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A349530
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Least positive integer m such that the n numbers k*(k^4+1) (k=1..n) are pairwise distinct modulo m^2.
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2
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1, 3, 3, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 25, 25, 25, 25, 25
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OFFSET
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1,2
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COMMENTS
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Conjecture 1: Suppose that 5^a < sqrt(n) <= 5^(a+1). Then a(n) = 3*5^a if sqrt(n) <= sqrt(3)*5^a, and a(n) = 5^(a+1) otherwise.
Conjecture 2: Let f(n) be the least positive integer m such that the n numbers 18k*(k^4+1) (k=1..n) are pairwise distinct modulo m^2. Then f(n) is the least power of 5 not smaller than sqrt(n), except for 25 < n <= 45 (and in this case f(n) = 19).
Conjecture 3: Let n be a positive integer not among 26, 27, 28, 626, 627, 628, 629, 630, and define D(n) as the least positive integer m such that the n numbers k*(k^4+1) (k=1..n) are pairwise distinct modulo m. If 5^a < n <= 3*5^a, then D(n) = 3*5^a. If 3*5^a < n <= 5^(a+1), then D(n) = 5^(a+1).
We have verified the above conjectures for n up to 10^5.
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LINKS
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EXAMPLE
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a(2) = 3 since the two numbers 1*(1^4+1) = 2 and 2*(2^4+1) = 34 are distinct modulo 3^2, but they are congruent modulo each of 1^2 and 2^2.
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MATHEMATICA
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f[k_]:=f[k]=k*(k^4+1);
U[m_, n_]:=U[m, n]=Length[Union[Table[Mod[f[k], m^2], {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, 80}]; 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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