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 1-3 of 3 results.

A254733 a(n) is the least k > n such that n divides k^3.

Original entry on oeis.org

2, 4, 6, 6, 10, 12, 14, 10, 12, 20, 22, 18, 26, 28, 30, 20, 34, 24, 38, 30, 42, 44, 46, 30, 30, 52, 30, 42, 58, 60, 62, 36, 66, 68, 70, 42, 74, 76, 78, 50, 82, 84, 86, 66, 60, 92, 94, 60, 56, 60, 102, 78, 106, 60, 110, 70, 114, 116, 118, 90, 122, 124, 84
Offset: 1

Views

Author

Peter Kagey, Feb 06 2015

Keywords

Comments

A073353(n) <= a(n) <= 2*n. Any prime that divides n must also divide a(n), and because n divides (2*n)^3.

Examples

			a(8) = 10 because 8 divides 10^3, but 8 does not divide 9^3.

Links

Crossrefs

Cf. A073353 (similar, with k^n).
Cf. A254732 (similar, with k^2), A254734 (similar, with k^4).
Cf. A019555 (similar without the restriction that a(n) > n).

Programs

  • Mathematica
    lkn[n_]:=Module[{k=n+1},While[PowerMod[k,3,n]!=0,k++];k]; Array[lkn,70] (* Harvey P. Dale, Nov 23 2024 *)
  • PARI
    a(n)=for(k=n+1,2*n,if(k^3%n==0,return(k)))
vector(100,n,a(n)) \\ Derek Orr, Feb 07 2015
  • Ruby
    def a(n)
  (n+1..2*n).find { |k| k**3 % n == 0 }
end

Formula

a(n) = n + A019555(n).

A254734 a(n) is the least k > n such that n divides k^4.

Original entry on oeis.org

2, 4, 6, 6, 10, 12, 14, 10, 12, 20, 22, 18, 26, 28, 30, 18, 34, 24, 38, 30, 42, 44, 46, 30, 30, 52, 30, 42, 58, 60, 62, 36, 66, 68, 70, 42, 74, 76, 78, 50, 82, 84, 86, 66, 60, 92, 94, 54, 56, 60, 102, 78, 106, 60, 110, 70, 114, 116, 118, 90, 122, 124, 84
Offset: 1

Views

Author

Peter Kagey, Feb 07 2015

Keywords

Comments

A073353(n) <= a(n) <= 2*n. Any prime that divides n must also divide a(n), and because n divides (2*n)^4.
a(n) = 2*n iff n is squarefree (A005117). - Robert Israel, Feb 08 2015

Examples

			a(16) = 18 because 16 divides 18^4, but 16 does not divide 17^4.

Links

Crossrefs

Cf. A005117 (squarefree).
Cf. A073353 (similar, with k^n).
Cf. A254732 (similar, with k^2), A254733 (similar, with k^3).

Programs

  • Maple
    f:= proc(n) local k;
     for k from n+1 do if (k^4/n)::integer then return k fi od:
end proc:
seq(f(n), n=1..100); # Robert Israel, Feb 08 2015
  • Mathematica
    lk[n_]:=Module[{k=n+1},While[PowerMod[k,4,n]!=0,k++];k]; Array[lk,70] (* Harvey P. Dale, Nov 22 2015 *)
  • PARI
    a(n)=for(k=n+1,2*n,if(k^4%n==0,return(k)))
vector(100,n,a(n)) \\ Derek Orr Feb 07 2015
  • Python
    def A254734(n):
    k = n + 1
    while pow(k, 4, n):
        k += 1
    return k # Chai Wah Wu, Feb 15 2015
  • Ruby
    def a(n)
  (n+1..2*n).find { |k| k**4 % n == 0 }
end

A272327 Table read by antidiagonals: T(n, k) is the least i > n such that n divides i^k (n > 0, k > 0).

Original entry on oeis.org

2, 4, 2, 6, 4, 2, 8, 6, 4, 2, 10, 6, 6, 4, 2, 12, 10, 6, 6, 4, 2, 14, 12, 10, 6, 6, 4, 2, 16, 14, 12, 10, 6, 6, 4, 2, 18, 12, 14, 12, 10, 6, 6, 4, 2, 20, 12, 10, 14, 12, 10, 6, 6, 4, 2, 22, 20, 12, 10, 14, 12, 10, 6, 6, 4, 2, 24, 22, 20, 12, 10, 14, 12, 10, 6
Offset: 1

Views

Author

Peter Kagey, Apr 25 2016

Keywords

Comments

T(n, k) = 2*n for squarefree n.

Examples

			a(1) = T(1, 1) = 2  because 1 divides 2^1
a(2) = T(2, 1) = 4  because 2 divides 4^1
a(3) = T(1, 2) = 2  because 1 divides 2^2
a(4) = T(3, 1) = 6  because 3 divides 6^1
a(5) = T(2, 2) = 4  because 2 divides 4^2
a(6) = T(1, 3) = 2  because 1 divides 2^3
a(7) = T(4, 1) = 8  because 4 divides 8^1
a(8) = T(3, 2) = 6  because 3 divides 6^2
a(9) = T(2, 3) = 4  because 2 divides 4^3
a(10) = T(1, 4) = 2 because 1 divides 2^4
Triangle begins:
   2  2 2 2 2 2
   4  4 4 4 4
   6  6 6 6
   8  6 6
  10 10
  12

Links

Crossrefs

Cf. A254732 (second column), A254733 (third column), A254734 (fourth column), A073353 (main diagonal).

Programs

  • Mathematica
    Table[Function[m, SelectFirst[Range[m + 1, 10^3], Divisible[#^k, m] &]][n - k + 1], {n, 12}, {k, n}] // Flatten (* Michael De Vlieger, Apr 25 2016, Version 10 *)
Showing 1-3 of 3 results.

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