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A349459
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Least positive integer m such that the n numbers k^2*(k^2-1) (k=1..n) are pairwise distinct modulo m.
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4
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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