A059929 -id:A059929 - cp's OEIS frontend

A121801 Expansion of 2x^2(3-x)/((1+x)(1-3x+x^2)).

0, 6, 10, 32, 78, 210, 544, 1430, 3738, 9792, 25630, 67106, 175680, 459942, 1204138, 3152480, 8253294, 21607410, 56568928, 148099382, 387729210, 1015088256, 2657535550, 6957518402, 18215019648, 47687540550, 124847601994

Offset: 1

Roger L. Bagula and Gary W. Adamson, Aug 27 2006

a(n) = the area of an irregular quadrilateral with vertices at the points (L(n),L(n+2)), (F(n+2),F(n+3)), (F(n+3),F(n+2)) and (L(n+2),L(n)), with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, Jun 16 2014

a(n+1) appears also as the fourth component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, where F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-A248161(n), A079472(n+1), A059929(n), a(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given. - Wolfdieter Lang, Nov 01 2014

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

List([1..30], n-> 2*(Lucas(1,-1,2*n+1)[2] +4*(-1)^n)/5 ); # G. C. Greubel, Jul 22 2019

I:=[0,6,10]; [n le 3 select I[n] else 2*Self(n-1)+2*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Nov 02 2014

[2*(Lucas(2*n+1) +4*(-1)^n)/5: n in [1..30]]; // G. C. Greubel, Jul 22 2019

c[i_, k_] := Floor[Mod[i/2^k, 2]] b[i_, k_] := If[c[i, k] == 0 && c[ i, k + 1] == 0, 0, If[c[i, k] == 1 && c[i, k + 1] == 1, 0, 1]] n = 4 - 1; M = Table[If[Sum[b[i, k]*b[j, k], {k, 0, n}] == 0, 1, 0], {j, 0, n}, {i, 0, n}] v[1] = {0, 1, 2, 3} v[n_] := v[n] = M.v[n - 1] a = Table[Floor[v[n][[1]]], {n, 1, 50}] Det[M - x*IdentityMatrix[4]] Factor[%] aaa = Table[x /. NSolve[Det[M - x*IdentityMatrix[4]] == 0, x][[n]], {n, 1, 4}] Abs[aaa] a1 = Table[N[a[[n]]/a[[n - 1]]], {n, 7, 50}]
CoefficientList[Series[2*x*(3-x)/((1+x)*(1-3*x+x^2)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Jun 16 2014 *)
LinearRecurrence[{2,2,-1},{0,6,10},30] (* Harvey P. Dale, Jan 06 2015 *)
With[{F=Fibonacci}, Table[2*(F[n]*F[n+1] +(-1)^n), {n,30}]] (* G. C. Greubel, Jul 22 2019 *)

concat(0,Vec(2*(3-x)/((1+x)*(1-3*x+x^2))+O(x^30))) \\ Charles R Greathouse IV, Sep 25 2012

vector(30, n, f=fibonacci; 2*(f(n)*f(n+1)+(-1)^n) ) \\ G. C. Greubel, Jul 22 2019

[2*(lucas_number2(2*n+1,1,-1) +4*(-1)^n)/5 for n in (1..30)] # G. C. Greubel, Jul 22 2019

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = -2*A121646(n+1).
G.f.: 2*x^2*(3-x)/((1+x)*(x^2-3*x+1)) (see name).
From Wolfdieter Lang, Nov 01 2014: (Start)
G.f.: (-10 + 8/(1+x) + 2*(1+x)/(1-3*x+x^2))/5 (partial fraction decomposition).
a(n) = (8*(-1)^n + 2*(F(2*(n+1)) + F(2*n)))/5 for n >= 1. a(0) = 0.
(End)
a(n) = 2*(Fibonacci(n)*Fibonacci(n+1) + (-1)^n). - G. C. Greubel, Jul 22 2019

Edited by the Associate Editors of the OEIS, Aug 18 2009

A236428 a(n) = F(n+1)^2 + F(n+1)*F(n) - F(n)^2, where F = A000045.

