A022300 -id:A022300 - cp's OEIS frontend

A025512 Index of n-th 2 in A022300.

3, 5, 8, 11, 13, 14, 16, 19, 22, 23, 25, 26, 28, 31, 33, 34, 37, 40, 41, 43, 44, 46, 49, 52, 53, 55, 58, 60, 61, 64, 67, 68, 70, 71, 73, 76, 78, 79, 82, 85, 87, 88, 90, 91, 94, 96, 97, 99, 102, 103, 105, 106, 109, 112, 114, 115, 117, 118, 121, 123, 124, 126, 129, 132, 133, 135

Offset: 1

David W. Wilson

nonn

Robert Israel, Table of n, a(n) for n = 1..10000

N:= 10^3: # to get all terms <= N
B:= Vector(N):
B[1..4]:= <1,1,2,1>:
m:= 4: t:= 2:
for n from 1 while m < N do
  t:= 3-t;
  B[m]:= t;
  if B[n] = 2 and m+1 < N then
     B[m+1]:= t; m:= m+2
  else m:= m+1
  fi
od:
select(t -> B[t]=2, [$1..N]); # Robert Israel, Nov 02 2016

A025513 Exactly half of first a(n) terms of A022300 are 1's (not known to be infinite).

Original entry on oeis.org

5904, 5986, 6050, 6068, 6074, 6076, 6078, 6080, 6084, 6086, 6092, 6094, 6096, 6098, 6100, 6102, 6104, 6106, 6108, 6114, 6116, 6118, 6120, 6124, 6126, 6136, 6138, 6142, 6144, 6146, 6148, 6154, 6156, 6158, 6160, 6162, 6164, 6166, 6168, 6170, 6174, 6176

Offset: 1

David W. Wilson

nonn

Even numbers n such that A025512(n/2) <= n and A025512(n/2+1) > n. - Robert Israel, Nov 02 2016

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A022300, A025512.

N:= 10000: # to use A022300(1..N)
B:= Vector(N):
B[1..4]:= <1,1,2,1>:
m:= 4: t:= 2:
for n from 1 while m < N do
  t:= 3-t;
  B[m]:= t;
  if B[n] = 2 and m+1 < N then
     B[m+1]:= t; m:= m+2
  else m:= m+1
  fi
od:
S:= ListTools:-PartialSums(convert(B,list)):
select(t -> S[t] = 3/2*t, [$1..nops(S)]); # Robert Israel, Nov 02 2016

A022301 Index of n-th 1 in A022300.

Original entry on oeis.org

1, 2, 4, 6, 7, 9, 10, 12, 15, 17, 18, 20, 21, 24, 27, 29, 30, 32, 35, 36, 38, 39, 42, 45, 47, 48, 50, 51, 54, 56, 57, 59, 62, 63, 65, 66, 69, 72, 74, 75, 77, 80, 81, 83, 84, 86, 89, 92, 93, 95, 98, 100

Offset: 0

Clark Kimberling

nonn

A022302 Least k such that first k terms of A022300 contain n more 1's than 2's.

Original entry on oeis.org

1, 2, 7, 10, 21, 66, 213, 246, 273, 454, 673, 700, 727, 934, 2451, 2766, 2799, 3006, 3099, 3396, 234665, 234758, 234977, 235004, 235013, 434216, 436951, 436978, 437005, 437128, 437155, 437708, 437735, 437762, 438017, 438044, 438145, 438314, 438341

Offset: 0

Clark Kimberling

nonn

A025515 Least k such that first k terms of A022300 contain n more 2's than 1's.

Original entry on oeis.org

6077, 6110, 6131, 6386, 6413, 7160, 7361, 8436, 8975, 9008, 9071, 11626, 11653, 11822, 11849, 11876, 12545, 12714, 12741, 12768, 13019, 13118, 13145, 13172, 13277, 13358, 16661, 26214, 26241, 26268, 26475, 51382, 51639, 51786, 52173, 52466, 52753

Offset: 1

David W. Wilson

nonn

A022303 The sequence a of 1's and 2's starting with (1,2,1) such that a(n) is the length of the (n+2)nd run of a.

Original entry on oeis.org

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, 1, 1, 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, 1, 1, 2, 1, 2, 2, 1, 2, 1, 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

Offset: 1

Clark Kimberling

nonn

			a(1) =1, so the 3rd run has length 1, so a(4) must be 2.
a(2) = 2, so the 4th run has length 2, so a(5) = 2 and a(6) = 1.
a(3) = 1, so the 5th run has length 1, so a(7) = 2.
a(4) = 2, so the 6th run has length 2, so a(8) = 1 and a(9) = 1.
Globally, the runlength sequence of a is 1,1,1,2,1,2,2,1,2,2,1,1,2,1,...., and deleting the first two terms leaves a = A022303.

