A337929 -id:A337929 - cp's OEIS frontend

A033888 a(n) = Fibonacci(4*n).

0, 3, 21, 144, 987, 6765, 46368, 317811, 2178309, 14930352, 102334155, 701408733, 4807526976, 32951280099, 225851433717, 1548008755920, 10610209857723, 72723460248141, 498454011879264, 3416454622906707, 23416728348467685, 160500643816367088, 1100087778366101931, 7540113804746346429

Offset: 0

N. J. A. Sloane

(x,y)=(a(n),a(n+1)) are solutions of (x+y)^2/(1+xy)=9, the other solutions are in A033890. - Floor van Lamoen, Dec 10 2001
Sequence A033888 provides half of the solutions to the equation 5*x^2 + 4 is a square. The other half are found in A033890. Lim_{n->infinity} a(n)/a(n-1) = phi^4 = (7+3*sqrt(5))/2. - Gregory V. Richardson, Oct 13 2002
Fibonacci numbers divisible by 3. - Reinhard Zumkeller, Aug 20 2011

			G.f. = 3*x + 21*x^2 + 144*x^3 + 987*x^4 + 6765*x^5 + 46368*x^6 + ...

Nathaniel Johnston, Table of n, a(n) for n = 0..300
Piero Filipponi and Marco Bucci, On the Integrity of Certain Fibonacci Sums, The Fibonacci Quarterly, Vol. 32, No. 3 (1994), pp. 245-252.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A033890, A337928, A337929.

Fourth column of array A102310.

[ Fibonacci(4*n): n in [0..100]]; // Vincenzo Librandi, Apr 15 2011

A033888:=n->combinat[fibonacci](4*n): seq(A033888(n), n=0..30); # Wesley Ivan Hurt, Apr 26 2017

Table[Fibonacci[4*n], {n,0,14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
Table[Mod[Fibonacci[(4 n + 2)] , Fibonacci[(4 n + 1)]], {n, 1, 10}] (* Artur Jasinski, Nov 15 2011 (corrected by Iain Fox, Dec 18 2017) *)

numlib::fibonacci(n*4) $ n = 0..30; // Zerinvary Lajos, May 08 2008

a(n)=fibonacci(4*n) \\ Charles R Greathouse IV, Feb 03 2014

first(n) = Vec(3*x/(1 - 7*x + x^2) + O(x^n), -n) \\ Iain Fox, Dec 18 2017

a(n) = fibonacci(4*n + 2) % fibonacci(4*n + 1) \\ Iain Fox, Dec 18 2017

[lucas_number1(n,3,1)*lucas_number2(n,3,1) for n in range(0,21)] # Zerinvary Lajos, Jun 28 2008

[fibonacci(4*n) for n in range(0, 20)] # Zerinvary Lajos, May 15 2009

a(n) = 7*a(n-1) - a(n-2).
a(n) = ((7+3*sqrt(5))^(n-1) - (7-3*sqrt(5))^(n-1)) / ((2^(n-1))*sqrt(5)). - Gregory V. Richardson, Oct 13 2002
a(n) = Sum_{k=0..n} F(3*n-k)*binomial(n, k). - Benoit Cloitre, Jun 07 2004
a(n) = Lucas(2*n) * Lucas(n) * Fibonacci(n). - Ralf Stephan, Sep 25 2004
G.f.: 3*x/(1-7*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = 3*A004187(n). - R. J. Mathar, Sep 03 2010
a(n) = Fibonacci[(4*n + 2)] modulo Fibonacci[(4*n + 1)]. - Artur Jasinski, Nov 15 2011 (corrected by Iain Fox, Dec 18 2017)
a(n) = (A337929(n) + A337928(n)) / 2. - Flávio V. Fernandes, Feb 06 2021
E.g.f.: 2*exp(7*x/2)*sinh(3*sqrt(5)*x/2)/sqrt(5). - Stefano Spezia, Feb 07 2021
a(n) = Sum_{k>=0} Fibonacci(2*n*k)/Lucas(2*n)^k (Filipponi and Bucci, 1994). - Amiram Eldar, Jan 17 2022
From Peter Bala, May 22 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/3 (telescoping series: 3/(a(n) - 1/a(n)) = 1/A033890(n) + 1/A033890(n-1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) (telescoping product: ((a(n) + 1)/(a(n) - 1))^2 = (5 - 4/Fibonacci(2*n+1)^2)/(5 - 4/Fibonacci(2*n-1)^2) from which we get Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5 - 4/Fibonacci(2*n+1)^2). (End)

A337928 Numbers w such that (F(2n+1)^2, -F(2n)^2, -w) are primitive solutions of the Diophantine equation 2x^3 + 2y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

1, 5, 31, 209, 1429, 9791, 67105, 459941, 3152479, 21607409, 148099381, 1015088255, 6957518401, 47687540549, 326855265439, 2240299317521, 15355239957205, 105246380382911, 721369422723169, 4944339578679269, 33889007628031711, 232278713817542705

Offset: 0

XU Pingya, Sep 30 2020

			2*(F(5)^2)^3 + 2*(-F(4)^2)^3 + (-31)^3 = 2*(25)^3 + 2*(-9)^3 + (-31)^3 = 1, a(2) = 31.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A000578, A056573, A081018, A337929.

