A232616 Least positive integer m such that {2^k - k: k = 1,...,m} contains a complete system of residues modulo n.

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

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 Zhi-Wei Sun)

Zhi-Wei Sun, On a^n + b*n modulo m, preprint, arXiv:1312.1166 [math.NT], 2013-2014.

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^k-k, 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: A122991 A245512 A366351 * A125728 A351871 A276608

Adjacent sequences: A232613 A232614 A232615 * A232617 A232618 A232619

Keyword

nonn

Author

Zhi-Wei Sun, Nov 26 2013

Status

approved