The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A001654 Golden rectangle numbers: F(n)*F(n+1), where F(n) = A000045(n) (Fibonacci numbers). (Formerly M1606 N0628) 122
 0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, 1576239, 4126648, 10803704, 28284465, 74049690, 193864606, 507544127, 1328767776, 3478759200, 9107509825, 23843770274, 62423800998, 163427632719 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS a(n)/A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg (roberthappelberg(AT)yahoo.com), Jul 25 2005 Let phi be the golden ratio (cf. A001622). Then 1/phi = phi - 1 = Sum_{n>=1} (-1)^(n-1)/a(n), an alternating infinite series consisting solely of unit fractions. - Franz Vrabec, Sep 14 2005 a(n+2) is the Hankel transform of A005807 aerated. - Paul Barry, Nov 04 2008 A more exact name would be: Golden convergents to rectangle numbers. These rectangles are not actually golden (ratio of sides is not phi) but are golden convergents (sides are numerator and denominator of convergents in the continued fraction expansion of phi, whence ratio of sides converges to phi). - Daniel Forgues, Nov 29 2009 The Kn4 sums (see A180662 for definition) of the "Races with Ties" triangle A035317 lead to this sequence. - Johannes W. Meijer, Jul 20 2011 Numbers m such that m(5m+2)+1 or m(5m-2)+1 is a square. - Bruno Berselli, Oct 22 2012 In pairs, these numbers are important in finding binomial coefficients that appear in at least six places in Pascal's triangle. For instance, the pair (m,n) = (40, 104) finds the numbers binomial(n-1,m) = binomial(n,m-1). Two additional numbers are found on the other side of the triangle. The final two numbers appear in row binomial(n-1,m). See A003015. - T. D. Noe, Mar 13 2013 For n>1, a(n) is one-half the area of the trapezoid created by the four points (F(n),L(n)), (L(n),F(n)), (F(n+1), L(n+1), (L(n+1), F(n+1)) where F(n) = A000045(n) and L(n) = A000032(n). - J. M. Bergot, May 14 2014 [Note on how to calculate: take the two points (a,b) and (c,d) with a 1. - Reinhard Zumkeller, Sep 24 2015 Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Fibonacci numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017 REFERENCES R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., 58:2 (2020), 140-142. A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9. N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe) Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6. A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83. Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 17. Ömer Egecioglu, Elif Saygi, and Zülfükar Saygi, The Mostar index of Fibonacci and Lucas cubes, arXiv:2101.04740 [math.CO], 2021. Mentions this sequence. Shalosh B. Ekhad and Doron Zeilberger, Automatic Counting of Tilings of Skinny Plane Regions, arXiv preprint arXiv:1206.4864 [math.CO], 2012. S. Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120. Dale Gerdemann, Golden Ratio Base Digit Patterns for Columns of the Fibonomial Triangle, "Another interesting pattern is for Golden Rectangle Numbers A001654. I made a short video illustrating this pattern, along with other columns of the Fibonomial Triangle A010048". Jonny Griffiths and Martin Griffiths, Fibonacci-related sequences via iterated QRT maps, Fib. Q., 51 (2013), 218-227. James P. Jones and Péter Kiss, Representation of integers as terms of a linear recurrence with maximal index, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae, 25. (1998) pp. 21-37. See Lemma 4.1 p. 34. C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7. Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009. Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992 Prapanpong Pongsriiam, Integral Values of the Generating Functions of Fibonacci and Lucas Numbers, College Math. J., 48 (No. 2 2017), pp 97ff. M. Renault, Dissertation Wikipedia, Illustration of 273 as a golden rectangle number. R. G. Wilson v, Letter to N. J. A. Sloane, circa 1993 Index entries for linear recurrences with constant coefficients, signature (2,2,-1). FORMULA a(n) = A010048(n+1, 2) = Fibonomial(n+1, 2). a(n) = A006498(2*n-1). a(n) = a(n - 1) + A007598(n) = a(n - 1) + A000045(n)^2 = Sum_{j <= n} j*Fibonacci(j)^2. - Henry Bottomley, Feb 09 2001 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. G.f.: x/(1-2*x-2*x^2+x^3) = x/((1+x)(1-3*x+x^2)) (Simon Plouffe in his 1992 dissertation; see Comments to A055870), a(n) = 3*a(n-1) - a(n-2) - (-1)^n = -a(-1-n). 