A349459 Least positive integer m such that the n numbers k^2*(k^2-1) (k=1..n) are pairwise distinct modulo m.

1, 5, 7, 11, 13, 23, 23, 23, 23, 41, 41, 41, 41, 41, 41, 101, 101, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 107, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 229, 239, 239, 239, 383, 383, 383, 383, 383, 383, 383, 383, 401, 401, 557, 557, 557, 557, 557, 557, 557, 557, 557, 557, 557, 733, 733, 733, 733, 733, 733, 733

Offset: 1

Author

Zhi-Wei Sun, Nov 18 2021

Keywords

• nonn

Comments

Conjecture: For any integer n > 1, the term a(n) is the least prime p > 2*n with p dividing a^2 + b^2 - 1 for no 1 <= a < b <= n.

This has been verified for all n = 2..15000.

Note that a^2*(a^2-1)-b^2*(b^2-1) = (a-b)*(a+b)*(a^2+b^2-1). For any positive integers m and n > 1, if k^2*(k^2-1) (k=1..n) are pairwise distinct modulo m, then it is easy to see that m > 2*n.

Examples

			a(2) = 5 since the two numbers 1^2*(1^2-1)=0 and 2^2*(2^2-1) = 12 are distinct modulo 5, but they are congruent modulo each of 1,2,3,4.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.

Crossrefs

Cf. A000290, A208643.

Programs

Mathematica

f[k_]:=f[k]=k^2*(k^2-1);
U[m_,n_]:=U[m,n]=Length[Union[Table[Mod[f[k],m],{k,1,n}]]]
tab={};s=1;Do[m=s;Label[bb];If[U[m,n]==n,s=m;tab=Append[tab,s];Goto[aa]];m=m+1;Goto[bb];Label[aa],{n,1,70}];Print[tab]