OFFSET
1,1
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.
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).
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
Sum_{n>=1} 1/a(n) = sqrt(5)/2 - 1 = A176055 - 2. - Amiram Eldar, Mar 04 2021
EXAMPLE
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 + ...
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 *)
PROG
(Magma) [Fibonacci(2*n+1)*Fibonacci(2*n+3) : n in [1..30]]; // Wesley Ivan Hurt, Oct 16 2015
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jonathan Sondow, Oct 16 2015
STATUS
approved