A051792 -id:A051792 - cp's OEIS frontend

A053602 a(n) = a(n-1) - (-1)^n*a(n-2), a(0)=0, a(1)=1.

0, 1, 1, 2, 1, 3, 2, 5, 3, 8, 5, 13, 8, 21, 13, 34, 21, 55, 34, 89, 55, 144, 89, 233, 144, 377, 233, 610, 377, 987, 610, 1597, 987, 2584, 1597, 4181, 2584, 6765, 4181, 10946, 6765, 17711, 10946, 28657, 17711, 46368, 28657, 75025, 46368, 121393, 75025

Offset: 0

Michael Somos, Jan 17 2000

If b(0)=0, b(1)=1 and b(n) = b(n-1) + (-1)^n*b(n-2), then a(n) = b(n+3). - Jaume Oliver Lafont, Oct 03 2009
a(n) is the number of palindromic compositions of n-1 into parts of 1 and 2. a(7) = 5 because we have 2+2+2, 2+1+1+2, 1+2+2+1, 1+1+2+1+1, 1+1+1+1+1+1. - Geoffrey Critzer, Mar 17 2014
a(n) is the number of palindromic compositions of n into odd parts (the corresponding generating function follows easily from Theorem 1.2 of the Hoggatt et al. reference). Example: a(7) = 5 because we have 7, 1+5+1, 3+1+3, 1+1+3+1+1, 1+1+1+1+1+1+1. - Emeric Deutsch, Aug 16 2016
The ratio of a(n)/a(n-1) oscillates between phi-1 and phi+1 as n tends to infinity, where phi is golden ratio (A001622). - Waldemar Puszkarz, Oct 10 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Krithnaswami Alladi and V. E. Hoggatt, Jr. Compositions with Ones and Twos, Fibonacci Quarterly, 13 (1975), 233-239. - _Ron Knott_, Oct 29 2010
A. R. Ashrafi, J. Azarija, K. Fathalikhani, S. Klavzar, et al., Orbits of Fibonacci and Lucas cubes, dihedral transformations, and asymmetric strings, 2014.
V. E. Hoggatt, Jr., and Marjorie Bicknell, Palindromic compositions, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.
M. A. Nyblom, Counting Palindromic Binary Strings Without r-Runs of Ones, J. Int. Seq. 16 (2013) #13.8.7, P_2(n)
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1)

Cf. A000045, A001622, A051792, A079977.

I:=[0,1,1,2]; [n le 4 select I[n] else Self(n-2)+Self(n-4): n in [1..50]]; // Vincenzo Librandi Oct 10 2017

a[0] := 0: a[1] := 1: for n from 2 to 60 do a[n] := a[n-1]-(-1)^n*a[n-2] end do: seq(a[n], n = 0 .. 50); # Emeric Deutsch, Oct 09 2017

nn=50;CoefficientList[Series[x (1+x+x^2)/(1-x^2-x^4),{x,0,nn}],x] (* Geoffrey Critzer, Mar 17 2014 *)
LinearRecurrence[{0,1,0,1},{0,1,1,2},60] (* Harvey P. Dale, Nov 07 2016 *)
RecurrenceTable[{a[0]==0, a[1]==1, a[n]==a[n-1]-(-1)^n a[n-2]}, a, {n, 50}] (* Vincenzo Librandi, Oct 10 2017 *)

PARI
```
a(n)=fibonacci(n\2+n%2*2)
```

[fibonacci(n//2 + 2*(n%2)) for n in range(61)] # G. C. Greubel, Dec 06 2022

G.f.: x*(1 + x + x^2)/(1 - x^2 - x^4).
a(n) = a(n-2) + a(n-4).
a(2n) = F(n), a(2n-1) = F(n+1) where F() is Fibonacci sequence.
a(3-n) = A051792(n).
a(3)=1, a(4)=2, a(n+2) = a(n+1) + sign(a(n) - a(n+1))*a(n), n > 4. - Benoit Cloitre, Apr 08 2002
a(n) = A079977(n-1) + A079977(n-2) + A079977(n-3), n > 2. - Ralf Stephan, Apr 26 2003
a(0) = 0, a(1) = 1; a(2n) = a(2n-1) - a(2n-2); a(2n+1) = a(2n) + a(2n-1). - Amarnath Murthy, Jul 21 2005

A123231 Row sums of A123230.

