0,3

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)planetuniqa.at), Sep 14 2005

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.

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.

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.

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 (qntmpkt(AT)yahoo.com), Oct 10 2004

Equals the partial sums of squares of Fibonnaci 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

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); [Conjectured by S. Plouffe in his 1992 dissertation.]

(PARI) a(n)=fibonacci(n)*fibonacci(n+1)

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.

First differences of A064831. Cf. A079472.

Cf. A119283, A000071, A005968, A005969, A098531, A098532, A098533, A128697.

Adjacent sequences: A001651 A001652 A001653 this_sequence A001655 A001656 A001657

Sequence in context: A001674 A121331 A026270 this_sequence A062106 A061322 A004664

nonn

njas

Extended by Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Jun 27 2000