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

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

Offset: 1

Author

Jonathan Sondow, Oct 16 2015

Keywords

• nonn
• easy

Comments

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.

Examples

			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 + ...
		

References

• 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.

Links

• 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).

Crossrefs

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

Programs

Magma

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

Maple

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

Mathematica

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 *)

PARI

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

PARI

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

Formula

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