Original entry on oeis.org

1, 2, 1, 3, 2, 5, 3, 8, 5, 13, 8, 21, 13, 34, 21, 55, 34, 89, 55, 144, 89, 233, 144, 377, 233, 610, 377, 987, 610, 1597, 987, 2584, 1597, 4181, 2584, 6765, 4181, 10946, 6765, 17711, 10946, 28657, 17711, 46368, 28657, 75025, 46368, 121393, 75025, 196418

Offset: 1

Roger L. Bagula, Oct 06 2006

nonn

All terms are Fibonacci numbers A000045: a(2n-1) = Fibonacci(n), a(2n) = Fibonacci(n+2), a(2n-1) = a(2n+2). - Alexander Adamchuk, Oct 08 2006

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1).

Cf. A051792, A000045, A028242, A030451.

a:=[1,2,1,3];; for n in [5..50] do a[n]:=a[n-2]+a[n-4]; od; a; # Muniru A Asiru, Oct 12 2018

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1 + 2*x + x^3)/(1 - x^2 - x^4))); // G. C. Greubel, Oct 12 2018

seq(coeff(series(-x*(1+2*x+x^3)/(x^4+x^2-1),x,n+1), x, n), n = 1 .. 50); # Muniru A Asiru, Oct 12 2018

p[0, x] = 1; p[1, x] = x + 1; p[k_, x_] := p[k, x] = x*p[k - 1, x] + (-1)^(n + 1)p[k - 2, x]; Table[Sum[CoefficientList[p[n, x], x][[m]], {m, 1, n + 1}], {n, 0, 20}]
Rest[Flatten[Reverse/@Partition[Fibonacci[Range[30]],2,1]]] (* Harvey P. Dale, Mar 19 2013 *)

vector(50, n, fibonacci(3/4 -(-1)^(n+1)*3/4 +(n+1)/2)) \\ G. C. Greubel, Oct 12 2018

From Alexander Adamchuk, Oct 08 2006: (Start)
a(n) = Fibonacci(A028242(n+2)).
a(n) = Fibonacci(A030451(n+1)).
a(n) = Fibonacci(3/4 -(-1)^(n+1)*3/4 +(n+1)/2). (End)
a(n) = A053602(n+1) = A097594(n-5). - R. J. Mathar, Mar 08 2011
G.f. -x*(1+2*x+x^3) / ( -1+x^2+x^4 ). - R. J. Mathar, Mar 08 2011
a(n) = a(n-2) + a(n-4). - Muniru A Asiru, Oct 12 2018

More terms from Alexander Adamchuk, Oct 08 2006

A097594 a(n) = (a(n-1) mod a(n-2)) + a(n-2), a(0) = 3, a(1) = 2.

Original entry on oeis.org

2, 5, 3, 8, 5, 13, 8, 21, 13, 34, 21, 55, 34, 89, 55, 144, 89, 233, 144, 377, 233, 610, 377, 987, 610, 1597, 987, 2584, 1597, 4181, 2584, 6765, 4181, 10946, 6765, 17711, 10946, 28657, 17711, 46368, 28657, 75025, 46368, 121393, 75025, 196418, 121393, 317811, 196418, 514229, 317811, 832040, 514229

Offset: 0

Gerald McGarvey, Aug 29 2004

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1).

Cf. A000045, A051792, A053602.

[Fibonacci(3 +Floor(n/2) +2*(n mod 2)): n in [0..60]]; // G. C. Greubel, Dec 06 2022

LinearRecurrence[{0,1,0,1}, {2,5,3,8}, 60] (* G. C. Greubel, Dec 06 2022 *)

[fibonacci(3 +(n//2) + 2*(n%2)) for n in range(61)] # G. C. Greubel, Dec 06 2022

a(2n) = Fibonacci(n+4), a(2n+1) = Fibonacci(n+3).
a(n) = A053602(n+6).
a(n) = abs( A051792(n+11) ).
G.f.: (2 + 5*x + x^2 + 3*x^3)/(1 - x^2 - x^4). - G. C. Greubel, Dec 06 2022

A032090 "BHK" (reversible, identity, unlabeled) transform of 0,1,1,1...

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 4, 5, 10, 14, 26, 39, 69, 107, 183, 289, 484, 772, 1276, 2047, 3356, 5402, 8812, 14213, 23113, 37325, 60581, 97905, 158718, 256622, 415716, 672337, 1088662, 1760998, 2850646, 4611643, 7463885, 12075527

Offset: 1

Christian G. Bower

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
C. G. Bower, Transforms (2)
Index entries for linear recurrences with constant coefficients, signature (2,1,-3,1,-1,0,1).

For n>2, a(n) = A032089(n-1) + [n even], a(2n) = A032097(n-1).

