A130196 -id:A130196 - cp's OEIS frontend

A306366 For any sequence s of positive integers without infinitely many consecutive equal terms, let T(s) be the sequence obtained by replacing each run, say of k consecutive t's, with a run of t consecutive k's; this sequence corresponds to T(K) (where K denotes the Kolakoski sequence A000002).

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

Offset: 1

Rémy Sigrist, Feb 10 2019

If s is finite, then s and T(s) have the same sum.
Fixed points of T correspond to sequences where each run, say of t's, has t elements; A001650, A001670, A002024, A130196, A167817, A175944 and A213083 are fixed points of T.
When s has no consecutive equal terms, then T(s) is all 1's (A000012).
Apparently, T^4(K) = T^2(K) (where T^i denotes the i-th iterate of K).

			The first terms of the Kolakoski sequence are:
       +-----+     +--+  +-----+  +-----+     +--
       |     |     |  |  |     |  |     |     |
    +--+     +-----+  +--+     +--+     +-----+
    |#1|#2   |#3   |#4|#5|#6   |#7|#8   |#9   |#10 ...
    +--+-----+-----+--+--+-----+--+-----+-----+--
      1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, ...
.
The first terms of this sequence are:
       +-----+--+        +-----+  +-----+--
       |     .  |        |     |  |     .
    +--+     .  +-----+--+     +--+     .
    |#1|#2   .#3|#4   .#5|#6   |#7|#8   .#9  ...
    +--+-----+--+-----+--+-----+--+-----+--
      1, 2, 2, 2, 1, 1, 1, 2, 2, 1, 2, 2, 2, ...

Rémy Sigrist, PARI program for A306366

Cf. A000002, A000012, A001650, A001670, A002024, A130196, A167817, A175944, A213083.

PARI
```
See Links section.
```

a(n) = A000002(ceiling(2*n/3)) (conjectured). - Jon Maiga, Jan 24 2021

A368148 Square array A(n, k), n, k > 0, read and filled in the greedy way by upwards antidiagonals such that A(n, k) corresponds to the size of the connected component (relative to the Von Neumann neighborhood) of terms equal to A(n, k) including the position (n, k).

Original entry on oeis.org

1, 2, 2, 2, 1, 2, 1, 3, 3, 1, 2, 2, 3, 2, 2, 2, 1, 2, 2, 1, 2, 1, 3, 3, 1, 3, 3, 1, 2, 2, 3, 2, 2, 3, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1

Offset: 1

Rémy Sigrist, Dec 13 2023

The array is doubly periodic (see first formula) and consists of the following repeating 3 X 3 pattern with two components of 1 1's, two components of 2 2's and one component of 3 3's:
    +---+-------+
    | 1 | 2   2 |
    +---+---+---+
    | 2 | 1 | 3 |
    |   +---+   |
    | 2 | 3   3 |
    +---+-------+

			Array A(n, k) begins:
  n\k | 1  2  3  4  5  6  7  8  9 10
  ----+-----------------------------
    1 | 1  2  2  1  2  2  1  2  2  1
    2 | 2  1  3  2  1  3  2  1  3  2
    3 | 2  3  3  2  3  3  2  3  3  2
    4 | 1  2  2  1  2  2  1  2  2  1
    5 | 2  1  3  2  1  3  2  1  3  2
    6 | 2  3  3  2  3  3  2  3  3  2
    7 | 1  2  2  1  2  2  1  2  2  1
    8 | 2  1  3  2  1  3  2  1  3  2
    9 | 2  3  3  2  3  3  2  3  3  2
   10 | 1  2  2  1  2  2  1  2  2  1
.
We can chose A(1, 1) = 1.
A(2, 1) cannot equal 1; we chose A(2, 1) = 2.
Likewise we chose A(1, 2).
A(2, 2) cannot equal 2 as this would imply a component with 3 or more 2's.
So, by necessity, we chose A(3, 1) = A(1, 3) = 2.
We chose A(2, 2) = 1.
We chose A(4, 1) = 1.
A(3, 2) cannot equal 1 or 2; we chose A(3, 2) = 3.
Likewise we chose A(2, 3) = 3.
We chose A(1, 4) = 1.
A(5, 1) cannot equal 1; we chose A(5, 1) = 2.
A(4, 2) cannot equal 1 (or 3); we chose A(4, 2) = 2.
By necessity, A(3, 3) = 3.
etc.

Cf. A130196 (one-dimensional variant).

A(n,k) = { [1,2,2; 2,1,3; 2,3,3][1+(n-1)%3, 1+(k-1)%3] }

A(n+3, k) = A(n, k+3) = A(n, k).

A(n, k) = A(k, n).

A177347 Decimal expansion of (5+sqrt(85))/10.

Original entry on oeis.org

1, 4, 2, 1, 9, 5, 4, 4, 4, 5, 7, 2, 9, 2, 8, 8, 7, 3, 1, 0, 0, 0, 2, 2, 7, 4, 2, 8, 1, 7, 6, 2, 7, 9, 3, 1, 5, 7, 2, 4, 6, 8, 0, 5, 0, 4, 8, 7, 2, 2, 4, 6, 4, 0, 0, 8, 0, 0, 7, 7, 5, 2, 2, 0, 5, 4, 4, 2, 6, 7, 1, 0, 2, 6, 8, 0, 1, 8, 7, 5, 4, 6, 0, 7, 6, 7, 8, 9, 4, 0, 9, 0, 7, 9, 3, 2, 8, 0, 5, 6, 4, 9, 4, 0, 3

Offset: 1

Klaus Brockhaus, May 07 2010

Continued fraction expansion of (5+sqrt(85))/10 is A130196.

