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A001654
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Golden rectangle numbers: F(n)*F(n+1), where F(n) = A000045(n) (Fibonacci numbers).
(Formerly M1606 N0628)
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74
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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; internal format)
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OFFSET
| 0,3
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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..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
a(n+2) is the Hankel transform of A005807 aerated. [From Paul Barry (pbarry(AT)wit.ie), Nov 04 2008]
Contribution from Daniel Forgues, Nov 29 2009: (Start)
A more exact name would be: Golden convergents 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 of the continued fraction expansion of Phi, whence
ratio of sides converges to Phi.) (End)
The Kn4 sums, see A180662 for the definition of these sums, of the ‘Races with Ties’ triangle A035317 lead to this sequence. [Johannes W. Meijer, Jul 20 2011]
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REFERENCES
| A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9.
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 17.
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).
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LINKS
| T. D. Noe, Table of n, a(n) for n=0..200
S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
M. Renault, Dissertation
Wikipedia, Illustration of 273 as a golden rectangle number.
Index entries for two-way infinite sequences
Index entries for sequences related to linear recurrences with constant coefficients, signature (2,2,-1).
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FORMULA
| a(n)= A010048(n+1, 2)= fibonomial(n+1, 2).
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
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/(1-2x-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).
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 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
From Daniel Forgues, Nov 29 2009: (Start)
With Phi(n) = [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^(2n+1))/5] = Floor[(1/2) + (Phi^(2n+1))/5], n >= 0. (End)
a(n)=2*a(n-1)+2*a(n-2)-a(n-3) a(1)=1 a(2)=2 a(3)=6 [From Sture Sjostedt (sture.sjostedt(AT)spray.se), Feb 06 2010]
a(n) = (A002878(n)-(-1)^n)/5. [From 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,..) [From Tomasz Ordowski (TomORDO(AT)gmail.com), Nov 04 2010]
a(n)= (fibonacci(n+2)^2-fibonacci(n-1)^2)/4 [From Gary Detlefs, Dec 03 2010]
Let d(n)=Mod(n,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
Closed-form without leading zero ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5. Closed-form with leading zero ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n+2*(-1)^n)/10. - Tim Monahan, Jul 11 2011
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MAPLE
| with (combinat):a:=n->fibonacci(n)*fibonacci(n+1): seq(a(n), n=0..28); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 07 2007
A001654:=1/(z+1)/(z**2-3*z+1); [S. Plouffe in his 1992 dissertation, omitting a(0).]
with(combinat): seq(1/4*(fibonacci(n+2)^2-fibonacci(n-1)^2), n=0..27) [From Gary Detlefs (gdetlefs(at)aol.com) Dec 03 2010]
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MATHEMATICA
| q=0; lst={}; Do[f=Fibonacci[n]; AppendTo[lst, f*q]; q=f, {n, 5!}]; lst [From Vladimir Orlovsky, Jul 21 2009]
LinearRecurrence[{2, 2, -1}, {0, 1, 2}, 100] (* From Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
Times@@@Partition[Fibonacci[Range[0, 30]], 2, 1] (* From Harvey P. Dale, Aug 18 2011 *)
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PROG
| (PARI) A001654(n)=fibonacci(n)*fibonacci(n+1)
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CROSSREFS
| Cf. A010048, A001655-A001658. A006498(2n-1)=a(n).
Bisection of A006498, A070550, A080239. Cf. A079472, A080145.
First differences of A064831. Partial sums of A007598.
Cf. A119283, A000071, A005968, A005969, A098531, A098532, A098533, A128697.
Sequence in context: A121331 A026270 A172399 * A062106 A206000 A061322
Adjacent sequences: A001651 A001652 A001653 * A001655 A001656 A001657
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KEYWORD
| nonn,easy,changed
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| Extended by Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Jun 27 2000
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