CoefficientList[Series[- x (x^6 + x^5 - x^4 + 2 x^3 - 2 x^2 - x + 1)/((x - 1) (x^2 + x - 1) (x^4 + x^2 - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 19 2013 *)
LinearRecurrence[{2,1,-3,1,-1,0,1},{0,1,1,1,2,2,4,5},40] (* Harvey P. Dale, Mar 31 2019 *)

G.f.: -x^2*(x^6+x^5-x^4+2*x^3-2*x^2-x+1) / ((x-1)*(x^2+x-1)*(x^4+x^2-1)). [Colin Barker, Dec 07 2012]

2*a(n) = 2+A000045(n-1) - |A051792(n+5)|, n>1. - R. J. Mathar, Mar 24 2023

A345347 Find the largest k with F(k) <= n, where F(k) is the k-th Fibonacci number. a(n) = F(k+2) + n.

Original entry on oeis.org

1, 4, 7, 11, 12, 18, 19, 20, 29, 30, 31, 32, 33, 47, 48, 49, 50, 51, 52, 53, 54, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 199, 200, 201, 202, 203, 204

Offset: 0

Peter Munn, Jun 14 2021

The terms consist of 1 together with numbers that appear in row m of the Wythoff array (A035513) if m is in the sequence.
a(0) = 1, otherwise a(n) is the number whose Zeckendorf representation is "10" followed by the Zeckendorf representation of n.
If we define an extended Zeckendorf representation to be the Zeckendorf representation with "01" appended, then the numbers in the sequence are exactly those whose extended representation starts 101... . This extended representation is a valid Fibonacci base representation if we specify the rightmost digit to have weight F(0) = 0.
Equivalently, for positive integer m, find the largest k with F(k) <= m, where F(k) is the k-th Fibonacci number. m is in the sequence if and only if m >= F(k) + F(k-2).
Numbers given to rabbits on Rabbit 1's branch of the generation tree described in the A035513 examples.
Equivalently, take the positive integers in turn, placing runs of them alternatively into 2 sets, with run lengths from A053602/A051792 (self-interleaved Fibonacci sequence) as follows:
   set A: 1   0   1   1   2   3   5 ...
   set B:   1   1   2   3   5   8 ...
The sequence lists the numbers in set A.

			The initial Fibonacci numbers are F(0)..F(5) = 0, 1, 1, 2, 3, 5.
For n = 0, the largest k with F(k) <= 0 is k = 0, so F(k+2) = F(2) = 1, so a(0) = 1 + 0 = 1.
For n = 1, the largest k with F(k) <= 1 is k = 2, so F(k+2) = F(4) = 3, so a(1) = 3 + 1 = 4.
For n = 4, the largest k with F(k) <= 4 is k = 4, so F(k+2) = F(6) = 8, so a(4) = 8 + 4 = 12.
In the paragraph that follows we use the Wythoff array-based definition from the start of the comments.
Every positive integer appears once (only) in the Wythoff array. 0 is not positive, so does not appear in the array, so is not in the sequence. 1 is in the sequence by definition. 2 appears in Wythoff row 0, and 0 is not in the sequence, so 2 is not in the sequence. 4 appears in Wythoff row 1, and 1 is in the sequence, so 4 is in the sequence.

Encyclopedia of Mathematics, Zeckendorf representation.
OEIS Wiki, Zeckendorf representation.
N. J. A. Sloane, Classic Sequences.

Cf. A000045, A035513, A051792, A053602, A108852.

Appears to be column 1 of A194030.

kmax=12;Flatten[Table[Range[Fibonacci[k]+Fibonacci[k-2],Fibonacci[k+1]-1],{k,2,kmax}]] (* Paolo Xausa, Jan 02 2022 *)
A108852[n_]:=1+Floor[Log[GoldenRatio,1+n*Sqrt[5]]];
nterms=100;Table[n+Fibonacci[1+A108852[n]],{n,0,nterms-1}](* Paolo Xausa, Jan 02 2022 *)

a(n) = my(k=0); while(fibonacci(k)<=n, k=k+1); n+fibonacci(k+1)

a(n) = A000045(A108852(n)+1) + n.

Union_{k >= 2} {m : F(k)+F(k-2) <= m < F(k+1)}, where F(k) = A000045(k).

cp's OEIS Frontend

A053602 a(n) = a(n-1) - (-1)^n*a(n-2), a(0)=0, a(1)=1.

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A123231 Row sums of A123230.

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A097594 a(n) = (a(n-1) mod a(n-2)) + a(n-2), a(0) = 3, a(1) = 2.

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A032090 "BHK" (reversible, identity, unlabeled) transform of 0,1,1,1...

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A345347 Find the largest k with F(k) <= n, where F(k) is the k-th Fibonacci number. a(n) = F(k+2) + n.

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