A232898 Least positive integer m such that {C(2k,k) + k: k = 1,...,m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

1, 2, 7, 5, 10, 12, 9, 24, 31, 22, 59, 25, 27, 30, 42, 56, 123, 66, 57, 72, 84, 78, 73, 132, 136, 57, 99, 80, 129, 211, 170, 226, 121, 170, 126, 129, 238, 218, 157, 132, 348, 198, 388, 103, 171, 166, 247, 181, 205, 352, 194, 136, 430, 226, 117, 224, 237, 292, 364, 241

Offset: 1

Author

Zhi-Wei Sun, Dec 02 2013

Keywords

• nonn

Comments

Conjecture: (i) Let n be any positive integer. Then 0 < a(n) <= n^2/2 + 3. Also, {C(2k,k) - k: k = 1, ..., [n^2/2] + 15} contains a complete system of residues modulo n, where [.] is the floor function.

(ii) For any integer n > 2, neither C(2n,n) + n nor C(2n,n) - n has the form x^m with m > 1.

Examples

			a(2) = 2 since C(2*1,1) + 1 = 3 is odd and C(2*2,2) + 2 = 8 is even.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..250 from Zhi-Wei Sun)

Crossrefs

Cf. A000984, A232616, A232862, A232894.

Programs

Mathematica

L[m_,n_]:=Length[Union[Table[Mod[Binomial[2k,k]+k,n],{k,1,m}]]]
Do[Do[If[L[m,n]==n,Print[n," ",m];Goto[aa]],{m,1,n^2/2+3}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,60}]