cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A349530 Least positive integer m such that the n numbers k*(k^4+1) (k=1..n) are pairwise distinct modulo m^2.

Original entry on oeis.org

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
Offset: 1

Views

Author

Zhi-Wei Sun, Nov 21 2021

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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]

A356976 Least positive integer m such that the numbers k^3 + 3*k (k = 1..n) are pairwise distinct modulo m.

Original entry on oeis.org

1, 3, 3, 7, 15, 15, 19, 27, 27, 39, 39, 39, 61, 61, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243, 243
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 07 2022

Keywords

Comments

Conjecture 1: If n is at least 15, then a(n) is the least power of 3 not smaller than 3*n.
Conjecture 2: For each positive integer n, the least positive integer m such that those numbers 2*k^3 + k (k = 1..n) are pairwise distinct modulo m, is just the least power of 2 not smaller than n.
Conjecture 3: For any positive integer n, the least positive integer m such that those numbers 2*k^3 - 4*k (k = 1..n) are pairwise distinct modulo m, is just the least power of 3 not smaller than n.
Conjecture 4: For each positive integer n not equal to 4, the least positive integer m such that those numbers 16*k^3 - 8*k (k = 1..n) are pairwise distinct modulo m, is just the least power of 3 not smaller than n.
The author formulated Conjectures 1-4 on Nov. 16, 2021, and verified them for n up to 10^5.

Examples

			a(2) = 3, for, 1^3 + 3*1 = 4 and 2^3 + 3*2 = 14 are incongruent modulo 3, but congruent modulo 1 and 2.
		

Crossrefs

Programs

  • Mathematica
    f[k_]:=f[k]=k^3+3*k;
    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, 80}]; Print[tab]
Showing 1-2 of 2 results.