A130747 A self-referential sequence related to Mancala solitaire (see comment).

1, 1, 2, 1, 3, 1, 4, 2, 5, 1, 6, 1, 7, 3, 8, 2, 9, 1, 10, 4, 11, 1, 12, 2, 13, 5, 14, 3, 15, 1, 16, 6, 17, 1, 18, 2, 19, 7, 20, 4, 21, 1, 22, 8, 23, 3, 24, 1, 25, 9, 26, 5, 27, 2, 28, 10, 29, 1, 30, 1, 31, 11, 32, 6, 33, 4, 34, 12, 35, 3, 36, 2, 37, 13, 38, 7, 39, 1, 40, 14, 41

Offset

1,3

Comments

To build the sequence, start from:

1,_,2,_,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...

At the n-th step use the rule: " fill a(n)-th hole with a(n) " (holes are numbered from 1 at each step)

So step 1 is "fill first hole with 1", giving:

1,1,2,_,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...

Since a(2)=1, step 2 is still "fill first hole with 1", giving:

1,1,2,1,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...

Since a(3)=2, step 3 is "fill second hole with 2", giving:

1,1,2,1,3,_,4,2,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...

Since a(4)=1, step 4 is "fill first hole with 1", giving:

1,1,2,1,3,1,4,2,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...

Since a(5)=3, step 5 is "fill third hole with 3", giving:

1,1,2,1,3,1,4,2,5,_,6,_,7,3,8,_,9,_,10,_,11,_,12,_,...

Iterating the process indefinitely yields:

1,1,2,1,3,1,4,2,5,1,6,1,7,3,8,2,9,1,10,4,11,1,12,2,13,5,...

Indices where 1's occur are n=1,2,4,6,10,... which are the smallest number of stones in Mancala solitaire which make use of the n-th hole. If f(k) denotes this sequence then lim_{k->oo} k^2/f(k) = Pi.

Ordinal transform of A028920. - Benoit Cloitre, Aug 03 2007

Although A028920 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

a(A002491(n)) = 1. - Reinhard Zumkeller, Jun 23 2009

A082447(n) = number of ones <= n. - Reinhard Zumkeller, Jul 01 2009

From Benoit Cloitre, Jul 17 2022: (Start)

Another way (less self-referent) to construct the sequence.

Step 1: Let's start from the integers separated by a hole:

1,_,2,_,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...

Step 2: Put integers in the holes leaving 2 holes between each integer giving:

1,*1*,2,_,3,_,4,*2*,5,_,6,_,7,*3*,8,_,9,_,10,*4*,11,_,12,_,...

Step 3: Put integers in the holes leaving 3 holes between each integer giving:

1,1,2,*1*,3,_,4,2,5,_,6,_,7,3,8,*2*,9,_,10,4,11,_,12,_,...

Step 4: Put integers in the holes leaving 4 holes between each integer giving:

1,1,2,1,3,*1*,4,2,5,_,6,_,7,3,8,2,9,_,10,4,11,_,12,*2*,...

Iterating the process yields the sequence

1, 1, 2, 1, 3, 1, 4, 2, 5, 1, 6, 1, 7, 3, 8, 2, 9, 1, 10, 4, 11, 1, 12, 2,... (End)

References

B. Cloitre, Pi in a hole, in preparation, 2007

Y. David, On a sequence generated by a sieving process, Riveon Lematematika, 11(1957), 26-31.

S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.4.7.

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

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

D. Betten, Kalahari and the Sequence "Sloane No. 377", Annals Discrete Math., 37, 51-58, 1988.

P. Erdõs and E. Jabotinsky, On sequences of integers generated by a sieving process, I, Indagationes Math., 20, 115-123, 1958.

P. Erdõs and E. Jabotinsky, On sequences of integers generated by a sieving process, II, Indagationes Math., 20, 124-128, 1958.

Mathematica

max = 100; A130747 = Flatten[ Transpose[ {Range[max], Table[0, {max}]}]]; Do[ hole = Last[ Position[ A130747, 0, 1, A130747[[n]] ]]; A130747[[hole]] = A130747[[n]], {n, 1, max}]; A130747 (* Jean-François Alcover, Dec 08 2011 *)

Crossrefs

Cf. A002491.

Sequence in context: A078898 A246277 A260739 * A370822 A055440 A250028

Adjacent sequences: A130744 A130745 A130746 * A130748 A130749 A130750

Keyword

nice,nonn

Author

Benoit Cloitre, Jul 12 2007

Status

approved