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Search: id:A126723
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| A126723 |
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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. |
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+0
1
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| 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
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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(1) In a message to seqfans(AT)seqfan.net (Dec 19, 2006), Max Alekseyev proved the following: suppose that N = c(1)F(1) - c(2)F(2) + c(3)F(3) - ..., where F(i) are Fibonaccci 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.
(2) 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.
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EXAMPLE
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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.
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CROSSREFS
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Cf. A000045, A000201.
Sequence in context: A090622 A035696 A035630 this_sequence A090030 A080159 A144299
Adjacent sequences: A126720 A126721 A126722 this_sequence A126724 A126725 A126726
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KEYWORD
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nonn,tabf
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AUTHOR
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Clark Kimberling (ck6(AT)evansville.edu), Dec 23 2006
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