A186679 -id:A186679 - cp's OEIS frontend

A081714 a(n) = F(n)*L(n+1) where F=Fibonacci and L=Lucas numbers.

0, 3, 4, 14, 33, 90, 232, 611, 1596, 4182, 10945, 28658, 75024, 196419, 514228, 1346270, 3524577, 9227466, 24157816, 63245987, 165580140, 433494438, 1134903169, 2971215074, 7778742048, 20365011075, 53316291172, 139583862446, 365435296161, 956722026042

Offset: 0

Ralf Stephan, Apr 03 2003

Also convolution of Fibonacci and Lucas numbers.
For n>2, a(n) represents twice the area of the triangle created by the three points (L(n-3), L(n-2)), (L(n-1), L(n)) and (F(n+3), F(n+2)) where L(k)=A000032(k) and F(k)=A000045(k). - J. M. Bergot, May 20 2014
For n>1, a(n) is the remainder when F(n+3)*F(n+4) is divided by F(n+1)*F(n+2). - J. M. Bergot, May 24 2014

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

Cf. A000045, A000204.

List([0..30], n -> Fibonacci(n)*(Fibonacci(n+2)+Fibonacci(n))); # G. C. Greubel, Jan 07 2019

[Fibonacci(n)*Lucas(n+1): n in [0..30]]; // Vincenzo Librandi, Sep 08 2012

with(combinat): F:=n-> fibonacci(n): L:= n-> F(n+1)+F(n-1):
a:= n-> F(n)*L(n+1): seq(a(n), n=0..30);

Fibonacci[Range[0,50]]*LucasL[Range[0,50]+1] (* Vladimir Joseph Stephan Orlovsky, Mar 17 2011*)

my(x='x+O('x^51));for(n=0,50,print1(polcoeff(serconvol(Ser((1+2*x)/(1-x-x*x)),Ser(x/(1-x-x*x))),n)", "))

a(n)=fibonacci(n)*(fibonacci(n+2)+fibonacci(n))

a(n) = round((-(-1)^n+(2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/sqrt(5))) \\ Colin Barker, Sep 28 2016

[fibonacci(n)*(fibonacci(n+2)+fibonacci(n)) for n in (0..30)] # G. C. Greubel, Jan 07 2019

G.f.: x*(3-2*x)/((1+x)*(1-3*x+x^2)).
a(n) = A122367(n) - (-1)^n. - R. J. Mathar, Jul 23 2010
a(n) = (L(n+1)^2 - F(2*n+2))/2 = ( A001254(n+1) - A001906(n+1) )/2. - Gary Detlefs, Nov 28 2010
a(n+1) = - A186679(2*n+1). - Reinhard Zumkeller, Feb 25 2011
a(n) = A035513(1,n-1)*A035513(2,n-1). - R. J. Mathar, Sep 04 2016
a(n)+a(n+1) = A005248(n+1). - R. J. Mathar, Sep 04 2016
a(n) = (-(-1)^n+(2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n)) / sqrt(5)). - Colin Barker, Sep 28 2016

Simpler definition from Michael Somos, Mar 16 2004

A075193 Expansion of (1-2*x)/(1+x-x^2).

Original entry on oeis.org

1, -3, 4, -7, 11, -18, 29, -47, 76, -123, 199, -322, 521, -843, 1364, -2207, 3571, -5778, 9349, -15127, 24476, -39603, 64079, -103682, 167761, -271443, 439204, -710647, 1149851, -1860498, 3010349, -4870847, 7881196, -12752043, 20633239, -33385282, 54018521, -87403803, 141422324

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Sep 07 2002

"Inverted" Lucas numbers:

The g.f. is obtained inserting 1/x into the g.f. of Lucas sequence and dividing by x. The closed form is a(n)=(-1)^n*a^(n+1)+(-1)^n*b^(n+1), where a=golden ratio and b=1-a, so that a(n)=(-1)^n*L(n+1), L(n)=Lucas numbers.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (-1,1).

Cf. A000032, A000045, A001906, A039834, A186679.

a075193 n = a075193_list !! n
a075193_list = 1 : -3 : zipWith (-) a075193_list (tail a075193_list)
-- Reinhard Zumkeller, Sep 15 2015

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1-2*x)/(1+x-x^2))); // Marius A. Burtea, Jan 12 2020

a:= n-> (Matrix([[1, -2]]). Matrix([[-1, 1], [1, 0]])^(n))[1, 1]:
seq(a(n), n=0..45); # Alois P. Heinz, Jul 31 2008

CoefficientList[Series[(1 - 2z)/(1 + z - z^2), {z, 0, 40}], z]

a(n) = -a(n-1)+a(n-2), a(0)=1, a(1)=-3.
a(n) = term (1,1) in the 1x2 matrix [1,-2] * [-1,1; 1,0]^n. - Alois P. Heinz, Jul 31 2008
a(n) = A186679(n)+A186679(n-2) for n>1. - Reinhard Zumkeller, Feb 25 2011
a(n) = A039834(n+1)-2*A039834(n). - R. J. Mathar, Sep 27 2014
a(n) = (-1)^(n-1)*A001906(n)/A000045(n). - Taras Goy, Jan 12 2020
E.g.f.: exp(-(1+sqrt(5))*x/2)*(3 + sqrt(5) - 2*exp(sqrt(5)*x))/(1 + sqrt(5)). - Stefano Spezia, Jan 12 2020

A116697 a(n) = -a(n-1) - a(n-3) + a(n-4).

