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.

A279073 Smallest positive number whose residues modulo the first n primes are all different.

Original entry on oeis.org

1, 2, 3, 5, 8, 87, 116, 129, 129, 129, 129, 129, 129, 202, 202, 202, 202, 202, 6753, 7769, 14614, 14614, 16574, 30777, 30777, 30777, 30777, 30777, 90878, 99483, 99483, 99483, 99483, 99483, 99483, 99483, 99483, 680384, 845662, 1719404, 1787204, 1787204, 1787204
Offset: 1

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Author

Jon E. Schoenfield, Jan 12 2017

Keywords

Comments

This sequence begins like the Fibonacci sequence. Are any terms beyond a(5) = 8 also Fibonacci numbers?
From Jon E. Schoenfield, Jan 15 2017: (Start)
a(n) = min_{k : A279086(k) >= n}.
For the smallest positive number having exactly n distinct residues modulo p before the first repeated residue occurs as p runs through the primes, see A279074. (E.g., a(n)=129 for n=8..13, but A279074(n)=129 only at n=13.) (End)

Examples

			For n=6, the first n primes are {2, 3, 5, 7, 11, 13}; 87 mod {2, 3, 5, 7, 11, 13} = {1, 0, 2, 3, 10, 9} (all different), and this does not occur for any k < 87, so a(6) = 87.
For n=8, 129 mod (each of the first n primes) gives {1, 0, 4, 3, 8, 12, 10, 15} (all different), and this does not occur for any k < 129, so a(8) = 129. Additionally, 129 mod p for each of the next 5 primes p gives {14, 13, 5, 18, 6} (all different from the first eight residues and from each other), so 129 is also a(9)..a(13). (This run of identical terms stops at n=13, since 129 mod prime(14) = 129 mod 43 = 0 = 129 mod prime(2).)
		

Crossrefs

Programs

  • Mathematica
    f[k_, m_] := Mod[k, #] & /@ Prime[Range[m]]; lst = {1};
    f[n_] := Module[{k = Last[lst]},While[Sort[f[k, n]] != Union[f[k, n]], k++]; AppendTo[lst, k]];f /@ Range[30]; Rest[lst] (* Ivan N. Ianakiev, Jan 17 2017 *)
  • PARI
    a(n) = {k = 1; ok = 0; while (!ok, vp = vector(n, j, k % prime(j)); if (#vecsort(vp,,8) == n, ok = 1, k++);); k;} \\ Michel Marcus, Jan 22 2017