			(5+sqrt(85))/10 = 1.42195444572928873100...

Cf. A010536 (decimal expansion of sqrt(85)), A130196 (repeat 1, 2, 2).

RealDigits[(5+Sqrt[85])/10,10,120][[1]] (* Harvey P. Dale, Dec 31 2011 *)

A267068 a(n) = (n+1) / A189733(n).

Original entry on oeis.org

1, 2, 3, 2, 5, 1, 7, 2, 3, 2, 11, 1, 13, 2, 3, 2, 17, 1, 19, 2, 3, 2, 23, 1, 25, 2, 3, 2, 29, 1, 31, 2, 3, 2, 35, 1, 37, 2, 3, 2, 41, 1, 43, 2, 3, 2, 47, 1, 49, 2, 3, 2, 53, 1, 55, 2, 3, 2, 59, 1, 61, 2, 3, 2, 65, 1, 67, 2, 3, 2

Offset: 0

Paul Curtz, Jan 10 2016

nonn

A189733(n) is the denominator of an autosequence of the first kind (the main diagonal is A000004).

Cf. A000004, A047238, A047245, A052901, A130196, A189733.

CoefficientList[Series[(1 + 2 x + 3 x^2 + 2 x^3 + 5 x^4 + x^5 + 5 x^6 - 2 x^7 - 3 x^8 - 2 x^9 + x^10 - x^11)/((1 - x)^2 (1 + x)^2 (1 - x + x^2)^2 (1 + x + x^2)^2), {x, 0, 69}], x] (* or *)
b[m_, n_] := b[m, n] = Which[m == n, 0, n == m + 1, (-1)^(n + 1)/n, n > m, b[m, n - 1] + b[m + 1, n - 1], n < m, b[m - 1, n + 1] - b[m - 1, n]]; Table[(n + 1)/Denominator@ b[0, n], {n, 0, 69}] (* Michael De Vlieger, Jan 15 2016, Jean-François Alcover at A189733 *)

a(2n+1) = A130196(n+1).
A052901(n+2) = period 3: 2, 3, 2  is at rank A047245(n+1) = 1, 2, 3, 7, 8, 9, ... .
Conjectures from Colin Barker, Jan 10 2016: (Start)
a(n) = 2*a(n-6) - a(n-12) for n>11.
G.f.: (1+2*x+3*x^2+2*x^3+5*x^4+x^5+5*x^6-2*x^7-3*x^8-2*x^9+x^10-x^11) / ((1-x)^2*(1+x)^2*(1-x+x^2)^2*(1+x+x^2)^2).
(End)
a(3n) + a(3n+1) + a(3n+2) = A047238(n+3).

A367705 Coefficients of expansion of (1 + 5x + 11x^2 + 5x^3 + 7x^4 + x^5)/(1 - x^3)^2 in powers of x.

Original entry on oeis.org

1, 5, 11, 7, 17, 23, 13, 29, 35, 19, 41, 47, 25, 53, 59, 31, 65, 71, 37, 77, 83, 43, 89, 95, 49, 101, 107, 55, 113, 119, 61, 125, 131, 67, 137, 143, 73, 149, 155, 79, 161, 167, 85, 173, 179, 91, 185, 191, 97, 197, 203, 103, 209, 215, 109, 221, 227, 115, 233, 239

Offset: 0

Philippe Deléham, Nov 27 2023

Based on an idea of Pierre CAMI.

Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A006369, A007310, A016921, A017581, A017653, A130196.

CoefficientList[Series[(1 + 5*x + 11*x^2 + 5*x^3 + 7*x^4 + x^5)/(1 - x^3)^2, {x, 0, 60}], x] (* or *)
LinearRecurrence[{0, 0, 2, 0, 0, -1}, {1, 5, 11, 7, 17, 23}, 60] (* Amiram Eldar, Nov 28 2023 *)

a(n) = 3*A006369(n) + A130196(n).
a(n) = A007310(A006369(n) + 1).
a(n) = 2*a(n-3) - a(n-6) for n >= 6.
a(3*n) = 6*n+1, a(3*n+1) = 12*n+5, a(3*n+2) = 12*n+11.
Sum_{n>=0} (-1)^n/a(n) = ((2+sqrt(2))*Pi + sqrt(3)*log(7+4*sqrt(3)) + sqrt(6)*log(5-2*sqrt(6)))/12. - Amiram Eldar, Nov 28 2023

cp's OEIS Frontend

A306366 For any sequence s of positive integers without infinitely many consecutive equal terms, let T(s) be the sequence obtained by replacing each run, say of k consecutive t's, with a run of t consecutive k's; this sequence corresponds to T(K) (where K denotes the Kolakoski sequence A000002).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A368148 Square array A(n, k), n, k > 0, read and filled in the greedy way by upwards antidiagonals such that A(n, k) corresponds to the size of the connected component (relative to the Von Neumann neighborhood) of terms equal to A(n, k) including the position (n, k).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula

A177347 Decimal expansion of (5+sqrt(85))/10.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A267068 a(n) = (n+1) / A189733(n).

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

Formula

A367705 Coefficients of expansion of (1 + 5*x + 11*x^2 + 5*x^3 + 7*x^4 + x^5)/(1 - x^3)^2 in powers of x.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A367705 Coefficients of expansion of (1 + 5x + 11x^2 + 5x^3 + 7x^4 + x^5)/(1 - x^3)^2 in powers of x.