A349459 - OEIS

%I #13 Nov 21 2021 01:16:13

%S 1,5,7,11,13,23,23,23,23,41,41,41,41,41,41,101,101,107,107,107,107,

%T 107,107,107,107,107,107,107,223,223,223,223,223,223,223,223,223,223,

%U 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

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

%C 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.

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

%C 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.

%H Zhi-Wei Sun, <a href="/A349459/b349459.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2013.02.003">On functions taking only prime values</a>, J. Number Theory 133(2013), no.8, 2794-2812.

%e 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.

%t f[k_]:=f[k]=k^2*(k^2-1);

%t U[m_,n_]:=U[m,n]=Length[Union[Table[Mod[f[k],m],{k,1,n}]]]

%t 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]

%Y Cf. A000290, A208643.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Nov 18 2021