A135992 -id:A135992 - cp's OEIS frontend

A135994 First differences of A135992.

0, 2, -1, 6, -3, 16, -8, 42, -21, 110, -55, 288, -144, 754, -377, 1974, -987, 5168, -2584, 13530, -6765, 35422, -17711, 92736, -46368, 242786, -121393, 635622, -317811, 1664080, -832040, 4356618, -2178309, 11405774, -5702887, 29860704, -14930352, 78176338

Offset: 0

Paul Curtz, Mar 03 2008

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

Cf. A000045, A000032, A135992.

Differences[Flatten[{Last[#],First[#]}&/@Partition[Fibonacci[ Range[ 40]],2]]] (* or *) LinearRecurrence[{0,3,0,-1},{0,2,-1,6},40] (* Harvey P. Dale, Sep 16 2013 *)
Table[(LucasL[n] - (-1)^n Fibonacci[n + 3])/2, {n, 0, 40}] (* Vladimir Reshetnikov, Sep 24 2016 *)

a(n) = 3*a(n-2) - a(n-4) for n>3. G.f.: -x*(x-2) / ((x^2-x-1)*(x^2+x-1)). [Colin Barker, Feb 02 2013]
From Vladimir Reshetnikov, Sep 24 2016: (Start)
a(n) = Sum_{k=1..n} (-1)^(k+1) * Fibonacci(k) * Lucas(n-k).
a(n) = (Lucas(n) - (-1)^n * Fibonacci(n+3))/2, where Fibonacci(n) = A000045(n), Lucas(n) = A000032(n). (End)

More terms from Colin Barker, Feb 02 2013

A108362 Pair reversal of Fibonacci numbers.

Original entry on oeis.org

1, 0, 2, 1, 5, 3, 13, 8, 34, 21, 89, 55, 233, 144, 610, 377, 1597, 987, 4181, 2584, 10946, 6765, 28657, 17711, 75025, 46368, 196418, 121393, 514229, 317811, 1346269, 832040, 3524578, 2178309, 9227465, 5702887, 24157817, 14930352, 63245986, 39088169, 165580141

Offset: 0

Paul Barry, May 31 2005

Here Fibonacci numbers are swapped in pairs, beginning with the pair (F(0),F(1)) changed in (F(1),F(0)). Similar to A135992, which starts switching F(1) and F(2). - Giuseppe Coppoletta, Mar 04 2015

			a(6) = Fibonacci(7) = 13, a(7) = Fibonacci(6) = 8.

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

Cf. A000045, A001906, A135992.

a:= n-> (<<0|1>, <1|1>>^(n+(-1)^n))[1,2]:
seq(a(n), n=0..40);  # Alois P. Heinz, Sep 27 2023

Flatten[Reverse/@Partition[Fibonacci[Range[0,40]],2]] (* or *) LinearRecurrence[{0,3,0,-1},{1,0,2,1},40] (* Harvey P. Dale, Sep 09 2015 *)
Table[((-1)^n Fibonacci[n] + LucasL[n])/2, {n, 0, 40}] (* Vladimir Reshetnikov, Sep 24 2016 *)

Vec((1-x^2+x^3)/(1-3*x^2+x^4) + O(x^50)) \\ Michel Marcus, Mar 04 2015

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

G.f.: (1-x^2+x^3)/(1-3x^2+x^4).
a(n) = 3*a(n-2) - a(n-4) for n>3 with a(0)=1, a(1)=0, a(2)=2, a(3)=1.
a(n) = (sqrt(5)/2-1/2)^n * ((-1)^n/2-sqrt(5)/10)+(sqrt(5)/2+1/2)^n * (sqrt(5)*(-1)^n/10+1/2).
From Giuseppe Coppoletta, Mar 04 2015: (Start)
a(2n) = A000045(2n+1), a(2n+1) = A000045(2n).
a(2n) = a(2n-1) + 2*a(2n-2), a(2n+1) = (a(2n) + a(2n-1))/2. (End)
a(n) = ((-1)^n * Fibonacci(n) + Lucas(n))/2. - Vladimir Reshetnikov, Sep 24 2016

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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A135994 First differences of A135992.

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A108362 Pair reversal of Fibonacci numbers.

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

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