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.

Previous Showing 11-14 of 14 results.

A270645 The sequence a of 1's and 2's starting with (2,1,1,1) such that a(n) is the length of the (n+2)nd run of a.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Apr 06 2016

Keywords

Comments

See A270641 for a guide to related sequences.

Examples

			a(1) = 1, so the 3rd run has length 2, so a(5) must be 2 and a(6) = 2.
a(2) = 2, so the 4th run has length 1, so a(7) = 1 and a(8) = 2.
a(3) = 1, so the 5th run has length 1, so a(9) = 1 and a(10) = 2.
Globally, the runlength sequence is 1,3,2,1,1,1,2,2,1,2,1,2,2,1,1,..., and deleting the first 2 terms leaves the same sequence.
		

Crossrefs

Programs

  • Mathematica
    a = {2,1,1,1}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n,   200}]; a  (* Peter J. C. Moses, Apr 01 2016 *)

A270646 The sequence a of 1's and 2's starting with (2,2,1,1) such that a(n) is the length of the (n+2)nd run of a.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Apr 07 2016

Keywords

Comments

See A270641 for a guide to related sequences.
a(1) = 2, so the 3rd run has length 2, so a(5) must be 2 and a(6) = 1.
a(2) = 2, so the 4th run has length 2, so a(7) = 1 and a(8) = 1.
a(3) = 1, so the 5th run has length 1, so a(9) = 2 and a(10) = 1.
Globally, the runlength sequence of a is 2,2,2,2,1,1,2,2,1,1,2,1,2,2,1,1,2,..., and deleting the first 2 terms leaves a = A270646.

Crossrefs

Programs

  • Mathematica
    a = {2,2,1,1}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n,   200}]; a  (* Peter J. C. Moses, Apr 01 2016 *)

A270647 The sequence a of 1's and 2's starting with (2,2,1,2) such that a(n) is the length of the (n+3)rd run of a.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Apr 07 2016

Keywords

Comments

See A270641 for a guide to related sequences.

Examples

			a(1) = 2, so the 4th run has length 2, so a(5) must be 1 and a(6) = 1.
a(2) = 2, so the 5th run has length 2, so a(7) = 2 and a(8) = 2.
a(3) = 1, so the 6th run has length 1, so a(9) = 1 and a(10) = 2.
Globally, the runlength sequence of a is 2,1,1,2,2,1,2,1,1,2,2,1,2,2,1,2,1,1,2,..., and deleting the first 3 terms leaves a = A270647.
		

Crossrefs

Programs

  • Mathematica
    a = {2,2,1,2}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n,   200}]; a  (* Peter J. C. Moses, Apr 01 2016 *)

Formula

Conjecture: a(n) = A270643(n+1). - R. J. Mathar, Jun 21 2025

A270648 The sequence a of 1's and 2's starting with (2,2,2,2) such that a(n) is the length of the (n+1)st run of a.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Apr 07 2016

Keywords

Comments

See A270641 for a guide to related sequences.

Examples

			a(1) = 2, so the 2nd run has length 2, so a(5) must be 1 and a(6) = 1.
a(2) = 2, so the 3rd run has length 2, so a(7) = 2 and a(8) = 2.
a(3) = 2, so the 4th run has length 2, so a(9) = 1 and a(10) = 1.
Globally, the runlength sequence of a is 4,2,2,2,2,1,1,2,2,1,1,2,2,1,2,1,1,2,..., and deleting the first term leaves a = A270648.
		

Crossrefs

Programs

  • Mathematica
    a = {2,2,2,2}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n,   200}]; a  (* Peter J. C. Moses, Apr 01 2016 *)
Previous Showing 11-14 of 14 results.