A167817 -id:A167817 - cp's OEIS frontend

A167816 Numerator of x(n) = x(n-1) + x(n-2), x(0)=0, x(1)=1/3; denominator=A167817.

0, 1, 1, 2, 1, 5, 8, 13, 7, 34, 55, 89, 48, 233, 377, 610, 329, 1597, 2584, 4181, 2255, 10946, 17711, 28657, 15456, 75025, 121393, 196418, 105937, 514229, 832040, 1346269, 726103, 3524578, 5702887, 9227465, 4976784, 24157817, 39088169, 63245986, 34111385

Offset: 0

Reinhard Zumkeller, Nov 13 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Fibonacci number
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 7, 0, 0, 0, -1).

Cf. A000045, A167808.

[0,1,1] cat [Numerator(Fibonacci(n)/Fibonacci(2*n-4)): n in [3..40]]; // Vincenzo Librandi, Jun 28 2016

Numerator[LinearRecurrence[{1,1},{0,1/3},40]] (* Harvey P. Dale, Dec 07 2014 *)
LinearRecurrence[{0, 0, 0, 7, 0, 0, 0, -1},{0, 1, 1, 2, 1, 5, 8, 13},39] (* Ray Chandler, Aug 03 2015 *)

a(n) = (a(n-1)*A093148(n+2) + a(n-2)*A093148(n+1))/A093148(n-1) for n>1.
a(4*n) = A004187(n) = (a(4*n-1) + a(4*n-2))/3;
a(4*n+1) = A033889(n) = 3*a(4*n-1) + a(4*n-2);
a(4*n+2) = A033890(n) = a(4*n-1) + 3*a(4*n-2);
a(4*n+3) = A033891(n) = a(4*n-1) + a(4*n-2).
Numerator of Fibonacci(n) / Fibonacci(2n-4) for n>=3. - Gary Detlefs, Dec 20 2010

Definition corrected by D. S. McNeil, May 09 2010

A177344 Decimal expansion of (33+sqrt(2805))/66.

Original entry on oeis.org

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

Offset: 1

Klaus Brockhaus, May 06 2010

Continued fraction expansion of (33+sqrt(2805))/66 is A167817.

			(33+sqrt(2805))/66 = 1.30245834405244608895...

Cf. A177345 (decimal expansion of sqrt(2805)), A167817 (repeat 1, 3, 3, 3).

RealDigits[(33+Sqrt[2805])/66,10,120][[1]] (* Harvey P. Dale, Mar 29 2025 *)

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).

Original entry on oeis.org

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

cp's OEIS Frontend

A167816 Numerator of x(n) = x(n-1) + x(n-2), x(0)=0, x(1)=1/3; denominator=A167817.

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A177344 Decimal expansion of (33+sqrt(2805))/66.

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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).

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