A135992 -id:A135992

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A135994

First differences of A135992.

+20
0

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

LINKS

Table of n, a(n) for n=0..37.

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

FORMULA

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)

MATHEMATICA

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 *)

CROSSREFS

Cf. A000045, A000032, A135992.

KEYWORD

sign,easy

AUTHOR

Paul Curtz, Mar 03 2008

EXTENSIONS

More terms from Colin Barker, Feb 02 2013

STATUS

approved

A108362

Pair reversal of Fibonacci numbers.

+10
2

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

COMMENTS

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

LINKS

Table of n, a(n) for n=0..40.

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

FORMULA

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

EXAMPLE

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

MAPLE

a:= n-> (<<0|1>, <1|1>>^(n+(-1)^n))[1, 2]:

seq(a(n), n=0..40); # Alois P. Heinz, Sep 27 2023

MATHEMATICA

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 *)

PROG

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

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

CROSSREFS

Cf. A000045, A001906, A135992.

KEYWORD

easy,nonn

AUTHOR

Paul Barry, May 31 2005

STATUS

approved

A138123

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

+10
2

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,3

COMMENTS

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.

LINKS

Table of n, a(n) for n=1..60.

FORMULA

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)

EXAMPLE

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.

KEYWORD

nonn

AUTHOR

Paul Curtz, May 04 2008

EXTENSIONS

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

STATUS

approved

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