A028920 Pit harvesting sequence for winning solitaire Tchoukaillon (or Mancala).

1, 2, 1, 3, 1, 4, 1, 2, 1, 5, 1, 6, 1, 2, 1, 3, 1, 7, 1, 2, 1, 8, 1, 4, 1, 2, 1, 3, 1, 9, 1, 2, 1, 10, 1, 5, 1, 2, 1, 3, 1, 11, 1, 2, 1, 4, 1, 12, 1, 2, 1, 3, 1, 6, 1, 2, 1, 13, 1, 14, 1, 2, 1, 3, 1, 4, 1, 2, 1, 5, 1, 7, 1, 2, 1, 3, 1, 15, 1, 2, 1, 16, 1, 4, 1, 2, 1, 3, 1, 8, 1, 2, 1, 6, 1, 5, 1, 2, 1, 3, 1, 17, 1

Offset

0,2

Comments

From Benoit Cloitre, Mar 09 2007: (Start)

The sequence can be constructed as follows using parentheses (NP means "term not in parentheses"):

Start from the positive integers:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,...

Step 1: put the least NP "1" in parentheses and every 2 terms giving:

(1),2,(3),4,(5),6,(7),8,(9),10,(11),12,(13),14,(15),16,(17),18,(19),...

Step 2: put the least NP "2" in 2 parentheses and every 3 NP giving:

(1),((2)),(3),4,(5),6,(7),((8)),(9),10,(11),12,(13),((14)),(15),16,(17),...

so that between 2 consecutives ((x)) there are 2 NP.

Step 3: put the least NP "4" in 3 parentheses and every 4 NP giving:

(1),((2)),(3),(((4))),(5),6,(7),((8)),(9),10,(11),12,(13),((14)),(15),(((16))),...

so that between 2 consecutives (((x))) there are 3 NP.

Step 4: put the least NP "6" in 4 parentheses and every 5 NP giving:

(1),((2)),(3),(((4))),(5),((((6)))),(7),((8)),(9),10,(11),12,(13),((14)),(15),(((16))),...

so that between 2 consecutives ((((x)))) there are 4 NP.

Iterating the process indefinitely yields:

(1),((2)),(3),(((4))),(5),((((6)))),(7),((8)),(9),(((((10))))),(11),...

Count the parentheses:

1,2,1,3,1,4,1,2,1,5,1,... - this is the sequence.

(End)

From Benoit Cloitre, Jul 26 2007: (Start)

A simpler way to construct the sequence: start from

1,_,1,_,1,_,1,_,1,_,1,_,1,_,1,... where 1's are spaced by one hole;

fill first hole with 2 and leave 2 holes between two 2's giving

1,2,1,_,1,_,1,2,1,_,1,_,1,2,1,...;

fill new first hole with 3 and leave 3 holes between two 3's giving

1,2,1,3,1,_,1,2,1,_,1,_,1,2,1,3...;

iterating the process indefinitely yields the sequence.

(End)

Ordinal transform of A130747. - Benoit Cloitre, Aug 03 2007

Although this sequence and A130747 are not fractal sequences (according to Kimberling's definition), we say they are "mutual fractal sequences" since the ordinal transform of one gives the other. - Benoit Cloitre, Aug 03 2007

The smallest n with a(n) = k is circa k^2/Pi.

The element n >= 0 occurs in this sequence with limiting density 1/(n*(n+1)).

Links

L. K. Mitchell, Table of n, a(n) for n=0..3280

Franklin T. Adams-Watters, Doubly Fractal Sequences and ordinal transform

D. M. Broline and Daniel E. Loeb, The combinatorics of Mancala-Type games: Ayo, Tchoukaillon and 1/Pi, arXiv:math/9502225 [math.CO], 1995; J. Undergrad. Math. Applic., vol. 16 (1995), pp. 21-36.

N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).

Index entries for sequences generated by sieves

Formula

a(2n+1) = 1 + A104706(n+1), a(2n) = 1. - Benoit Cloitre, Mar 09 2007

The sieve of A007952 processes n in the a(n)-th pass. a(A007952(n)) = n+1.

Mathematica

n = 15; Fold[If[Length@Position[#1, 0] > 0, ReplacePart[#1, First /@ Partition[Position[#1, 0], #2 + 1, #2 + 1, {1, 1}] -> #2], #1] &,  Flatten@Array[{1, 0} &, n], Range[2, 2 n]] (* Birkas Gyorgy, Feb 26 2011 *)

Prog

(C++)
int A028920(int n) {
  for (int m = 1; ; m++) {
    if (n%(m+1) == 0)
      return m;
    n = n*m/(m+1);
  }
} /* David W. Wilson, Feb 25 2010 */
(PARI) a(n) = {ok = 0; m = 1; while (!ok, if ((n%(m+1) == 0), ok = 1, n = n*m\(m+1); m++); ); m; } \\ Michel Marcus, Dec 06 2015

Crossrefs

Cf. A002491, A007952, A028931, A028932, A028933, A130747.

Sequence in context: A128267 A289508 A364448 * A260738 A055396 A363486

Adjacent sequences: A028917 A028918 A028919 * A028921 A028922 A028923

Keyword

nonn

Author

David W. Wilson

Extensions

Additional comments from David W. Wilson, Feb 25 2010

Status

approved