A349537 Least positive integer m such that the n numbers 33*k^2*(k^3+1) (k = 1..n) are pairwise distinct modulo m.
1, 4, 7, 7, 13, 13, 13, 13, 13, 31, 41, 41, 61, 61, 61, 61, 61, 61, 61, 73, 101, 137, 137, 137, 137, 137, 137, 137, 137, 233, 233, 233, 233, 233, 233, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 349, 547, 547, 547, 547, 547, 547, 547, 547, 547, 547, 859, 859, 859, 859, 859, 859
Offset: 1
Keywords
Examples
a(2) = 4 since 33*1^2*(1^3+1) = 66 and 33*2^2*(2^3+1) = 1188 are incongruent modulo 4, but they are congruent modulo each of 1, 2 and 3.
Links
- Zhi-Wei Sun, On functions taking only prime values, J. Number Theory 133(2013), no.8, 2794-2812.
- Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.
- Quan-Hui Yang and Lilu Zhao, On a conjecture of Sun involving powers of three, arXiv:2111.02746 [math.NT], 2021.
Programs
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Mathematica
f[k_]:=f[k]=33*k^2*(k^3+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]
Comments