

A025480


a(2n) = n, a(2n+1) = a(n).


44



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
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OFFSET

0,5


COMMENTS

These are the Grundy values or nimvalues 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. See Levine 2004/2006 for more about this.  N. J. A. Sloane, Aug 14 2016
When n>0 is written as (2k+1)*2^j then k=A000265(n1) and j=A007814(n), so: when n is written as (2k+1)*2^j1 then k=A025480(n) and j=A007814(n+1), when n>1 is written as (2k+1)*2^j+1 then k=A025480(n2) and j=A007814(n1).  Henry Bottomley, Mar 02 2000
According to the comment from Deuard Worthen (see Example section), this may be regarded as a triangle where row r=1,2,3... has length 2^(r1) and values T[r,2k1]=T[r1,k], T[r,2k]=2^(r1)+k1, i.e., previous row gives first, 3rd, 5th, ... term and 2nd, 4th, ... terms are numbers 2^(r1),...,2^r1 (i.e., those following the last one from the previous row).  M. F. Hasler, May 03 2008
Let StB be a SternBrocot tree hanging between (pseudo)fractions Left and Right, then StB(1) = mediant(Left,Right) and for n>1: StB(n) = if a(n1)<>0 and a(n)<>0 then mediant(StB(a(n1)),StB(a(n))) else if a(n)=0 then mediant(StB(a(n1)),Right) else mediant(Left,StB(a(n1))), where mediant(q1,q2) = ((numerator(q1)+numerator(q2)) / (denominator(q1)+denominator(q2))).  Reinhard Zumkeller, Dec 22 2008
This sequence is the unique fixed point of the function (a(0), a(1), a(2), ...) > (0, a(0), 1, a(1), 2, a(2), ...) which interleaves the nonnegative integers between the elements of a sequence.  Cale Gibbard (cgibbard(AT)gmail.com), Nov 18 2009
The following relation holds: 2^A007814(n+1)*(2*A025480(n)+1)=A001477(n+1). (See functions hd,tl and cons in [Paul Tarau 2009].)  Paul Tarau (paul.tarau(AT)gmail.com), Mar 21 2010
Also the number of remaining survivors in a Josephus problem after the person originally first in line has been eliminated (see A225381).  Marcus Hedbring, May 18 2013
A fractal sequence  see Levine 2004/2006.  N. J. A. Sloane, Aug 14 2016
From David James Sycamore, Apr 29 2020 : (Start)
One of a family of fractal sequences, S_k; defined as follows for k>=2: a(k*n) = n, a(k*n+r) = a((k1)*n + (r1)), r = 1..(k1). S_2 is A025480; S_3 gives: a(3*n) = n, a(3*n + 1) = a(2*n), a(3*n + 2) = a(2*n + 1), which is A263390 (whose name is not expressed entirely in this form, but it is the same sequence).
Sequences corresponding to higher k values do not appear to be recorded in oeis. The subsequence of all non zero terms is A131987. Thus A025480 has a proper subsequence identical to itself and another proper subsequence different from itself, which is also fractal (End).
Similar to but different from A108202.  N. J. A. Sloane, Nov 26 2020


REFERENCES

L. Levine, Fractal sequences and restricted Nim, Ars Combin. 80 (2006), 113127.


LINKS

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Josef EschgfÃ¤ller, Andrea Scarpante, Dichotomic random number generators, arXiv:1603.08500 [math.CO], 2016.
R. Hinze, Concrete stream calculus: An extended study, J. Funct. Progr. 20 (56) (2010) 463535, doi, Section 3.2.4.
L. Levine, Fractal sequences and restricted Nim, arXiv:math/0409408 [math.CO], 2004.
R. Stephan, Some divideandconquer sequences ...
R. Stephan, Table of generating functions
Paul Tarau, A Groupoid of Isomorphic Data Transformations, Calculemus 2009, 8th International Conference, MKM 2009, pp. 170185, Springer, LNAI 5625.
Index entries for sequences related to the Josephus Problem


FORMULA

a(n) = (A000265(n+1)1)/2 = ((n+1)/A006519(n+1)1)/2.
a(n) = A153733(n)/2.  Reinhard Zumkeller, Dec 31 2008
a(3*n + 1) = A173732(n).  Reinhard Zumkeller, Apr 29 2012
a((2*n+1)*2^p1) = n, p >= 0 and n >= 0.  Johannes W. Meijer, Jan 24 2013
a(n) = n  A225381(n).  Marcus Hedbring, May 18 2013
G.f.: 1/(1x) + Sum_{k>=0} x^(2^k1)/(12*x^2^(k+1)+x^2^(k+2)).  Ralf Stephan, May 19 2013
a(n) = A049084(A181363(n+1)).  Reinhard Zumkeller, Mar 22 2014
a(n) = floor( n/ 2^A001511(n+1) ).  Adam Shelly, Mar 05 2019
Recursion: a(0) = 0; a(n + 1) = a(a(n)) if a(n) is a first occurrence of a term, else a(n + 1) = n  a(n1).  David James Sycamore, Apr 29 2020


EXAMPLE

From Deuard Worthen (deuard(AT)raytheon.com), Jan 27 2006: (Start)
The sequence can be constructed as a triangle as:
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 interleave the next 2^m numbers in the previous row. (End)
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.  Reinhard Zumkeller, Dec 22 2008


MAPLE

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:
nmax := 83: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 0 to ceil(nmax/(p+2))+1 do a((2*n+1)*2^p1) := n od: od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Jan 24 2013
A025480 := proc(n)
option remember ;
if type(n, 'even') then
n/2 ;
else
procname((n1)/2) ;
end if;
end proc: # R. J. Mathar, Jul 16 2020


MATHEMATICA

a[n_] := a[n] = If[OddQ@n, a[(n  1)/2], n/2]; Table[ a[n], {n, 0, 83}] (* Robert G. Wilson v, Mar 30 2006 *)
Table[BitShiftRight[n, IntegerExponent[n, 2] + 1], {n, 100}] (* IWABUCHI Yu(u)ki, Oct 13 2012 *)


PROG

(PARI) a(n)={while(n%2, n\=2); n\2} \\ M. F. Hasler, May 03 2008
(PARI) A025480(n)=n>>valuation(n*2+2, 2) \\ M. F. Hasler, Apr 12 2012
(Haskell)
import Data.List
interleave xs ys = concat . transpose $ [xs, ys]
a025480 = interleave [0..] a025480
 _Cale Gibbard_, Nov 18 2009:
(Haskell) Cf. comments by Worthen and Hasler.
import Data.List (transpose)
a025480 n k = a025480_tabf !! n !! k
a025480_row n = a025480_tabf !! n
a025480_tabf = iterate (\xs > concat $
transpose [xs, [length xs .. 2 * length xs  1]]) [0]
a025480_list = concat $ a025480_tabf
 Reinhard Zumkeller, Apr 29 2012
(Sage)
A025480 = lambda n: odd_part(n+1)//2
[A025480(n) for n in (0..83)] # Peter Luschny, May 20 2014


CROSSREFS

a(n) = A003602(n)1.
The Yprojection of A075300.
Cf. A108202, A138002, A000265, A003602, A103391, A153733, A220466, A225381, A131987, A263390.
Sequence in context: A334594 A062778 A108202 * A088002 A030109 A208571
Adjacent sequences: A025477 A025478 A025479 * A025481 A025482 A025483


KEYWORD

easy,nonn,nice,tabf,hear,changed


AUTHOR

David W. Wilson


EXTENSIONS

Edited by M. F. Hasler, Mar 16 2018


STATUS

approved