Original entry on oeis.org

1, 1, -2, 2, -2, 5, -9, 13, -20, 34, -56, 89, -143, 233, -378, 610, -986, 1597, -2585, 4181, -6764, 10946, -17712, 28657, -46367, 75025, -121394, 196418, -317810, 514229, -832041, 1346269, -2178308, 3524578, -5702888

Offset: 0

Creighton Dement, Feb 23 2006

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

Cf. A000045, A001519, A006498, A056594, A115008, A116698, A116699, A128535.

Cf. A186679 (first differences).

a116697 n = a116697_list !! n
a116697_list = [1,1,-2,2]
               ++ (zipWith (-) a116697_list
                               $ zipWith (+) (tail a116697_list)
                                             (drop 3 a116697_list))
a128535_list = 0 : (map negate $ map a116697 [0,2..])
a001519_list = 1 : map a116697 [1,3..]
a186679_list = zipWith (-) (tail a116697_list) a116697_list
a128533_list = map a186679 [0,2..]
a081714_list = 0 : (map negate $ map a186679 [1,3..])
a075193_list = 1 : -3 : (zipWith (+) a186679_list $ drop 2 a186679_list)
-- Reinhard Zumkeller, Feb 25 2011

A116697:= func< n | (-1)^Floor((n+1)/2)*(1+(-1)^n)/2 -(-1)^n*Fibonacci(n) >;
[A116697(n): n in [0..50]]; // G. C. Greubel, Jun 08 2025

LinearRecurrence[{-1,0,-1,1},{1,1,-2,2},40] (* Harvey P. Dale, Nov 04 2011 *)

def A116697(n): return (-1)^(n//2)*((n+1)%2) - (-1)^n*fibonacci(n)
print([A116697(n) for n in range(51)]) # G. C. Greubel, Jun 08 2025

G.f.: (1 + 2*x - x^2 + x^3)/((1 + x^2)*(1 + x - x^2)).
a(2*n+1) = A000045(2*n+1) = A001519(n).
a(2*n) = - A128535(n+1). - Reinhard Zumkeller, Feb 25 2011
a(n) = A056594(n) - (-1)^n*A000045(n). - Bruno Berselli, Feb 26 2011
E.g.f.: cos(x) + (2/sqrt(5))*exp(-x/2)*sinh(sqrt(5)*x/2). - G. C. Greubel, Jun 08 2025

A128533 a(n) = F(n)*L(n+2) where F=Fibonacci and L=Lucas numbers.

Original entry on oeis.org

0, 4, 7, 22, 54, 145, 376, 988, 2583, 6766, 17710, 46369, 121392, 317812, 832039, 2178310, 5702886, 14930353, 39088168, 102334156, 267914295, 701408734, 1836311902, 4807526977, 12586269024, 32951280100, 86267571271, 225851433718, 591286729878

Offset: 0

Axel Harvey, Mar 08 2007

Generally, F(n)*L(n+k) = F(2*n + k) + F(k)*(-1)^(n+1): if k = 0 then sequence is A001906, if k = 1 it is A081714.

			a(4) = 54 because F(4)*L(6) = 3*18.
G.f. = 4*x + 7*x^2 + 22*x^3 + 54*x^4 + 145*x^5 + 376*x^6 + 988*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001906, A081714, A128534, A128535.

List([0..30], n -> Fibonacci(2*(n+1)) + (-1)^(n+1)); # G. C. Greubel, Jan 07 2019

[Fibonacci(n)*Lucas(n+2): n in [0..30]]; // Vincenzo Librandi, Feb 20 2013

with(combinat); A128533:=n->fibonacci(2*n+2)+(-1)^(n+1); seq(A128533(k),k=0..50); # Wesley Ivan Hurt, Oct 19 2013

LinearRecurrence[{2,2,-1}, {0,4,7}, 40] (* Vincenzo Librandi, Feb 20 2013 *)
a[n_]:= Fibonacci[2n+2] -(-1)^n; (* Michael Somos, May 26 2014 *)

vector(30, n, n--; fibonacci(2*(n+1)) + (-1)^(n+1)) \\ G. C. Greubel, Jan 07 2019

[fibonacci(2*(n+1)) + (-1)^(n+1) for n in (0..30)] # G. C. Greubel, Jan 07 2019

a(n) = F(2*(n+1)) + (-1)^(n+1), assuming F(0) = 0 and L(0) = 2.
From R. J. Mathar, Apr 16 2009: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: x*(4-x)/((1+x)*(x^2-3*x+1)). (End)
a(n) = A186679(2*n). - Reinhard Zumkeller, Feb 25 2011
a(-n) = - A128535(n). - Michael Somos, May 26 2014
0 = a(n)*(+4*a(n) + a(n+1) - 17*a(n+2)) + a(n+1)*(-14*a(n+1) + a(n+2)) + a(n+2)*(+4*a(n+2)) for all n in Z. - Michael Somos, May 26 2014

More terms from Vincenzo Librandi, Feb 20 2013

cp's OEIS Frontend

A081714 a(n) = F(n)*L(n+1) where F=Fibonacci and L=Lucas numbers.

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A075193 Expansion of (1-2*x)/(1+x-x^2).

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A116697 a(n) = -a(n-1) - a(n-3) + a(n-4).

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A128533 a(n) = F(n)*L(n+2) where F=Fibonacci and L=Lucas numbers.

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GAP

Magma

Maple

Mathematica

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