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%I #14 May 17 2017 20:10:19
%S 1,1,3,0,0,1,4,1,0,1,6,0,1,0,0,1,8,0,0,0,0,1,9,1,0,0,0,1,11,0,0,1,0,1,
%T 12,1,0,1,0,1,14,0,1,0,1,0,0,1,16,0,0,0,1,0,0,1,17,1,0,0,1,0,0,1,19,0,
%U 1,0,0,0,0,1,21,0,0,0,0,0,0,1,22,1,0,0,0,0,0,1,24,0,0,1,0,0,0,1,25,1,0,1
%N Successive rows of coefficients c(0), c(1), c(2),... for the greedy-algorithm representation of a positive integer n: n = c(0)/x + c(1)/x^2 + c(2)/x^3 + ..., where x = (1+sqrt(5))/2.
%C _Max Alekseyev_ (see link below) proved the following: suppose that N = c(1)F(1) - c(2)F(2) + c(3)F(3) - ..., where F(i) are Fibonacci numbers and each coefficient c(i) is either 0 or 1 with no adjacent unit coefficients. Then these coefficients are exactly those produced by the greedy algorithm: N = c(0)/x + c(1)/x^2 + c(2)/x^3 + ... . It follows that there are only finitely many nonzero terms and that the representation is unique for the stated properties.
%C c(0)=Floor(N*x) (as in A000201, the lower Wythoff sequence). Thus as N*x-c(0) is the fractional part {N*x} of N*x, we have {N*x} represented as a sum of finitely many fractions 1/x^k.
%H Max Alekseyev, <a href="/A126723/a126723_1.txt">Re: Representations found by the greedy algorithm</a>, SeqFan Mailing List, Dec 19 2006
%e First five rows:
%e 1 1
%e 3 0 0 1
%e 4 1 0 1
%e 6 0 1 0 0 1
%e 8 0 0 0 0 1
%e Row 4 matches 6 = 6/x + 0/x^2 + 1/x^3 + 0/x^4 + 0/x^5 + 1/x^6.
%Y Cf. A000045, A000201.
%K nonn,tabf
%O 1,3
%A _Clark Kimberling_, Dec 23 2006
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