Original entry on oeis.org

1, 1, 5, 11, 31, 79, 209, 545, 1429, 3739, 9791, 25631, 67105, 175681, 459941, 1204139, 3152479, 8253295, 21607409, 56568929, 148099381, 387729211, 1015088255, 2657535551, 6957518401, 18215019649, 47687540549, 124847601995, 326855265439, 855718194319

Offset: 0

Richard R. Forberg, Jan 25 2014

(a(n) + a(n+1))/2 = F(2n+2).
a(n) = -a(-n-1), using the negative Fibonacci values.
First differences equal 2*A059929.
Partial sums equal A192873.
Unlike Fibonacci, the divisibility of a(n) by the primes is quite limited, specifically to p = 5, 11, 19, 31, 59, 71, 79, 109, ... where those after 5 are only a subset of primes congruent to {1,4} mod 5.
Values of a(n) mod p, for all primes p exhibit repeating pattern cycles of length k = (p-1)/m or (p+1)/m (except p = 5), based on whether p is congruent to {1,4} mod 5 or {2,3} mod 5. For p = 5, k = 2p = 10. Only the slightest similarity exists here with Fibonacci: there are formulas like this for a cycle length k, but for Fibonacci those are "divisibility cycles" for prime p, not the "pattern cycles" on mod p, and the m values differ for many primes, creating different cycle lengths for the same p.
a(n) has the property: a(k/2 + i) mod p + a(k/2 - 1 - i) mod p = p or 0, for all primes p, and all i 0 <= i <= k/2, in every cycle of length k. Thus, when plotted, the lower and upper halves of a every cycle have an inverted (i.e., flipped) symmetry.
For some primes (e.g., 13, 17, 37, 53, 61, 89, 97) each half-cycle (of length k/2) is internally symmetric (i.e., the second quarter-cycle is a mirror image of the first quarter cycle, and the fourth is a mirror image of the third, on each side of some value at k/4), while the flipped symmetry still holds for the upper and lower halves. See example for p = 61, with k = 30 in pdf file below.
No such symmetries on mod p, of either type, exist for Fibonacci.
a(n) is also (apart from sign) the determinant of a 2 X 2 matrix of squares of successive Fibonacci numbers: a(n) = (-1)^(n)*(F(n+2)^2*F(n-1)^2 -F(n)^2*F(n+1)^2). - R. M. Welukar, Aug 30 2014
For n>1 a(n) is the ceiling of the maximum area of a quadrilateral having sides of length in increasing order F(n), F(n+1), L(n), and L(n+1) with L(n)=A000032(n). - J. M. Bergot, Jan 19 2016
For n>1 a(n) is the numerator of the continued fraction [1, 1, ... 1, 2, 1, 1, ... 1, 2] with n-2 1's before each 2. - Greg Dresden and Kevin Zhanming Zheng, Aug 16 2020
a(n) is the number of edge covers in the rocket graph R_{3,n+1,n}. A rocket graph R_{m,i,j} is a m-cycle with two paths attached to adjacent vertices of the cycle, which have lengths i and j respectively. This is similar to a tadpole graph but with two tails. - Bridget Rozema, Oct 09 2024

Colin Barker, Table of n, a(n) for n = 0..1000
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Richard R. Forberg, Plot of a(n) mod 61
Bridget Rozema and Maisie Smith, Edge Covers of Unions of Path and Cycle Graphs, 2024.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A000045, A059929, A192873.

Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.

[Fibonacci(n+1)^2+Fibonacci(n+1)*Fibonacci(n)- Fibonacci(n)^2: n in [0..30]]; // Vincenzo Librandi, Jan 20 2016

F:=Fibonacci; [F(n+1)^2+F(n)*F(n-1): n in [0..30]]; // Bruno Berselli, Feb 15 2017

a[n_] := Fibonacci[n+1]^2 + Fibonacci[n+1]*Fibonacci[n] - Fibonacci[n]^2; Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Feb 27 2014 *)
LinearRecurrence[{2, 2, -1}, {1, 1, 5}, 40] (* Vincenzo Librandi, Jan 20 2016 *)

F=fibonacci;
a(n)=F(n+1)^2 + F(n+1)*F(n) - F(n)^2;
vector(33,n,a(n-1)) \\ Joerg Arndt, Feb 23 2014

