A208494 Least integer m>1 such that those k! mod m with k=1,...,n are pairwise distinct.

2, 2, 3, 7, 10, 13, 13, 13, 31, 37, 37, 37, 61, 61, 61, 83, 83, 83, 127, 127, 127, 127, 127, 179, 179, 179, 179, 179, 193, 193, 193, 193, 193, 193, 193, 193, 277, 277, 277, 277, 277, 277, 383, 383, 383, 383, 383, 479, 479, 479, 479, 479, 479, 479, 541, 541, 541, 541, 541, 541, 541, 541, 541, 641, 641, 641, 641, 641, 641, 641, 641, 641, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 877, 1013, 1013, 1013, 1013, 1013, 1013, 1013, 1013, 1013, 1013, 1013

Offset: 1

Author

Zhi-Wei Sun, Feb 27 2012

Keywords

• nonn
• nice

Comments

On Feb 27 2012 Zhi-Wei Sun conjectured that a(n) is a prime with the only exception a(5)=10, and that a(n) does not exceed n^2/2 for all n=2,3,4,... He also conjectured that max{n>0: 1!,...,n! are pairwise incongruent mod p} is asymptotically equivalent to sqrt(p), where p is an odd prime.

He guessed that if we replace k! in the definition of a(n) by (-1)^k*k! then a(n) is a prime with the only exception a(3)=6. If we replace k! in the definition of a(n) by (2k)! or (-1)^k*(2k)!, then Zhi-Wei Sun conjectured that a(n) will take only prime values.

He also has similar conjectures involving (r*k)! or (-1)^k*(r*k)! with r>2.

Examples

			For n=5 we have a(5)=10 since 1!=1, 2!=2, 3!=6, 4!=24 and 5!=120 are pairwise incongruent mod 10 but not pairwise incongruent modulo any of 2,3,...,9.
		

Links

• Zhi-Wei Sun and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 400 terms from Sun)
• Zhi-Wei Sun, A function taking only prime values, a message to Number Theory List, Feb 21 2012.
• Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.

Crossrefs

Cf. A000142, A207982.

Programs

Mathematica

R[n_,i_]:=Union[Table[Mod[k!,i],{k,1,n}]]
Do[Do[If[Length[R[n,i]]==n,Print[n," ",i];Goto[aa]],{i,2,Max[n^2,2]}];
Print[n];Label[aa];Continue,{n,1,1000}]

PARI

has(n,m)=my(t=1); #Set(vector(n,i,t=(t*i)%m))==n
a(n,last=2)=while(!has(n,last), last++); last
t=2;vector(100,n,t=a(n,t)) \\ Charles R Greathouse IV, Jul 31 2016