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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 (list; graph; refs; listen; history; text; internal format)
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.

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: A077389 A122991 A245512 * A125728 A276608 A173660

Adjacent sequences:  A232613 A232614 A232615 * A232617 A232618 A232619

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Nov 26 2013

STATUS

approved

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Last modified January 16 23:44 EST 2019. Contains 319206 sequences. (Running on oeis4.)