Table[(2*Fibonacci[2n+1]^6 - 2*Fibonacci[2n]^6 - 1)^(1/3), {n, 0, 21}]
Table[(Fibonacci[2n+1]*Fibonacci[2n+2]- Fibonacci[2n]^2), {n, 0, 21}] (* Wolfgang Berndt, May 26 2023 *)
LinearRecurrence[{8,-8,1},{1,5,31},30] (* Harvey P. Dale, Dec 17 2023 *)

Vec((1 - 3*x - x^2) / ((1 - x)*(1 - 7*x + x^2)) + O(x^20)) \\ Colin Barker, Oct 01 2020

a(n) = (2*F(2*n+1)^6 - 2*F(2*n)^6 - 1)^(1/3).
From Colin Barker, Oct 01 2020: (Start)
G.f.: (1 - 3*x - x^2) / ((1 - x)*(1 - 7*x + x^2)).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) for n>2.
(End)
a(n) = 2*A081018(n) + 1. - Hugo Pfoertner, Oct 01 2020
a(n) = A064170(n+2) + A033888(n). - Flávio V. Fernandes, Jan 10 2021
a(n) = F(2*n+1)*F(2*n+2) - F(2*n)^2. - Wolfgang Berndt, May 26 2023
a(2*n-1) = 5 + 6*Sum_{k=1..n-1} F(8*k+1), a(2*n) = 1 + 6*Sum_{k=1..n} F(8*k-3). - XU Pingya, Jun 09 2024

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

A354337 a(n) is the integer w such that (L(2n)^2, -L(2n + 1)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

Original entry on oeis.org

19, 149, 1039, 7139, 48949, 335519, 2299699, 15762389, 108037039, 740496899, 5075441269, 34787591999, 238437702739, 1634276327189, 11201496587599, 76776199786019, 526231901914549, 3606847113615839, 24721697893396339, 169445038140158549, 1161393569087713519

Offset: 1

XU Pingya, Jun 20 2022

Subsequence of A017377.

			2*(L(4)^2)^3 + 2*(-L(5)^2)^3 + (149)^3 = 2*(49)^3 + 2*(-121)^3 + (149)^3 = 125, a(2) = 149.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000032, A002878, A005248, A017377, A081017, A089508, A092521, A288913.

Cf. A337929, A354336.

LinearRecurrence[{7,-1},{19,149},21]-1 + LucasL[2*Range[21]-3]^2

a(n) = (125 - 2*A005248(n)^6 + 2*A002878(n)^6)^(1/3).
a(n) = Lucas(4*n+2) + Lucas(4n-1) - 3 = 2*A056914(n)-3 = 15*A092521(n) + A288913(n-1).
a(n) = 2*A081017(n) - 1.
a(n) = 10*A089508(n) + 9.
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: x*(19 - 3*x - x^2)/((1 - x)*(1 - 7*x + x^2)). - Stefano Spezia, Jun 22 2022

A356717 a(n) is the integer w such that (c(n)^2, -d(n)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+3) + (-1)^n * F(n-2) and F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

1, 29, 59, 241, 445, 1691, 3089, 11629, 21211, 79745, 145421, 546619, 996769, 3746621, 6831995, 25679761, 46827229, 176011739, 320958641, 1206402445, 2199883291, 8268805409, 15078224429, 56675235451, 103347687745, 388457842781, 708355589819, 2662529664049

Offset: 1

XU Pingya, Aug 24 2022

			For n=3, 2 * ((F(5) - F(0))^2)^3 + 2 * (-(F(6) - F(1))^2)^3 + 59^3 = 2 * 25^3 - 2 * 49^3 + 59^3 = 1331, a(3) = 59.

Index entries for linear recurrences with constant coefficients, signature (1,7,-7,-1,1).

Cf. A000045, A081016, A081018, A089270, A228208, A237132.

Cf. also A337929, A354337, A356716.

