A262342 - OEIS

%I #91 Feb 16 2025 08:33:27

%S 10,65,442,3026,20737,142130,974170,6677057,45765226,313679522,

%T 2149991425,14736260450,101003831722,692290561601,4745030099482,

%U 32522920134770,222915410843905,1527884955772562,10472279279564026,71778070001175617,491974210728665290,3372041405099481410

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

%C 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...."

%C 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.

%C 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.

%D W. W. Rouse Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13th edition, Dover, 1987, p. 85.

%D Henry E. Dudeney, 536 Puzzles and Curious Problems, Scribner, reprinted 1970, Problems 352-353 and their answers.

%D Martin Gardner, Mathematics, Magic and Mystery, Dover, 1956, Chap. 8.

%D Edward Wakeling, Rediscovered Lewis Carroll Puzzles, Dover, 1995, p. 12.

%D David Wells, The Penguin Book of Curious and Interesting Puzzles, Penguin, 1997, Puzzle 143.

%H Colin Barker, <a href="/A262342/b262342.txt">Table of n, a(n) for n = 1..1000</a>

%H Margherita Barile, <a href="https://mathworld.wolfram.com/DissectionFallacy.html">Dissection Fallacy</a>, MathWorld.

%H A. F. Horadam, <a href="http://www.jstor.org/stable/2689091">Fibonacci Sequences and a Geometrical Paradox</a>, Math. Mag., Vol. 35, No. 1 (1962), pp. 1-11.

%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibpuzzles2.html#jigsaw1">Fibonacci jigsaw puzzles</a>, 2014.

%H Santosh Kumar, <a href="http://www.jstor.org/stable/2688592">On Fibonacci Sequences and a Geometrical Paradox</a>, Math. Mag., Vol. 37, No. 4 (1964), pp. 221-223.

%H Oskar Schlömilch, <a href="https://archive.org/details/zeitschriftfrma20runggoog/page/n171/mode/2up">Ein geometrisches Paradoxon</a>, Zeitschrift für Mathematik und Physik, Vol. 13 (1868), p. 162.

%H Torsten Sillke, <a href="https://www.math.uni-bielefeld.de/~sillke/PUZZLES/jigsaw-paradox.html">Jigsaw paradox</a>, 2004.

%H David Singmaster, <a href="http://rmm.ludus-opuscula.org/PDF_Files/Singmaster_Vanish_10_21(1_2014)_low.pdf">Vanishing area puzzles</a>, Recreational Math. Mag., Vol. 1 (2014), pp. 10-21.

%H Warren Weaver, <a href="http://www.jstor.org/stable/2302608">Lewis Carroll and a Geometrical Paradox</a>, American Math. Monthly, Vol. 45, No. 4 (1938), pp. 234-236.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Cassini_and_Catalan_identities">Cassini and Catalan identities</a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Fibonacci_number">Fibonacci number</a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Missing_square_puzzle">Missing square puzzle</a>; also see External Links.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-8,1).

%F a(n) = Fibonacci(2n+1) * Fibonacci(2n+3) = Fibonacci(2n+2)^2 + 1 for n > 0.

%F From _Colin Barker_, Oct 17 2015: (Start)

%F a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).

%F G.f.: -x*(2*x^2-15*x+10) / ((x-1)*(x^2-7*x+1)).

%F (End)

%F 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

%F 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

%F a(n) = A064170(n+3). - _Alois P. Heinz_, Oct 17 2015

%F 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

%F a(n) = (A337928(n+1) - A337929(n+1)) / 2. - _Flávio V. Fernandes_, Feb 06 2021

%F Sum_{n>=1} 1/a(n) = sqrt(5)/2 - 1 = A176055 - 2. - _Amiram Eldar_, Mar 04 2021

%e F(3) * F(5) = 2 * 5 = 10 = 3^2 + 1 = F(4)^2 + 1, so a(1) = 10.

%e G.f. = 10*x + 65*x^2 + 442*x^3 + 3026*x^4 + 20737*x^5 + 142130*x^6 + 974170*x^7 + ...

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

%t Table[Fibonacci[2 n + 1] Fibonacci[2 n + 3], {n, 22}]

%t LinearRecurrence[{8,-8,1},{10,65,442},30] (* _Harvey P. Dale_, Aug 06 2024 *)

%o (Magma) [Fibonacci(2*n+1)*Fibonacci(2*n+3) : n in [1..30]]; // _Wesley Ivan Hurt_, Oct 16 2015

%o (PARI) Vec(-x*(2*x^2-15*x+10)/((x-1)*(x^2-7*x+1)) + O(x^30)) \\ _Colin Barker_, Oct 17 2015

%o (PARI) a(n) = fibonacci(2*n+1) * fibonacci(2*n+3) \\ _Altug Alkan_, Oct 17 2015

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

%K nonn,easy,changed

%O 1,1

%A _Jonathan Sondow_, Oct 16 2015