Vec((x^2-x+1)/((x+1)*(x^2-3*x+1)) + O(x^100)) \\ Colin Barker, Dec 20 2014

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

a(n) = A001654(n) + A226205(n+1).
G.f.: (x^2 - x + 1)/((x + 1)*(x^2 - 3*x + 1)). - Joerg Arndt, Feb 23 2014
a(n) = (2*Lucas(2*n+1) + 3*(-1)^n)/5. - Ralf Stephan, Feb 27 2014
a(n) = 2*a(n-1) + 2*a(n-2)-a(n-3). - Colin Barker, Dec 20 2014
a(n) = F(n-1)*F(n+2) + F(n)*F(n+1). - J. M. Bergot, Dec 20 2014
a(n) = 2*F(n)*F(n+1) + (-1)^n. - Bruno Berselli, Oct 30 2015
a(n) = F(2*n+1) - F(n-1)^2 +(-1)^n for n>0. - J. M. Bergot, Jan 19 2016
a(n) = (2^(-n)*(3*(-2)^n-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 28 2016
a(n) = F(n+1)^2 + F(n)*F(n-1). See also A099016, tenth formula. - Bruno Berselli, Feb 15 2017
2*a(n) = L(n)*L(n+1) - F(n)*F(n+1), where L = A000032. - Bruno Berselli, Sep 27 2017

A248161 Expansion of (2-x+x^2)/((1+x)(1-3x+x^2)).

Original entry on oeis.org

2, 3, 11, 26, 71, 183, 482, 1259, 3299, 8634, 22607, 59183, 154946, 405651, 1062011, 2780378, 7279127, 19056999, 49891874, 130618619, 341963987, 895273338, 2343856031, 6136294751, 16065028226, 42058789923, 110111341547

Offset: 0

Wolfdieter Lang, Nov 01 2014

The negative of this sequence provides the first component of the square of [F(n), F(n+1), F(n+2), F(n+3)], for n >= 0, where F(n) = A000045(n), in the Clifford algebra Cl_2 over Euclidean 2-space. The whole quartet of sequences for this square is [-a(n), A079472(n+1), A059929(n), A121801(n+1)]. See the Oct 15 2014 comment in A147973 where also a reference is given.

G. C. Greubel, Table of n, a(n) for n = 0..1000
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.

Cf. A000045, A059929, A079472, A121801, A248156.

[-(Fibonacci(n)^2 +Fibonacci(n+1)^2 + Fibonacci(n+2)^2 - Fibonacci(n+3)^2): n in [0..30]]; // Vincenzo Librandi, Nov 01 2014

CoefficientList[Series[(2 - x + x^2)/((1 + x) (1 - 3 x + x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Nov 01 2014 *)
With[{F=Fibonacci}, Table[F[2*n+2] +F[n]*F[n+1] +(-1)^n, {n,0,40}]] (* G. C. Greubel, May 30 2025 *)

def A248161(n): return fibonacci(2*n+2) +fibonacci(n)*fibonacci(n+1) +(-1)^n
print([A248161(n) for n in range(41)]) # G. C. Greubel, May 30 2025

a(n) = F(n+3)^2 - (F(n)^2 + F(n+1)^2 + F(n+2)^2), F(n) = A000045(n).
a(n) = (6*F(2*n+2) + F(2*n) + 4*(-1)^n)/5, with the Fibonacci numbers F = A000045.
O.g.f.: (2-x+x^2)/((1+x)*(1-3*x+x^2)) = (4/(1+x) + (x+6)/(1-3*x+x^2))/5.
From G. C. Greubel, May 30 2025: (Start)
a(n) = Fibonacci(2*n+2) + Fibonacci(n)*Fibonacci(n+1) + (-1)^n.
E.g.f.: (1/5)*(exp(3*x/2)*(6*cosh(sqrt(5)*x/2) + 4*sqrt(5)*sinh(sqrt(5)*x/2)) + 4*exp(-x)). (End)

Typo in formula fixed by Vincenzo Librandi, Nov 01 2014

A292612 a(n) = F(n)^2 + 4(-1)^n = F(n+3)F(n-3), where F = A000045.

Original entry on oeis.org

4, -3, 5, 0, 13, 21, 68, 165, 445, 1152, 3029, 7917, 20740, 54285, 142133, 372096, 974173, 2550405, 6677060, 17480757, 45765229, 119814912, 313679525, 821223645, 2149991428, 5628750621, 14736260453, 38580030720, 101003831725, 264431464437, 692290561604, 1812440220357

Offset: 0

Bruno Berselli, Sep 20 2017

This is the case k=3 of the identity F(n)^2 - F(k)^2*(-1)^(n+k) = F(n+k)*F(n-k), known also as Catalan's identity.

