 14, 10, 12, 18, 29, 13, 33, 22, 40, 19, 38, 18, 58, 21, 36, 58, 75, 26, 60, 66, 40, 64, 195, 53, 87, 36, 158, 67, 130, 37, 133, 94, 90, 42, 95, 42, 105, 112, 112, 140, 247, 51, 122, 94, 119, 120, 311, 54, 126, 90, 184, 223, 264, 61
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OFFSET

1,2


COMMENTS

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.
Conjectures:
(i) a(n) < 2*(prime(n)1) for all n > 0.
(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.


LINKS

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..700 from ZhiWei Sun)
ZhiWei Sun, On a^n + b*n modulo m, preprint, arXiv:1312.1166.


EXAMPLE

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.


MATHEMATICA

L[m_, n_]:=Length[Union[Table[Mod[2^kk, n], {k, 1, m}]]]
Do[Do[If[L[m, n]==n, Print[n, " ", m]; Goto[aa]], {m, 1, n^2}];
Print[n, " ", 0]; Label[aa]; Continue, {n, 1, 60}]


CROSSREFS

Cf. A000079, A000325, A231201, A231725, A232398, A232548, A232862.
Sequence in context: A077389 A122991 A245512 * A125728 A276608 A173660
Adjacent sequences: A232613 A232614 A232615 * A232617 A232618 A232619


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Nov 26 2013


STATUS

approved



