%I A001654 M1606 N0628
%S A001654 0,1,2,6,15,40,104,273,714,1870,4895,12816,33552,87841,229970,602070,
%T A001654 1576239,4126648,10803704,28284465,74049690,193864606,507544127,
%U A001654 1328767776,3478759200,9107509825,23843770274,62423800998,163427632719
%N A001654 Golden rectangle numbers: F(n)F(n+1), where F() = Fibonacci numbers A000045.
%C A001654 a(n)/A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg 
               (roberthappelberg(AT)yahoo.com), Jul 25 2005
%C A001654 Let phi be the golden ratio (cf. A001622). Then 1/phi=phi-1=Sum_{n=1..inf} 
               (-1)^(n-1)/a(n), an alternating infinite series consisting solely 
               of unit fractions. - Franz Vrabec (franz.vrabec(AT)aon.at), Sep 14 
               2005
%C A001654 a(n+2) is the Hankel transform of A005807 aerated. [From Paul Barry (pbarry(AT)wit.ie), 
               Nov 04 2008]
%C A001654 Contribution from Daniel Forgues (squid(AT)zensearch.com), Nov 29 2009: 
               (Start)
%C A001654 A more exact name would be: Golden convergents rectangle numbers
%C A001654 These rectangles are not actually Golden (ratio of sides is not Phi)
%C A001654 but are Golden convergents (sides are numerator and denominator
%C A001654 of convergents of the continued fraction expansion of Phi, whence
%C A001654 ratio of sides converges to Phi.) (End)
%D A001654 A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of 
               combinatorial proof, M.A.A. 2003, id. 9.
%D A001654 A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
%D A001654 A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci 
               Association, San Jose, CA, 1972, p. 17.
%D A001654 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 
               (includes this sequence).
%D A001654 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 
               Academic Press, 1995 (includes this sequence).
%H A001654 T. D. Noe, <a href="b001654.txt">Table of n, a(n) for n=0..200</a>
%H A001654 S. Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/MasterThesis.pdf">
               Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures</
               a>, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 
               1992.
%H A001654 S. Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/FonctionsGeneratrices.pdf">
               1031 Generating Functions and Conjectures</a>, Universit\'{e} du 
               Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
%H A001654 M. Renault, <a href="http://www.math.temple.edu/~renault/fibonacci/thesis.html">
               Dissertation</a>
%H A001654 Wikipedia, <a href="http://en.wikipedia.org/wiki/Image:FibonacciBlocks.png">
               Illustration of 273 as a golden rectangle number</a>.
%H A001654 <a href="Sindx_Tu.html#2wis">Index entries for two-way infinite sequences</
               a>
%H A001654 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%F A001654 a(n)= A010048(n+1, 2)= fibonomial(n+1, 2).
%F A001654 a(n) = a(n - 1) + A007598(n) = a(n - 1) + A000045(n)^2 = sum_j[Fib(j)^2] 
               over j <= n - Henry Bottomley (se16(AT)btinternet.com), Feb 09 2001
%F A001654 For n>0, 1-1/a(n+1)=sum(k=1, n, 1/F(k)/F(k+2)) where F(k) is the k-th 
               Fibonacci number. - Benoit Cloitre, Aug 31, 2002.
%F A001654 G.f.: x/(1-2x-2x^2+x^3) = x/((1+x)(1-3x+x^2)) (see Comments to A055870), 
               a(n)=3a(n-1)-a(n-2)-(-1)^n=-a(-1-n).
%F A001654 Let M = the 3 X 3 matrix [1 2 1 / 1 1 0 / 1 0 0]; then a(n) = the center 
               term in M^n *[1 0 0]. E.g. a(5) = 40 since M^5 * [1 0 0] = [64 40 
               25]. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 10 2004
%F A001654 Equals the partial sums of squares of Fibonacci numbers. The proof is 
               easy. Start from a square (1*1)On the right side, draw another square 
               (1*1).On the above side draw a square ((1+1)*(1+1). On the left side, 
               draw a square ((1+2)*(1+2)) and so one. You get a rectangle (F(n)*F(1+n)) 
               which contains all the squares of side F(1), F(2),. . . F(n) - Philippe 
               LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 19 2007
%F A001654 Contribution from Daniel Forgues (squid(AT)zensearch.com), Nov 29 2009: 
               (Start)
%F A001654 With Phi(n) = [1+sqrt(5)]/2 as the Golden ratio, the following formula 
               gives
%F A001654 EXACT values (not just approximations!) of a(n) for n >= 0:
%F A001654 a(n) = Round[(Phi^(2n+1))/5] = Floor[(1/2) + (Phi^(2n+1))/5], n >= 0. 
               (End)
%p A001654 with (combinat):a:=n->fibonacci(n)*fibonacci(n+1): seq(a(n), n=0..28); 
               - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 07 2007
%p A001654 A001654:=1/(z+1)/(z**2-3*z+1); [Conjectured by S. Plouffe in his 1992 
               dissertation.]
%t A001654 q=0;lst={};Do[f=Fibonacci[n];AppendTo[lst,f*q];q=f,{n,5!}];lst [From 
               Vladimir Orlovsky (4vladimir(AT)gmail.com), Jul 21 2009]
%o A001654 (PARI) a(n)=fibonacci(n)*fibonacci(n+1)
%Y A001654 Cf. A010048, A001655-A001658. A006498(2n-1)=a(n).
%Y A001654 Bisection of A006498, A070550, A080239. Cf. A079472, A080145.
%Y A001654 First differences of A064831. Partial sums of A007598.
%Y A001654 First differences of A064831. Cf. A079472.
%Y A001654 Cf. A119283, A000071, A005968, A005969, A098531, A098532, A098533, A128697.
%Y A001654 Sequence in context: A001674 A121331 A026270 this_sequence A062106 A061322 
               A004664
%Y A001654 Adjacent sequences: A001651 A001652 A001653 this_sequence A001655 A001656 
               A001657
%K A001654 nonn,new
%O A001654 0,3
%A A001654 N. J. A. Sloane (njas(AT)research.att.com).
%E A001654 Extended by Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), 
               Jun 27 2000