Colin Barker, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Catalan's Identity.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A001622, A005248: Lucas(2*n), A001654: F(n)*F(n+1).

Cf. A007598 (k=0), A059929 (k=1, without initial 1), A192883 (k=2, without initial -1), this sequence (k=3).

List([0..10^2],n ->Fibonacci(n)^2+4*(-1)^n); # Muniru A Asiru, Sep 26 2017

[Fibonacci(n)^2+4*(-1)^n: n in [0..40]];

with(combinat,fibonacci):  A292612:=seq(fibonacci(n)^2+4*(-1)^n, n=0..10^2); # Muniru A Asiru, Sep 26 2017

Table[Fibonacci[n]^2 + 4 (-1)^n, {n, 0, 40}]

for(n=0, 40, print1(fibonacci(n)^2+4*(-1)^n", "));

Vec((4-11*x+3*x^2)/((1+x)*(1-3*x+x^2))+O(x^30)) \\ Colin Barker, Sep 20 2017

[fibonacci(n)^2+4*(-1)^n for n in range(40)]

G.f.: (4 - 11*x + 3*x^2)/((1 + x)*(1 - 3*x + x^2)).
a(n) = a(-n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = 4*A001654(n+1) - 11*A001654(n) + 3*A001654(n-1) with A001654(-1)=0.
5*a(n) = Lucas(2*n) + 18*(-1)^n. Note that Lucas(2*n) + r*(-1)^n is divisible by 5 for r = -2, 3, -7, 8, -12, 13, -17, 18, -22, 23, -27, ... = (-1/4)*(3 + 5*(2*m+1)*(-1)^m) = (-1)^m*A047221(m). On the other hand, a(n) is divisible by 5 when n is a member of A047221.
a(n) = (1/5)*(18*(-1)^n + ((3-sqrt(5))/2)^n + ((3+sqrt(5))/2)^n). - Colin Barker, Sep 20 2017
Sum_{n>=4} 1/a(n) = 143/960. - Amiram Eldar, Oct 05 2020
Sum_{n>=4} (-1)^n/a(n) = 3/(4*phi) - 407/960, where phi is the golden ratio (A001622). - Amiram Eldar, Oct 06 2020

A260259 a(n) = F(n)*F(n+1) - (-1)^n, where F = A000045.

Original entry on oeis.org

-1, 2, 1, 7, 14, 41, 103, 274, 713, 1871, 4894, 12817, 33551, 87842, 229969, 602071, 1576238, 4126649, 10803703, 28284466, 74049689, 193864607, 507544126, 1328767777, 3478759199, 9107509826, 23843770273, 62423800999, 163427632718, 427859097161, 1120149658759

Offset: 0

Bruno Berselli, Oct 31 2015

Primes in sequence for n = 1, 3, 5, 6, 9, 24, 42, 48, 53, 71, 86, 102, 138, 182, 302, 438, 506, 926, ...

Bruno Berselli, Table of n, a(n) for n = 0..500
A. Bremner, R. Høibakk, D. Lukkassen, Crossed ladders and Euler’s quartic, Annales Mathematicae et Informaticae, 36 (2009) pp. 29-41. See p. 33.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

First bisection of A111569.
Cf. A226205: numbers of the form F(n)*F(n+1)+(-1)^n.
Cf. A000045, A001654, A003482, A059929, A089508 (first bisection, without -1), A206351.

[Fibonacci(n)*Fibonacci(n+1)-(-1)^n: n in [0..30]];

with(combinat): A260259:=n->fibonacci(n)*fibonacci(n+1)-(-1)^n: seq(A260259(n), n=0..50); # Wesley Ivan Hurt, Feb 04 2017

Table[Fibonacci[n] Fibonacci[n + 1] - (-1)^n, {n, 0, 30}]

makelist(fib(n)*fib(n+1)-(-1)^n,n,0,30);

for(n=0, 30, print1(fibonacci(n)*fibonacci(n+1)-(-1)^n", "));

a(n) = round((2^(-1-n)*(-3*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Sep 29 2016

Vec(-(1-4*x+x^2)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

[fibonacci(n)*fibonacci(n+1)-(-1)^n for n in (0..30)]

