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A014742 Expansion of (1+x^2)/(1 - 2*x - 2*x^2 + x^3).

%I A014742 #74 Feb 16 2025 08:32:33
%S A014742 1,2,7,17,46,119,313,818,2143,5609,14686,38447,100657,263522,689911,
%T A014742 1806209,4728718,12379943,32411113,84853394,222149071,581593817,
%U A014742 1522632382,3986303327,10436277601,27322529474,71531310823,187271402993,490282898158,1283577291479
%N A014742 Expansion of (1+x^2)/(1 - 2*x - 2*x^2 + x^3).
%C A014742 Let M = a triangle with (1,1,1,3,3,5,5,7,7,...) as the left border and (0,1,2,3,4,5,...) as all other columns. A014742 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - _Gary W. Adamson_, Jul 26 2010
%C A014742 For n >= 1, a(n) is the ratio of L/h (rounded down), where L = length of short sides of parallelogram appearing in dissection fallacy of square F(n+2) X F(n+2), F(n) is Fibonacci number, and h = perpendicular distance between the long sides LL. The first differences of A069921 give L^2. See illustration. - _Kival Ngaokrajang_, Jun 27 2014
%C A014742 From _Wolfdieter Lang_, Jul 15 2014: (Start)
%C A014742 The preceding comment is a conjecture using a(n) = floor(LL(n)*L(n)) with LL(n) = sqrt(F(n+2)^2 + F(n)^2) and L(n) = LL(n-1), n >= 1 (its author agreed with this in an email). See also, e.g., the Koshy reference for the dissection fallacy, sect. 6, 100 - 108.
%C A014742 The proof of the conjecture uses first the identity (LL(n)*LL(n-1))^2 - a(n)^2 = + 1 with a(n) = F(n-1)*F(n) + F(n+1)*F(n+2) (see the formula section for a(n)). This identity is due to the factorization of the left-hand side which is A(n)^2 with A(n) = F(n)*F(n+1) - F(n-1)*F(n+2). But A(n) = (-1)^(n+1) is a special instance of a well known Fibonacci identity (Koshy, p. 88, Nr. 19 for h=-1, k=2, F(-1) = 1). Now one has (LL(n)*LL(n-1))^2 = 1 + a(n)^2, that is LL(n)*LL(n-1) = sqrt(1 + a(n)^2). Because a(n) < sqrt(1 + a(n)^2) <  a(n) + 1 (just square both inequalities using a(n) > 0) one has now proved that floor(LL(n)*LL(n-1)) = a(n), n >= 1. (End)
%C A014742 a(n) = numerator(Re(C(n))), with the complex sequence C(n) defined in the name of A069921. - _Wolfdieter Lang_, Jul 16 2014
%D A014742 T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley & Sons, 2001.
%H A014742 Colin Barker, <a href="/A014742/b014742.txt">Table of n, a(n) for n = 0..1000</a>
%H A014742 Kival Ngaokrajang, <a href="/A014742/a014742_1.pdf">Illustration of initial terms</a>
%H A014742 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/DissectionFallacy.html">Dissection Fallacy</a>
%H A014742 Wikipedia, <a href="http://en.wikipedia.org/wiki/Missing_square_puzzle">Missing square puzzle</a>
%H A014742 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F A014742 a(n) = F(n-1)*F(n) + F(n+1)*F(n+2), where F = A000045. - _James R. Buddenhagen_, Jan 06 2009
%F A014742 From _Wolfdieter Lang_, Jul 15 2014: (Start)
%F A014742 G.f.: (1+x^2)/(1 - 2*x - 2*x^2 + x^3) = (1+x^2)/((1+x)*(1 - 3*x + x^2)) = (2/(1+x) + 3*(1+x)/(1 - 3*x + x^2))/5 (see the name).
%F A014742 a(n) = (2*(-1)^n + 3*(F(2*n) + F(2*(n+1))))/5,
%F A014742 a(n) = (2*(-1)^n + L(2*n-1) + L(2*n+3))/5 with L(n) = A000032(n) and L(-1) = -1. (End)
%F A014742 a(n) = 3*F(n)*F(n+1) + (-1)^n. - _Bruno Berselli_, Oct 30 2015
%F A014742 a(n) = (2^(-1-n)*((-1)^n*2^(2+n) - 3*(3-sqrt(5))^n*(-1+sqrt(5)) + 3*(1+sqrt(5))*(3+sqrt(5))^n))/5. - _Colin Barker_, Sep 29 2016
%F A014742 a(n) = Fibonacci(n + 2)^2 - 2*Fibonacci(n)^2. - _Detlef Meya_, Jun 29 2024
%F A014742 E.g.f.: exp(-x)*(2 + 3*exp(5*x/2)*(cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2)))/5. - _Stefano Spezia_, Jun 30 2024
%e A014742 a(2) = F(1)*F(2) + F(3)*F(4) = 1*1 + 2*3 = 7. - _James R. Buddenhagen_, Jan 06 2009
%p A014742 seq(combinat[fibonacci](n-1)*combinat[fibonacci](n)+combinat[fibonacci](n+1)*combinat[fibonacci](n+2), n=0..50); # will give first 50 terms - _James R. Buddenhagen_, Jan 06 2009
%t A014742 CoefficientList[Series[(1 + x^2)/(1 - 2*x - 2*x^2 + x^3), {x, 0, 30}], x] (* _Wesley Ivan Hurt_, Jun 27 2014 *)
%t A014742 LinearRecurrence[{2, 2, -1},{1, 2, 7},30] (* _Ray Chandler_, Sep 23 2015 *)
%o A014742 (PARI) Vec((1+x^2)/(1-2*x-2*x^2+x^3)+O(x^99)) \\ _Charles R Greathouse IV_, Sep 26 2012
%o A014742 (PARI) a(n) = round((2^(-1-n)*((-1)^n*2^(2+n)-3*(3-sqrt(5))^n*(-1+sqrt(5))+3*(1+sqrt(5))*(3+sqrt(5))^n))/5) \\ _Colin Barker_, Sep 29 2016
%Y A014742 Cf. A069921. - _Kival Ngaokrajang_, Jun 27 2014
%Y A014742 Cf. similar sequences of the type k*F(n)*F(n+1) + (-1)^n listed in A264080.
%Y A014742 Cf. A000032, A000045.
%K A014742 nonn,easy
%O A014742 0,2
%A A014742 _N. J. A. Sloane_
%E A014742 Buddenhagen's Jan 06 2009 entries adjusted for offset 0 by _Wolfdieter Lang_, Jul 15 2014