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.

A335898 a(n) = a(floor((n-1)/a(n-1))) + a(floor((n-2)/a(n-2))) with a(1) = a(2) = 1.

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 3, 3, 2, 3, 4, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 4, 5, 5, 4, 5, 6, 5, 5, 6, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 5, 6, 6, 4, 6, 6, 5, 6, 6, 6, 6, 6, 7, 6, 6, 8, 7, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 6, 7, 7, 6, 7
Offset: 1

Views

Author

Altug Alkan, following a suggestion from Andrew R. Booker, Jun 29 2020

Keywords

Comments

This sequence is a_1(n) where a_i(n) = Sum_{k=1..i+1} a_i(floor((n-k)/a_i(n-k))) with a_i(n) = 1 for n <= i+1.
Conjecture: This sequence hits every positive integer.

Crossrefs

Programs

  • Mathematica
    a[1] = a[2] = 1; a[n_] := a[n] = a[Floor[(n-1)/a[n-1]]] + a[Floor[(n-2)/a[n-2]]]; Array[a, 100] (* Amiram Eldar, Jun 29 2020 *)
  • PARI
    a=vector(10^2); a[1]=a[2]=1; for(n=3, #a, a[n]=a[(n-1)\a[n-1]]+a[(n-2)\a[n-2]]); a

A335901 a(n) = 2*a(floor((n-1)/a(n-1))) with a(1) = 1.

Original entry on oeis.org

1, 2, 2, 2, 4, 2, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 4, 8, 4, 4, 4, 4, 4, 8, 4, 8, 4, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8, 4, 8, 4, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8, 4, 8, 4, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8
Offset: 1

Views

Author

Altug Alkan, following a suggestion from Andrew R. Booker, Jun 29 2020

Keywords

Comments

Least k such that a(k) = 2^n are 1, 2, 5, 21, 169, 2705, ... (Conjecture: This sequence is A117261).

Crossrefs

Programs

  • Maple
    f:= proc(n) option remember;
      2*procname(floor((n-1)/procname(n-1))) end proc:
    f(1):= 1:
    map(f, [$1..105]); # Robert Israel, Jul 08 2020
  • Mathematica
    a[1] = 1; a[n_] := a[n] = 2 * a[Floor[(n-1)/a[n-1]]]; Array[a, 100] (* Amiram Eldar, Jun 29 2020 *)
  • PARI
    a=vector(10^3); a[1]=1; for(n=2, #a, a[n]=2*a[(n-1)\a[n-1]]); a

A335855 a(n) = a(a(n-1)) + a(floor(n/a(n-1))) with a(1) = a(2) = a(3) = a(4) = 2.

Original entry on oeis.org

2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8
Offset: 1

Views

Author

Altug Alkan, Jun 27 2020

Keywords

Crossrefs

Programs

  • Mathematica
    a[1] = a[2] = a[3] = a[4] = 2; a[n_] := a[n] = a[a[n-1]] + a[Floor[n / a[n-1]]]; Array[a, 100] (* Amiram Eldar, Jun 27 2020 *)
  • PARI
    a=vector(10^2); a[1]=a[2]=a[3]=a[4]=2; for(n=5, #a, a[n]=a[n\a[n-1]]+a[a[n-1]]); a
Showing 1-3 of 3 results.