G.f.: (-1 + 4*x - x^2)/((1 + x)*(1 - 3*x + x^2)).
a(n) = -a(-n-1) = 2*a(n-1) + 2*a(n-2) - a(n-3) for all n in Z.
a(n) = F(n+2)^2 - 2*F(n+1)^2.
a(n) = A059929(n) - A059929(n-1) with A059929(-1)=1.
a(n) = -A001654(n+1) + 4*A001654(n) - A001654(n-1).
a(n) = A206351((n+2)/2)-2 for even n; a(n) = A003482((n-1)/2)+2 for odd n.
Sum_{i>=0} 1/a(i) = .754301907697893871765121109686...
a(n) = (2^(-1-n)*(-3*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 29 2016

A088545 Quotient Fibonacci(5n)/(5Fibonacci(n)), where Fibonacci(n) = A000045(n).

Original entry on oeis.org

1, 11, 61, 451, 3001, 20801, 141961, 974611, 6675901, 45768251, 313671601, 2150012161, 14736206161, 101003973851, 692290189501, 4745031073651, 32522917584361, 222915417520961, 1527884938291801, 10472279325329251, 71778069881360701, 491974211042344811, 3372041404278257761

Offset: 1

Lekraj Beedassy, Nov 17 2003

The sequences {Fibonacci(k*n)/(Fibonacci(k)*Fibonacci(n)): n >= 1} are integral in the three cases k = 1 (A000012), k = 2 (A000032) and k = 5 (the present sequence). See Young, Section 4. - Peter Bala, Jan 09 2023

Chatchawan Panraksa and Aram Tangboonduangjit, On Some Arithmetic Properties of a Sequence Related to the Quotient of Fibonacci Numbers, Fibonacci Quart. 55 (2017), no. 1, 21-28.
Paul Thomas Young, p-adic congruences for generalized Fibonacci sequences, The Fibonacci Quarterly, Vol. 32, No. 1, 1994.

Cf. A000032, A000045, A001656, A103326.

a(n)=fibonacci(5*n)/(5*fibonacci(n)); \\ Joerg Arndt, Jul 16 2013

a(n) = 5*Fib(n)^2*(Fib(n)^2 + (-1)^n) + 1 = 5*A007598(n)*A059929(n+1) + 1.
a(n) = A103326(n) / 5.
G.f.: -x*(x^4-4*x^3-9*x^2+6*x+1) / ((x-1)*(x^2-7*x+1)*(x^2+3*x+1)). - Colin Barker, Jul 16 2013
The expansion of exp(Sum_{n >= 1} a(n)*x^n/n) = 1 + x + 6*x^2 + 26*x^3 + 151*x^4 + 851*x^5 + 5101*x^6 + ... has integral coefficients and is equal to G(x)^(1/5), where G(x) is the o.g.f. of A001656. See Young, Theorem 3. - Peter Bala, Jan 09 2023

A236165 a(n) = a(n-1) + a(n-2) + a(n-3), with a(0) = a(1) = 1, a(2) = 0.

Original entry on oeis.org

1, 1, 0, 0, 2, 3, 3, 5, 10, 16, 24, 39, 65, 105, 168, 272, 442, 715, 1155, 1869, 3026, 4896, 7920, 12815, 20737, 33553, 54288, 87840, 142130, 229971, 372099, 602069, 974170, 1576240, 2550408, 4126647, 6677057, 10803705, 17480760, 28284464, 45765226, 74049691

Offset: 0

Michael Somos, Jan 19 2014

			G.f. = 1 + x + 2*x^4 + 3*x^5 + 3*x^6 + 5*x^7 + 10*x^8 + 16*x^9 + ...

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. A000045, A059929, A226205.

I:=[1,1,0,0]; [n le 4 select I[n] else Self(n-1)+Self(n-3)+Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jan 20 2015

a[ n_] := Fibonacci[ Quotient[ n, 2] - 1] Fibonacci[ Quotient[ n, 2] + 1 + Mod[n, 2]];
LinearRecurrence[{1,0,1,1},{1,1,0,0},50] (* Harvey P. Dale, Jan 19 2015 *)
CoefficientList[Series[(1 - x^2 - x^3) / (1 - x - x^3 - x^4), {x, 0, 70}], x] (* Vincenzo Librandi, Jan 20 2015 *)

{a(n) = fibonacci( n\2 - 1 ) * fibonacci( n\2 + 1 + n%2 )};

G.f.: (1 - x^2 - x^3) / (1 - x - x^3 - x^4).
a(n+1)*a(n+3) = a(n)*a(n+2) + a(n+1)*a(n+2) for all n in Z.
a(n+1) + a(n-1) = A000045(n) for all n in Z.
a(2n) = A059929(n-1), a(2n-1) = A226205(n).