Table[(1331-2*((Fibonacci[n+2]+(-1)^n*Fibonacci[n-3]))^6+2*(Fibonacci[n+3]+(-1)^n*Fibonacci[n-2])^6)^(1/3), {n,28}]

a(n) = (1331 - 2 * A237132(n)^6 + 2 * A228208(n+1)^6)^(1/3).
a(n) = ((1-(-1)^n)/2) * (-5 + 14 * Sum_{k=1..n-1} Fibonacci(4*k-1) + 6 * Sum_{k=0..n-1} Fibonacci(4*k+1)) + ((1+(-1)^n)/2) * (-5 + 14 * Sum_{k=1..n} Fibonacci(4*k-1) + 6 * Sum_{k=0..n-1} Fibonacci(4*k+1)).
a(n) = ((1-(-1)^n)/2) * (-5 + 14 * A081018(n-1) + 6 * A081016(n-1)) + ((1+(-1)^n)/2) * (-5 + 14 * A081018(n) + 6 * A081016(n-1)).
From Stefano Spezia, Aug 25 2022: (Start)
G.f.: x*(1 + 28*x + 23*x^2 - 14*x^3 - 5*x^4)/((1 - x)*(1 - 3*x + x^2)*(1 + 3*x + x^2)).
a(n) = a(n-1) + 7*a(n-2) - 7*a(n-3) - a(n-4) + a(n-5) for n > 5. (End)

A338239 Values z of primitive solutions (x, y, z) to the Diophantine equation 2x^3 + 2y^3 + z^3 = 1.

Original entry on oeis.org

-1, 1, -5, 11, -17, 19, 29, -31, -37, -61, 79, -85, 113, -127, -143, 145, -209, 305, 361, -485, 487, 545, 647, 667, 811, -1091, -1151, 1153, -1235, -1429, -1525, 1597, 1699, -1793, -2249, 2251, -2533, 2627, -2677, 2977, -2981, 3089, -3295, 3739, -3887, 3889

Offset: 1

XU Pingya, Oct 18 2020

sign

Terms are arranged in order of increasing absolute value (if equal, the negative number comes first).

When x = (3*c)*t - (9*a)*t^4, y = (9*a)*t^4, z = c - (9*a)*t^3; a*x^3 + a*y^3 + c*z^3 = c^4. Let a = 2, c = 1, then 1 - 18*n^3 and 1 + 18*n^3 are terms of the sequence. Also, -A337928 and A337929 are subsequences.

			2*25^3 + 2*(-64)^3 + 79^3 = 2*164^3 + 2*(-167)^3 + 79^3 = 1, 79 is a term.

Beniamino Segre, On the rational solutions of homogeneous cubic equations in four variables, Math. Notae, 11 (1951), 1-68.

Cf. A000045, A000290, A000578, A003215, A004825, A004826, A047201, A050791, A130472, A195006, A337928, A337929.

Clear[t]
t = {};
Do[y = ((1 - 2x^3 - z^3)/2)^(1/3) /. (-1)^(1/3) -> -1;
 If[IntegerQ[y] && GCD[x, y, z] == 1, AppendTo[t, z]], {z, -4000, 4000}, {x, -Round[(Abs[1 + z^3]/6)^(1/2)], Round[(Abs[1 + z^3]/6)^(1/2)]}]
u = Union@t;
v = Table[(-1)^n*Floor[(n + 1)/2], {n, 0, 8001}];
Select[v, MemberQ[u, #] &]

cp's OEIS Frontend

A033888 a(n) = Fibonacci(4*n).

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A337928 Numbers w such that (F(2n+1)^2, -F(2n)^2, -w) are primitive solutions of the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

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

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A354337 a(n) is the integer w such that (L(2*n)^2, -L(2*n + 1)^2, w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

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A356717 a(n) is the integer w such that (c(n)^2, -d(n)^2, w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+3) + (-1)^n * F(n-2) and F(n) is the n-th Fibonacci number (A000045).

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A338239 Values z of primitive solutions (x, y, z) to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 1.

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A337928 Numbers w such that (F(2n+1)^2, -F(2n)^2, -w) are primitive solutions of the Diophantine equation 2x^3 + 2y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

A354337 a(n) is the integer w such that (L(2n)^2, -L(2n + 1)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

A356717 a(n) is the integer w such that (c(n)^2, -d(n)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 11^3, where c(n) = F(n+2) + (-1)^n * F(n-3), d(n) = F(n+3) + (-1)^n * F(n-2) and F(n) is the n-th Fibonacci number (A000045).

A338239 Values z of primitive solutions (x, y, z) to the Diophantine equation 2x^3 + 2y^3 + z^3 = 1.