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A035614
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Horizontal para-Fibonacci sequence: says which column of Wythoff array (starting column count at 0) contains n+1.
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14
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0, 1, 2, 0, 3, 0, 1, 4, 0, 1, 2, 0, 5, 0, 1, 2, 0, 3, 0, 1, 6, 0, 1, 2, 0, 3, 0, 1, 4, 0, 1, 2, 0, 7, 0, 1, 2, 0, 3, 0, 1, 4, 0, 1, 2, 0, 5, 0, 1, 2, 0, 3, 0, 1, 8, 0, 1, 2, 0, 3, 0, 1, 4, 0, 1, 2, 0, 5, 0, 1, 2, 0, 3, 0, 1, 6, 0, 1, 2, 0, 3
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listen;
history;
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internal format)
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OFFSET
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0,3
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COMMENTS
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This is probably the same as the "Fibonacci ruler function" mentioned by Knuth. - N. J. A. Sloane, Aug 03 2012
a(n) is the number of the trailing zeros in the Zeckendorf representation of (n+1) (A014417).
The asymptotic density of the occurrences of k is 1/phi^(k+2), where phi is the golden ratio (A001622).
The asymptotic mean of this sequence is phi. (End)
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REFERENCES
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D. E. Knuth, The Art of Computer Programming, Vol. 4A, Section 7.1.3, p. 82, solution to Problem 179.
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LINKS
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FORMULA
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The segment between the first M and the first M+1 is given by the segment before the first M-1.
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MATHEMATICA
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max = 81; wy = Table[(n-k)*Fibonacci[k] + Fibonacci[k+1]*Floor[ GoldenRatio*(n - k + 1)], {n, 1, max}, {k, 1, n}]; a[n_] := Position[wy, n][[1, 2]]-1; Table[a[n], {n, 1, max}] (* Jean-François Alcover, Nov 02 2011 *)
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PROG
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(Haskell)
(Python)
from sympy import fibonacci
def a122840(n): return len(str(n)) - len(str(int(str(n)[::-1])))
def a014417(n):
k=0
x=0
while n>0:
k=0
while fibonacci(k)<=n: k+=1
x+=10**(k - 3)
n-=fibonacci(k - 1)
return x
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CROSSREFS
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KEYWORD
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nonn,nice,easy
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AUTHOR
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STATUS
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approved
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