A262342 Area of Lewis Carroll's paradoxical F(2n+1) X F(2n+3) rectangle.

Original entry on oeis.org

10, 65, 442, 3026, 20737, 142130, 974170, 6677057, 45765226, 313679522, 2149991425, 14736260450, 101003831722, 692290561601, 4745030099482, 32522920134770, 222915410843905, 1527884955772562, 10472279279564026, 71778070001175617, 491974210728665290, 3372041405099481410

Offset: 1

Jonathan Sondow, Oct 16 2015

Warren Weaver (1938): "In a familiar geometrical paradox a square of area 8 X 8 = 64 square units is cut into four parts which may be refitted to form a rectangle of apparent area 5 X 13 = 65 square units.... Lewis Carroll generalized this paradox...."
Carroll cuts a F(2n+2) X F(2n+2) square into four parts, where F(n) is the n-th Fibonacci number. Two parts are right triangles with legs F(2n) and F(2n+2); two are right trapezoids three of whose sides are F(2n), F(2n+1), and F(2n+1). (Thus n > 0.) The paradox (or dissection fallacy) depends on Cassini's identity F(2n+1) * F(2n+3) = F(2n+2)^2 + 1.
For an extension of the paradox to a F(2n+1) X F(2n+1) square using Cassini's identity F(2n) * F(2n+2) = F(2n+1)^2 - 1, see Dudeney (1970), Gardner (1956), Horadam (1962), Knott (2014), Kumar (1964), and Sillke (2004). Sillke also has many additional references and links.

			F(3) * F(5) = 2 * 5 = 10 = 3^2 + 1 = F(4)^2 + 1, so a(1) = 10.
G.f. = 10*x + 65*x^2 + 442*x^3 + 3026*x^4 + 20737*x^5 + 142130*x^6 + 974170*x^7 + ...

W. W. Rouse Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13th edition, Dover, 1987, p. 85.
Henry E. Dudeney, 536 Puzzles and Curious Problems, Scribner, reprinted 1970, Problems 352-353 and their answers.
Martin Gardner, Mathematics, Magic and Mystery, Dover, 1956, Chap. 8.
Edward Wakeling, Rediscovered Lewis Carroll Puzzles, Dover, 1995, p. 12.
David Wells, The Penguin Book of Curious and Interesting Puzzles, Penguin, 1997, Puzzle 143.

Colin Barker, Table of n, a(n) for n = 1..1000
Margherita Barile, Dissection Fallacy, MathWorld.
A. F. Horadam, Fibonacci Sequences and a Geometrical Paradox, Math. Mag., Vol. 35, No. 1 (1962), pp. 1-11.
Ron Knott, Fibonacci jigsaw puzzles, 2014.
Santosh Kumar, On Fibonacci Sequences and a Geometrical Paradox, Math. Mag., Vol. 37, No. 4 (1964), pp. 221-223.
Oskar Schlömilch, Ein geometrisches Paradoxon, Zeitschrift für Mathematik und Physik, Vol. 13 (1868), p. 162.
Torsten Sillke, Jigsaw paradox, 2004.
David Singmaster, Vanishing area puzzles, Recreational Math. Mag., Vol. 1 (2014), pp. 10-21.
Warren Weaver, Lewis Carroll and a Geometrical Paradox, American Math. Monthly, Vol. 45, No. 4 (1938), pp. 234-236.
Wikipedia, Cassini and Catalan identities.
Wikipedia, Fibonacci number.
Wikipedia, Missing square puzzle; also see External Links.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A001519, A059929, A064170, A070550, A166516, A176055, A190018, A236165, A245306.

