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Search: id:A025480
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| A025480 |
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a(2n) = n, a(2n+1) = a(n). |
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+0
26
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| 0, 0, 1, 0, 2, 1, 3, 0, 4, 2, 5, 1, 6, 3, 7, 0, 8, 4, 9, 2, 10, 5, 11, 1, 12, 6, 13, 3, 14, 7, 15, 0, 16, 8, 17, 4, 18, 9, 19, 2, 20, 10, 21, 5, 22, 11, 23, 1, 24, 12, 25, 6, 26, 13, 27, 3, 28, 14, 29, 7, 30, 15, 31, 0, 32, 16, 33, 8, 34, 17, 35, 4, 36, 18, 37, 9, 38, 19, 39, 2, 40, 20, 41, 10
(list; graph; listen)
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OFFSET
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0,5
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COMMENT
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These are the nim-values for heaps of n beans in the game where you're allowed to take up to half of the beans in a heap. - R. K. Guy, Mar 30 2006
When n>0 is written as (2k+1)*2^j then k=A000265(n-1) and j=A007814(n), so: when n is written as (2k+1)*2^j-1 then k=A025480(n) and j=A007814(n+1), when n>1 is written as (2k+1)*2^j+1 then k=A025480(n-2) and j=A007814(n-1)
According to the comment from Deuard Worthen, this may be regarded as a triangle where row r=1,2,3... has length 2^(r-1) and values T[r,2k-1]=T[r-1,k], T[r,2k]=2^(r-1)+k-1, i.e. previous row gives 1st, 3rd, 5th... term and 2nd, 4th... terms are numbers 2^(r-1),...,2^r-1 (i.e. those following the last one from the previous row). - M. F. Hasler (www.univ-ag.fr/~mhasler), May 03 2008
Let StB be a Stern-Brocot tree hanging between (pseudo)fractions Left and Right, then StB(1) = mediant(Left,Right) and for n>1: StB(n) = if a(n-1)<>0 and a(n)<>0 then mediant(StB(a(n-1)),StB(a(n))) else if a(n)=0 then mediant(StB(a(n-1)),Right) else mediant(Left,StB(a(n-1))), where mediant(q1,q2) = ((numerator(q1)+numerator(q2)) / (denominator(q1)+denominator(q2))). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Dec 22 2008]
a(n) = A153733(n)/2. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Dec 31 2008]
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REFERENCES
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L. Levine, Fractal sequences and restricted Nim, Ars Comb., Ars Combin. 80 (2006), 113-127.
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LINKS
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N. J. A. Sloane, Table of n, a(n) for n = 0..10000
L. Levine, Fractal sequences and restricted Nim
R. Stephan, Some divide-and-conquer sequences ...
R. Stephan, Table of generating functions
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FORMULA
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a(n) = (A000265(n+1)-1)/2 = ((n+1)/A006519(n+1)-1)/2
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EXAMPLE
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Comment from Deuard Worthen (deuard(AT)raytheon.com), Jan 27 2006: This sequence can be constructed as a triangle, thus:
0
0 1
0 2 1 3
0 4 2 5 1 6 3 7
0 8 4 9 2 10 5 11 1 12 6 13 3 14 7 15
...
-at each stage we interpolate the next 2^m numbers in the previous row.
Left=0/1, Right=1/0: StB=A007305/A047679; Left=0/1, Right=1/1: StB=A007305/A007306; Left=1/3, Right=2/3: StB=A153161/A153162. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Dec 22 2008]
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MAPLE
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a:=array[0..10001]; M:=5000; for n from 0 to M do a[2*n]:=n; a[2*n+1]:=a[n]; od: for n from 0 to 2*M do lprint(n, a[n]); od:
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MATHEMATICA
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a[n_] := a[n] = If[OddQ@n, a[(n - 1)/2], n/2]; Table[ a[n], {n, 0, 83}] (from Robert G. Wilson v (rgwv(at)rgwv.com), Mar 30 2006)
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PROGRAM
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(PARI) A025480(n)={while(n%2, n\=2); n\2} - M. F. Hasler (www.univ-ag.fr/~mhasler), May 03 2008
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CROSSREFS
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The Y-projection of A075300.
a(n) = A003602(n)-1.
Cf. A108202.
Cf. A138002.
Sequence in context: A081171 A062778 A108202 this_sequence A088002 A030109 A058208
Adjacent sequences: A025477 A025478 A025479 this_sequence A025481 A025482 A025483
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KEYWORD
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easy,nonn,nice,tabf
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AUTHOR
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David W. Wilson (davidwwilson(AT)comcast.net)
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EXTENSIONS
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Additional comments from Henry Bottomley (se16(AT)btinternet.com), Mar 02 2000
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