A108362 -id:A108362 - cp's OEIS frontend

A135992 Positive Fibonacci numbers swapped in pairs.

1, 1, 3, 2, 8, 5, 21, 13, 55, 34, 144, 89, 377, 233, 987, 610, 2584, 1597, 6765, 4181, 17711, 10946, 46368, 28657, 121393, 75025, 317811, 196418, 832040, 514229, 2178309, 1346269, 5702887, 3524578, 14930352, 9227465, 39088169, 24157817

Offset: 1

Paul Curtz, Mar 03 2008

Analogous to A108362. It could be natural to define here too a(0) = 1 (swapping Fibonacci numbers from A212804). - Giuseppe Coppoletta, Mar 04 2015

			a(7) = Fibonacci(8) = 21, a(8) = Fibonacci(7) = 13.

Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1).

Cf. A000045, A001906, A108362, A212804.

a[1]:=1: a[2]:=1: for n from 2 to 20 do a[2*n-1]:=a[2*n-2]+2*a[2*n-3]: a[2*n]:=a[2*n-1]-a[2*n-3] end do: seq(a[n],n=1..40); # Emeric Deutsch, Mar 22 2008

Flatten[{Last[#],First[#]}&/@Partition[Fibonacci[Range[40]],2]] (* Harvey P. Dale, Sep 16 2013 *)
Table[(LucasL[n] - (-1)^n Fibonacci[n])/2, {n, 40}] (* Vladimir Reshetnikov, Sep 24 2016 *)

[fibonacci(n-(-1)^n) for n in range (1,39)] # Giuseppe Coppoletta, Mar 04 2015

From Emeric Deutsch, Mar 22 2008: (Start)
a(2n-1) = Fibonacci(2n), a(2n) = Fibonacci(2n-1).
a(2n-1) = a(2n-2) + 2*a(2n-3), a(2n) = a(2n-1) - a(2n-3), a(1)=a(2)=1. (End)
G.f.: (x*(1+x-x^3)) / ((x^2+x-1)*(x^2-x-1)). - R. J. Mathar, Mar 08 2011
a(n) = (Lucas(n) - (-1)^n * Fibonacci(n))/2. - Vladimir Reshetnikov, Sep 24 2016

More terms from Emeric Deutsch, Mar 22 2008

A138123 Antidiagonal sums of a triangle of coefficients of recurrences of the Fibonacci sequence.

Original entry on oeis.org

1, 1, 3, 0, 3, 0, 7, 1, 11, 0, 17, 0, 29, 1, 47, 0, 75, 0, 123, 1, 199, 0, 321, 0, 521, 1, 843, 0, 1363, 0, 2207, 1, 3571, 0, 5777, 0, 9349, 1, 15127, 0, 24475, 0, 39603, 1, 64079, 0, 103681, 0, 167761, 1, 271443, 0, 439203, 0, 710647, 1, 1149851, 0, 1860497, 0

Offset: 1

Paul Curtz, May 04 2008

nonn

Consider the irregular sparse triangle T(p,p) = A000204(p), T(p,2p)= -A033999(p)=(-1)^(p+1), T(p,m) =0 else; 1<=m<=2p, p>=1. Then a(n)=sum_{m=1..[2(n+1)/3]} T(1+n-m,m).
The T are coefficients in recurrences f(n)=sum_{m=1..2p} T(p,m)*f(n-m).
The recurrence for p=1, f(n)=f(n-1)+f(n-2), is satisfied by the Fibonacci sequence A000045. The recurrence for p=2, f(n)=3f(n-2)-f(n-4), is satisfied by A005013, A005247, A075091, A075270, A108362 and A135992.
Conjecture: The Fibonacci sequence F obeys all the recurrences: A000045(n)=F(n)= L(p)*F(n-p)-(-1)^p*F(n-2p), any p>0, L=A000204.
[Proof: conjecture is equivalent to the existence of a g.f. of F with denominator 1-L(p)x^p+(-1)^p*x^(2p). Since 1-x-x^2 is known to be a denominator of such a g.f. of A000045, the conjecture is that 1-L(p)*x^p+(-1)^p*x^(2p) can be reduced to 1-x-x^2. One finds: {1-L(p)*x^p+(-1)^p*x^(2p)}/(1-x-x^2) = sum{n=0..p-1}F(n+1)x^n-sum{n=0..p-2} (-1)^(n+p)F(n+1)x^(2p-n-2) is a polynomial with integer coefficients, which is proved by multiplication with 1-x-x^2 and via F(n)+F(n+1)=F(n+2) and L(n)=F(n-1)+F(n+1). - R. J. Mathar, Jul 10 2008].
Conjecture: The Lucas sequence L also obeys all the recurrences: L(n)= L(p)*L(n-p)-(-1)^p*L(n-2p), any p>0, L=A000204.

			The triangle T(p,m) with Lucas numbers on the diagonal starts
  1, 1;
  0, 3, 0,-1;
  0, 0, 4, 0, 0, 1;
  0, 0, 0, 7, 0, 0, 0,-1;
  0, 0, 0, 0,11, 0, 0, 0, 0, 1;
The antidiagonal sums are a(1)=1. a(2)=0+1=1. a(3)=0+3=3. a(4)=0+0+0=0. a(5)=0+0+4-1=3.

Row sums: Sum_{m=1..2p} T(p,m) = A098600(p).
Conjectures from Chai Wah Wu, Apr 15 2024: (Start)
a(n) = a(n-2) - a(n-3) + a(n-4) + a(n-5) + a(n-7) for n > 7.
G.f.: x*(-x^5 - 2*x^2 - x - 1)/((x + 1)*(x^2 - x + 1)*(x^4 + x^2 - 1)). (End)

Edited and extended by R. J. Mathar, Jul 10 2008

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A135992 Positive Fibonacci numbers swapped in pairs.

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