[Fibonacci(2*n+1)*Fibonacci(2*n+3) : n in [1..30]]; // Wesley Ivan Hurt, Oct 16 2015

with(combinat): A262342:=n->fibonacci(2*n+1)*fibonacci(2*n+3): seq(A262342(n), n=1..30); # Wesley Ivan Hurt, Oct 16 2015

Table[Fibonacci[2 n + 1] Fibonacci[2 n + 3], {n, 22}]
LinearRecurrence[{8,-8,1},{10,65,442},30] (* Harvey P. Dale, Aug 06 2024 *)

Vec(-x*(2*x^2-15*x+10)/((x-1)*(x^2-7*x+1)) + O(x^30)) \\ Colin Barker, Oct 17 2015

a(n) = fibonacci(2*n+1) * fibonacci(2*n+3) \\ Altug Alkan, Oct 17 2015

a(n) = Fibonacci(2n+1) * Fibonacci(2n+3) = Fibonacci(2n+2)^2 + 1 for n > 0.
From Colin Barker, Oct 17 2015: (Start)
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: -x*(2*x^2-15*x+10) / ((x-1)*(x^2-7*x+1)).
(End)
a(3*k-2) mod 2 = 0; a(3*k-1) mod 2 = 1; a(3*k) mod 2 = 0, k > 0. - Altug Alkan, Oct 17 2015
a(n) = A059929(2*n+1) = A070550(4*n+1) = A166516(2*n+2) = A190018(8*n) = A236165(4*n+4) = A245306(2*n+2). - Bruno Berselli, Oct 17 2015
a(n) = A064170(n+3). - Alois P. Heinz, Oct 17 2015
E.g.f.: (1/5)*((1/phi*r)*exp(b*x) + (phi^4/r)*exp(a*x) + 3*exp(x) - 10), where r = 2*phi+1, 2*a=7+3*sqrt(5), 2*b=7-3*sqrt(5). - G. C. Greubel, Oct 17 2015
a(n) = (A337928(n+1) - A337929(n+1)) / 2. - Flávio V. Fernandes, Feb 06 2021
Sum_{n>=1} 1/a(n) = sqrt(5)/2 - 1 = A176055 - 2. - Amiram Eldar, Mar 04 2021

A292696 a(n) = L(n)^2 - 5(-1)^n = L(n+1)L(n-1), where L = A000032.

Original entry on oeis.org

-1, 6, 4, 21, 44, 126, 319, 846, 2204, 5781, 15124, 39606, 103679, 271446, 710644, 1860501, 4870844, 12752046, 33385279, 87403806, 228826124, 599074581, 1568397604, 4106118246, 10749957119, 28143753126, 73681302244, 192900153621, 505019158604, 1322157322206, 3461452807999

Offset: 0

Bruno Berselli, Sep 21 2017

This is the case k=1 of the identity L(n)^2 + 5*F(k)^2*(-1)^(n+k) = L(n+k)*L(n-k), where F = A000045. See also the comment in A292612.

Steven Vajda, Fibonacci and Lucas Numbers, and the Golden Section: Theory and Applications, Dover Publications (2008), page 29 (the formula 20b, for h=-k, gives the identity in Comments section).

Colin Barker, Table of n, a(n) for n = 0..1000
Ron Knott, Fibonacci and Lucas Numbers (see 31st formula: the identity in the Comments section is the case i=-k).
Eric Weisstein's World of Mathematics, Lucas Number (see the formula n. 5).
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A001622, A001654, A292612.

Cf. A059929: Fibonacci(n+2)*Fibonacci(n).

a:=[-1,6,4];; for n in [4..10^3] do a[n]:= 2*a[n-1]+2*a[n-2]-a[n-3]; od; A292696:=a; # Muniru A Asiru, Oct 03 2017

Magma
```
[Lucas(n)^2-5*(-1)^n: n in [0..40]];
```

A292696:=proc(n) option remember:
if n=0 then -1 elif n=1 then 6 elif n=2 then 4 elif  n>=3 then 2*procname(n-1)+2*procname(n-2)-procname(n-3) fi; end:
seq(A292696(n),n=0..10^2); # Muniru A Asiru, Oct 03 2017

Table[LucasL[n]^2 - 5 (-1)^n, {n, 0, 40}]
LinearRecurrence[{2,2,-1},{-1,6,4},40] (* Harvey P. Dale, Oct 02 2018 *)

Vec(-(1 - 8*x + 6*x^2) / ((1 + x)*(1 - 3*x + x^2)) + O(x^30)) \\ Colin Barker, Sep 21 2017

[lucas(n)^2-5*(-1)^n for n in range(40)]

