A126723 - OEIS

A126723

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.

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, 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, 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

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OFFSET

1,3

COMMENTS

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(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.

LINKS

Table of n, a(n) for n=1..100.

Max Alekseyev, Re: Representations found by the greedy algorithm, SeqFan Mailing List, Dec 19 2006

EXAMPLE

First five rows:

1 1

3 0 0 1

4 1 0 1

6 0 1 0 0 1

8 0 0 0 0 1

Row 4 matches 6 = 6/x + 0/x^2 + 1/x^3 + 0/x^4 + 0/x^5 + 1/x^6.

CROSSREFS