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, Oct 10 2004 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 on. 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 With phi = (1+sqrt(5))/2 as the golden ratio, the following formula gives EXACT values (not just approximations!) of a(n) for n >= 0: a(n) = round((phi^(2*n+1))/5) = floor((1/2) + (phi^(2*n+1))/5), n >= 0. - Daniel Forgues, Nov 29 2009 a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), a(1)=1, a(2)=2, a(3)=6. - Sture Sjöstedt, Feb 06 2010 a(n) = (A002878(n) - (-1)^n)/5. - R. J. Mathar, Jul 22 2010 a(n) = 1/|F(n+1)/F(n) - F(n)/F(n-1)| where F(n) = Fibonacci numbers A000045. b(n) = F(n+1)/F(n) - F(n)/F(n-1): 1/1, -1/2, 1/6, -1/15, 1/40, -1/104, ...; c(n) = 1/b(n) = a(n)*(-1)^(n+1): 1, -2, 6, -15, 40, -104, ... (n=1,2,...). - Thomas Ordowski, Nov 04 2010 a(n) = (Fibonacci(n+2)^2 - Fibonacci(n-1)^2)/4. - Gary Detlefs, Dec 03 2010 Let d(n) = n mod 2, a(0)=0 and a(1)=1. For n > 1, a(n) = d(n) + 2*a(n-1) + Sum_{k=0..n-2} a(k). - L. Edson Jeffery, Mar 20 2011 From Tim Monahan, Jul 11 2011: (Start) a(n+1) = ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5. a(n) = ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n-2*(-1)^n)/10. (End) From Wolfdieter Lang, Jul 21 2012: (Start) a(n) = Sum_{k=0..n} F(k)^2, n >= 0. See the above statement and proof given by Philippe LALLOUET. a(n) = (2*A059840(n+2) - A027941(n))/3, n >= 0, with A059840(n+2) = Sum_{k=0..n} F(k)*F(k+2) and A027941(n) = A001519(n+1) - 1, n >= 0, where A001519(n+1) = F(2*n+1). (End) a(n) = (-1)^n * Sum_{k=0..n} (-1)^k*F(2*k), n >= 0. - Wolfdieter Lang, Aug 11 2012 a(-1-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2014 0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014 a(n) = (L(2*n+1) - (-1)^n)/5 with L(k) = A000032(k). - J. M. Bergot, Apr 15 2016 E.g.f.: ((3 + sqrt(5))*exp((5+sqrt(5))*x/2) - 2*exp((2*x)/(3+sqrt(5))+x) - 1 - sqrt(5))*exp(-x)/(5*(1 + sqrt(5))). - Ilya Gutkovskiy, Apr 15 2016 From Klaus Purath, Apr 24 2019: (Start) a(n) = A061646(n) - Fibonacci(n-1)^2. a(n) = (A061646(n+1) - A061646(n))/2. (End) a(n) = A226205(n+1) + (-1)^(n+1). - Flávio V. Fernandes, Apr 23 2020 Sum_{n>=1} 1/a(n) = A290565. - Amiram Eldar, Oct 06 2020 EXAMPLE G.f. = x + 2*x^2 + 6*x^3 + 15*x^4 + 40*x^5 + 104*x^6 + 273*x^7 + 714*x^8 + ... MAPLE with(combinat): A001654:=n->fibonacci(n)*fibonacci(n+1): seq(A001654(n), n=0..28); # Zerinvary Lajos, Oct 07 2007 MATHEMATICA LinearRecurrence[{2, 2, -1}, {0, 1, 2}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *) Times@@@Partition[Fibonacci[Range[0, 30]], 2, 1] (* Harvey P. Dale, Aug 18 2011 *) PROG (PARI) A001654(n)=fibonacci(n)*fibonacci(n+1); (PARI) b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j)); vector(30, n, b(n-1, 2))  \\ Joerg Arndt, May 08 2016 (Haskell) a001654 n = a001654_list !! n a001654_list = zipWith (*) (tail a000045_list) a000045_list -- Reinhard Zumkeller, Jun 08 2013 (Python) from sympy import fibonacci as F def a(n): return F(n)*F(n + 1) [a(n) for n in range(101)] # Indranil Ghosh, Aug 03 2017 (Python) from math import prod from gmpy2 import fib2 def A001654(n): return prod(fib2(n+1)) # Chai Wah Wu, May 19 2022 (Magma) I:=[0, 1, 2]; [n le 3 select I[n] else 2*Self(n-1) + 2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 17 2018 CROSSREFS Cf. A001655, A001656, A001657, A001658, A010048, A067962. Cf. A005968, A005969, A098531, A098532, A098533, A119283, A128697. Cf. A000071, A079472, A080145, A290565. Bisection of A006498, A070550, A080239. First differences of A064831. Partial sums of A007598. Sequence in context: A259399 A307128 A172399 * A062106 A206000 A316981 Adjacent sequences:  A001651 A001652 A001653 * A001655 A001656 A001657 KEYWORD nonn,easy AUTHOR EXTENSIONS Extended by Wolfdieter Lang, Jun 27 2000 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified September 25 12:12 EDT 2022. Contains 356984 sequences. (Running on oeis4.)