Clark Kimberling, Table of n, a(n) for n = 1..20000

Cf. A022300, A006928, A000002.

a = {1, 2}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n, 100}]; a (* Peter J. C. Moses, Apr 02 2016 *)

Clarified and augmented by Clark Kimberling, Apr 02 2016

A270641 The sequence a of 1's and 2's starting with (1,1,1,1) such that a(n) is the length of the (n+1)st run of a.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 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, 1, 2, 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, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2

Offset: 1

Clark Kimberling, Apr 05 2016

Guide to related sequences (with adjustments for initial terms):
1, 1, 1, 1;    a(n) = length of (n + 1)st run of a; A270641
1, 1, 1, 2;    a(n) = length of (n + 2)nd run of a; A270641
1, 1, 2, 1;    a(n) = length of (n + 3)rd run of a; A270641
1, 1, 2, 2;    a(n) = length of (n + 2)nd run of a; A270642
1, 2, 1, 1;    a(n) = length of (n + 3)rd run of a; A022300
1, 2, 1, 2;    a(n) = length of (n + 4)th run of a; A270641
1, 2, 2, 1;    a(n) = length of (n + 3)rd run of a; A270643
1, 2, 2, 2;    a(n) = length of (n + 2)nd run of a; A270644
2, 1, 1, 1;    a(n) = length of (n + 2)nd run of a; A270645
2, 1, 1, 2;    a(n) = length of (n + 3)rd run of a; A222300
2, 1, 2, 1;    a(n) = length of (n + 4)th run of a; A270641
2, 1, 2, 2;    a(n) = length of (n + 3)rd run of a; A000002 (Kolakoski)
2, 2, 1, 1;    a(n) = length of (n + 2)nd run of a; A270646
2, 2, 1, 2;    a(n) = length of (n + 3)rd run of a; A270647
2, 2, 2, 1;    a(n) = length of (n + 2)nd run of a; A270644
2, 2, 2, 2;    a(n) = length of (n + 1)st run of a; A270648

			a(1) = 1, so the 2nd run has length 1, so a(5) must be 2 and a(6) = 1.
a(2) = 1, so the 3rd run has length 1, so a(7) = 2.
a(3) = 1, so the 4th run has length 1, so a(8) = 1.
a(4) = 1, so the 5th run has length 1, so a(9) = 2.
a(5) = 2, so the 6th run has length 2, so a(10) = 2 and a(11) = 1.
Globally, the runlength sequence of a is 4,1,1,1,1,2,1,2,1,2,2,1,...., and deleting the first term leaves a = A270641.

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A000002, A006928, A022300,

a = {1, 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 *)

A270642 The sequence a of 1's and 2's starting with (1,1,2,2) such that a(n) is the length of the (n+2)nd run of a.

Original entry on oeis.org

1, 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, 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

Offset: 1

Clark Kimberling, Apr 06 2016

See A270641 for a guide to related sequences.

			a(1) = 1, so the 3rd run has length 1, so a(5) must be 1.
a(2) = 1, so the 4th run has length 1, so a(6) = 2 and a(7) = 1.
a(3) = 2, so the 5th run has length 2, so a(8) = 1 and a(9) = 2.
a(4) = 2, so the 6th run has length 2, so a(10) = 2 and a(11) = 1.
Globally, the runlength sequence is 2,2,1,1,2,2,1,2,1,1,2,2,1,2,..., and deleting the first 2 terms leaves the same sequence.

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A000002, A006928, A022300, A270641.

a = {1, 1, 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 *)

A270643 The sequence a of 1's and 2's starting with (1,2,2,1) such that a(n) is the length of the (n+3)rd run of a.

Original entry on oeis.org

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, 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

Offset: 1

Clark Kimberling, Apr 06 2016

See A270641 for a guide to related sequences.

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

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A000002, A006928, A022300, A270641.

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

Conjecture: a(n) = A270642(n+1). - R. J. Mathar, Jun 21 2025

A270644 The sequence a of 1's and 2's starting with (1,2,2,2) such that a(n) is the length of the (n+2)nd run of a.

Original entry on oeis.org

1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 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, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1

Offset: 1

Clark Kimberling, Apr 06 2016

See A270641 for a guide to related sequences.

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

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A000002, A006928, A022300, A270641.

a = {1, 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 *)

cp's OEIS Frontend

A025512 Index of n-th 2 in A022300.

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A025513 Exactly half of first a(n) terms of A022300 are 1's (not known to be infinite).

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A022301 Index of n-th 1 in A022300.

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A022302 Least k such that first k terms of A022300 contain n more 1's than 2's.

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A025515 Least k such that first k terms of A022300 contain n more 2's than 1's.

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A022303 The sequence a of 1's and 2's starting with (1,2,1) such that a(n) is the length of the (n+2)nd run of a.

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A270641 The sequence a of 1's and 2's starting with (1,1,1,1) such that a(n) is the length of the (n+1)st run of a.

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A270642 The sequence a of 1's and 2's starting with (1,1,2,2) such that a(n) is the length of the (n+2)nd run of a.

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A270643 The sequence a of 1's and 2's starting with (1,2,2,1) such that a(n) is the length of the (n+3)rd run of a.

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A270644 The sequence a of 1's and 2's starting with (1,2,2,2) such that a(n) is the length of the (n+2)nd run of a.

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