G.f.: (-1 + 8*x - 6*x^2)/((1 + x)*(1 - 3*x + x^2)).
a(n) = a(-n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = -A001654(n+1) + 8*A001654(n) - 6*A001654(n-1) with A001654(-1)=0.
a(n) = L(2*n) - 3*(-1)^n.
Sum_{i=0..n} a(i) = L(2*n+1) - (3*(-1)^n + 1)/2.
a(n) = 2^(-n)*(-3*(-2)^n + (3-sqrt(5))^n + (3+sqrt(5))^n). - Colin Barker, Sep 21 2017
From Peter Bala, Oct 14 2019: (Start)
Sum_{n >= 1} L(n)/a(n) = 3/2.
Sum_{n >= 1} (-1)^n*L(n)/a(n) = 1/2.  (End)
From Amiram Eldar, Oct 06 2020: (Start)
Sum_{n>=1} 1/a(n) = 1/2.
Sum_{n>=1} (-1)^n/a(n) = 7/10 - 2*phi/5, where phi is the golden ratio (A001622). (End)

A375803 a(n) = Fibonacci(n-1) * Fibonacci(n+1) * Fibonacci(2n-1) Fibonacci(2*n+1).

Original entry on oeis.org

0, 20, 195, 4420, 72624, 1347905, 23877840, 430583140, 7712000835, 138485573876, 2484341814240, 44584372180225, 800002107309600, 14355674602647860, 257600625681170499, 4622465972012379940, 82946715695078486160, 1488418904383171787585, 26708590219470770907120

Offset: 1

Amiram Eldar, Aug 29 2024

Amiram Eldar, Table of n, a(n) for n = 1..798
Hideyuki Ohtskua, proposer, Problem H-944, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 266.
Index entries for linear recurrences with constant coefficients, signature (13,104,-260,-260,104,13,-1).

Cf. A000045, A059929, A064170, A134944, A374149, A375804.

a[n_] := Fibonacci[n-1] * Fibonacci[n+1] * Fibonacci[2*n-1] * Fibonacci[2*n+1]; Array[a, 20]

a(n) = fibonacci(n-1) * fibonacci(n+1) * fibonacci(2*n-1) * fibonacci(2*n+1);

a(n) = A059929(n-1) * A059929(2*n-1) = A059929(n-1) * A064170(n+2).
Sum_{n>=2} (-1)^n/a(n) = (5*sqrt(5) - 11)/4 = A374149 - 11/2 = 10 * A134944 - 4 (Ohtskua, 2024).
G.f.: -x^2*(-20+65*x+195*x^2-84*x^3-13*x^4+x^5) / ( (1+x)*(x^2-3*x+1)*(x^2-18*x+1)*(x^2+7*x+1) ). - R. J. Mathar, Aug 30 2024

cp's OEIS Frontend

A121801 Expansion of 2*x^2*(3-x)/((1+x)*(1-3*x+x^2)).

Views

Author

Keywords

Comments

Links

Programs

GAP

Magma

Magma

Mathematica

PARI

PARI

Sage

Formula

Extensions

A236428 a(n) = F(n+1)^2 + F(n+1)*F(n) - F(n)^2, where F = A000045.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Magma

Mathematica

PARI

PARI

PARI

Formula

A248161 Expansion of (2-x+x^2)/((1+x)*(1-3*x+x^2)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

Extensions

A292612 a(n) = F(n)^2 + 4*(-1)^n = F(n+3)*F(n-3), where F = A000045.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

PARI

Sage

Formula

A260259 a(n) = F(n)*F(n+1) - (-1)^n, where F = A000045.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Maxima

PARI

PARI

PARI

Sage

Formula

A088545 Quotient Fibonacci(5*n)/(5*Fibonacci(n)), where Fibonacci(n) = A000045(n).

Views

A121801 Expansion of 2x^2(3-x)/((1+x)(1-3x+x^2)).

A248161 Expansion of (2-x+x^2)/((1+x)(1-3x+x^2)).

A292612 a(n) = F(n)^2 + 4(-1)^n = F(n+3)F(n-3), where F = A000045.

A088545 Quotient Fibonacci(5n)/(5Fibonacci(n)), where Fibonacci(n) = A000045(n).

A292696 a(n) = L(n)^2 - 5(-1)^n = L(n+1)L(n-1), where L = A000032.

A375803 a(n) = Fibonacci(n-1) * Fibonacci(n+1) * Fibonacci(2n-1) Fibonacci(2*n+1).