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.

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A332875 Sizes of maximal weakly increasing subsequences of A000002.

Original entry on oeis.org

3, 3, 3, 3, 3, 4, 2, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 4, 3, 3, 3, 3, 3, 3, 3, 2, 4, 3, 3, 3, 3, 3, 4, 2, 3, 3, 3, 3, 3, 3, 3, 4, 2, 3, 4, 3, 3, 3, 2, 4, 3, 2, 3, 4, 3, 3, 3, 2, 3, 4, 2, 3, 3, 3, 3, 3, 2, 4, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 2, 3, 3, 3, 4, 2, 3, 4, 3
Offset: 1

Views

Author

Gus Wiseman, Mar 08 2020

Keywords

Examples

			The weakly increasing subsequences begin: (1,2,2), (1,1,2), (1,2,2), (1,2,2), (1,1,2), (1,1,2,2), (1,2), (1,1,2), (1,2,2), (1,1,2), (1,1,2), (1,2,2), (1,2,2).
		

Crossrefs

The number of runs in the first n terms of A000002 is A156253.
The weakly decreasing version is A332273.

Programs

  • Mathematica
    kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],q[[-2]],Last[q]},{1,1,1},0,{1,1,2},1,{1,2,1},2,{1,2,2},0,{2,1,1},2,{2,1,2},2,{2,2,1},1,{2,2,2},1]]]
    kol[n_Integer]:=Nest[kolagrow,{1},n-1];
    Length/@Split[kol[40],#1<=#2&]

Formula

a(n) = A000002(2*n - 1) + A000002(2*n).

A156728 a(n) = abs(A054354(n)).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1
Offset: 1

Views

Author

Benoit Cloitre, Feb 14 2009

Keywords

Comments

This sequence is the image of the Kolakoski sequence A000002 under the morphism 1->1, 2->01. - Gabriele Fici, Aug 12 2013

Crossrefs

Programs

  • Mathematica
    a2 = {1, 2, 2}; Do[ a2 = Join[a2, {1 + Mod[n - 1, 2]}], {n, 3, 105}, {i, 1, a2[[n]]}]; a[n_] := Mod[a2[[n]] + a2[[n + 1]], 2]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Jun 18 2013 *)

Formula

a(n) = (v(n+1) - v(n) + 1)/2 where v(n) = A156253(n) - A156251(n).
a(n) = (A000002(n) + A000002(n+1)) mod 2.
a(n) = A156253(n+1) - A156253(n). - Alan Michael Gómez Calderón, Dec 20 2024

A321020 A hybrid of Kolakoski's sequence A000002 and Golomb's sequence A001462: if A001462(n) is odd replace it with 1, if even with 2.

Original entry on oeis.org

1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1
Offset: 1

Views

Author

N. J. A. Sloane, Nov 11 2018

Keywords

Comments

This is A000002 rewritten so the run lengths are given by A001462.
The companion sequence, A001462 rewritten so the run lengths are given by A000002, seems to be A156253.
Note that Kolakoski's sequence A000002 and Golomb's sequence A001462 have very similar definitions, although the asymptotic behavior of A001462 is well-understood, while that of A000002 is a mystery. The asymptotic behavior of the two hybrids A156253 and A321020 might be worth investigating.

Crossrefs

Programs

  • PARI
    a = vector(84, k, k); for (i=1, oo, for (j=1, a[i], a[n++] = i; print1 (2-(i%2) ", "); if (n==#a, break(2)))) \\ Rémy Sigrist, Nov 12 2018

A333229 First sums of the Kolakoski sequence A000002.

Original entry on oeis.org

3, 4, 3, 2, 3, 3, 3, 4, 3, 3, 4, 3, 2, 3, 3, 2, 3, 4, 3, 3, 3, 2, 3, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 3, 4, 3, 3, 4, 3, 2, 3, 3, 3, 4, 3, 3, 3, 2, 3, 3, 2, 3, 4, 3, 3, 4, 3, 2, 3, 3, 3, 4, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 3, 4, 3, 3, 3, 2, 3, 4, 3, 3, 4, 3, 2, 3, 3
Offset: 1

Views

Author

Gus Wiseman, Mar 18 2020

Keywords

Crossrefs

Positions of 3's are A054353.
Positions of 2's are A074262.
Positions of 4's are A074263.
The number of runs in the first n terms of A000002 is A156253(n).
Even-indexed terms are A332273 (without the first term).
Odd-indexed terms are A332875.

Programs

  • Mathematica
    kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],q[[-2]],Last[q]},{1,1,1},0,{1,1,2},1,{1,2,1},2,{1,2,2},0,{2,1,1},2,{2,1,2},2,{2,2,1},1,{2,2,2},1]]]
    kol[n_Integer]:=Nest[kolagrow,{1},n-1];
    Table[kol[n][[-1]]+kol[n+1][[-1]],{n,30}]

Formula

a(n) = A000002(n) + A000002(n + 1).

A350126 a(n) = (a(a(n-1)) mod 2) + a(n-2) with a(0) = 0 and a(1) = 1.

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 2, 4, 2, 5, 3, 5, 4, 5, 5, 6, 5, 7, 5, 8, 5, 9, 6, 9, 7, 9, 8, 9, 9, 10, 10, 11, 11, 12, 11, 13, 12, 13, 13, 14, 14, 15, 14, 16, 15, 16, 16, 17, 17, 18, 18, 19, 18, 20, 19, 20, 20, 21, 21, 22, 21, 23, 22, 23, 23, 24, 24, 25, 25, 26, 25, 27
Offset: 0

Views

Author

Jon Maiga, Dec 15 2021

Keywords

Comments

It appears that lim_{n->infinity} a(n)/n = 0.33960...
A similar formula is conjectured for A156253: (a(a(n-1)) mod 2) + a(n-2) + 1.

Crossrefs

Cf. A156253.

Programs

  • Mathematica
    a[0] = 0;
    a[1] = 1;
    a[n_] := a[n] = Mod[a[a[n - 1]], 2] + a[n - 2]
    Array[a, 100, 0]
  • Python
    a = [0, 1]
    [a.append(a[a[n-1]]%2 + a[n-2]) for n in range(2, 72)]
    print(a) # Michael S. Branicky, Dec 15 2021
Previous Showing 11-15 of 15 results.