A025480 a(2n) = n, a(2n+1) = a(n).

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

Offset: 0

Author

David W. Wilson

Keywords

• easy
• nonn
• nice
• tabf
• hear

Comments

These are the Grundy values or 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. 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 = a(n-1) and j = A007814(n), so: when n is written as (2k+1)*2^j-1 then k = a(n) and j = A007814(n+1), when n > 1 is written as (2k+1)*2^j+1 then k = a(n-2) and j = A007814(n-1). - Henry Bottomley, Mar 02 2000 [sequence id corrected by Peter Munn, Jun 22 2022]

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^(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, 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))). - 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

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((k-1)*n + (r-1)), r = 1..(k-1). 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.

The subsequence of all nonzero terms is A131987. (End)

Similar to but different from A108202. - N. J. A. Sloane, Nov 26 2020

This sequence can be otherwise defined in two alternative (but related) ways, with a(0)=0, as follows: (i) If a(n) is a novel term, then a(n+1) = a(a(n)); if a(n) has been seen before, most recently at a(m), then a(n+1) = n-m (as in A181391). (ii) As above for novel a(n), then if a(n) has been seen before, a(n+1) = smallest k < a(n) which is not already a term. - David James Sycamore, Jul 13 2021

From a binary perspective, the sequence can be seen as even,odd pairs where the odd value is the previous even value, dropping the rightmost bits up to and including the lowest zero bit, aka right-shifted past the lowest clear bit. E.g., (5)101 -> 1, (17)10001 -> (4)100, (29)11101 -> (7)111, (39)100111 -> (2)10. - Joe Nellis, Oct 09 2022

Examples

			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
		

References

• L. Levine, Fractal sequences and restricted Nim, Ars Combin. 80 (2006), 113-127.

Links

• N. J. A. Sloane, Table of n, a(n) for n = 0..10000
• Josef Eschgfäller and Andrea Scarpante, Dichotomic random number generators, arXiv:1603.08500 [math.CO], 2016.
• Ralf Hinze, Concrete stream calculus: An extended study, J. Funct. Progr. 20 (5-6) (2010) 463-535, doi, Section 3.2.4.
• L. Levine, Fractal sequences and restricted Nim, arXiv:math/0409408 [math.CO], 2004.
• Ralf Stephan, Some divide-and-conquer sequences ....
• Ralf Stephan, Table of generating functions.
• Paul Tarau, A Groupoid of Isomorphic Data Transformations, Calculemus 2009, 8th International Conference, MKM 2009, pp. 170-185, Springer, LNAI 5625.
• Index entries for sequences related to the Josephus Problem.

Crossrefs

The Y-projection of A075300.

Cf. A108202, A138002, A000265, A003602, A103391, A153733, A220466, A225381, A131987, A263390, A181391, A007814.

Programs

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

Maple

A025480 := proc(n)
    option remember ;
    if type(n,'even') then
        n/2 ;
    else
        procname((n-1)/2) ;
    end if;
end proc:
seq(A025480(n),n=0..100) ; # 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 *)

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

Python

def A025480(n): return n>>((~(n+1)&n).bit_length()+1) # Chai Wah Wu, Jul 13 2022

Sage

A025480 = lambda n: odd_part(n+1)//2
[A025480(n) for n in (0..83)] # Peter Luschny, May 20 2014

Formula

a(n) = A003602(n+1) - 1. [Corrected by Max Alekseyev, May 05 2022]

a(n) = (A000265(n+1)-1)/2 = ((n+1)/A006519(n+1)-1)/2.

a(n) = A153733(n)/2. - Reinhard Zumkeller, Dec 31 2008

2^A007814(n+1)*(2*a(n)+1) = n+1. (See functions hd, tl and cons in [Paul Tarau 2009].) - Paul Tarau (paul.tarau(AT)gmail.com), Mar 21 2010

a(3*n + 1) = A173732(n). - Reinhard Zumkeller, Apr 29 2012

a((2*n+1)*2^p-1) = 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/(1-x) + Sum_{k>=0} x^(2^k-1)/(1-2*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(n-1). - David James Sycamore, Apr 29 2020

a(n) * 2^(A007814(n+1)+1) + 2^A007814(n+1) - 1 = n (equivalent to the formula given in the comment by Paul Tarau). - Ruud H.G. van Tol, Apr 14 2023

Sum_{k=1..n} a(k) = n^2/6 + O(n). - Amiram Eldar, Aug 07 2023

Extensions

Edited by M. F. Hasler, Mar 16 2018