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

%I A130747 #35 Apr 25 2025 06:28:11
%S A130747 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,
%T A130747 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,
%U A130747 30,1,31,11,32,6,33,4,34,12,35,3,36,2,37,13,38,7,39,1,40,14,41
%N A130747 A self-referential sequence related to Mancala solitaire (see comment).
%C A130747 To build the sequence, start from:
%C A130747 1,_,2,_,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
%C A130747 At the n-th step use the rule: " fill a(n)-th hole with a(n) " (holes are numbered from 1 at each step)
%C A130747 So step 1 is "fill first hole with 1", giving:
%C A130747 1,1,2,_,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
%C A130747 Since a(2)=1, step 2 is still "fill first hole with 1", giving:
%C A130747 1,1,2,1,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
%C A130747 Since a(3)=2, step 3 is "fill second hole with 2", giving:
%C A130747 1,1,2,1,3,_,4,2,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
%C A130747 Since a(4)=1, step 4 is "fill first hole with 1", giving:
%C A130747 1,1,2,1,3,1,4,2,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
%C A130747 Since a(5)=3, step 5 is "fill third hole with 3", giving:
%C A130747 1,1,2,1,3,1,4,2,5,_,6,_,7,3,8,_,9,_,10,_,11,_,12,_,...
%C A130747 Iterating the process indefinitely yields:
%C A130747 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,...
%C A130747 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.
%C A130747 Ordinal transform of A028920. - _Benoit Cloitre_, Aug 03 2007
%C A130747 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
%C A130747 a(A002491(n)) = 1. - _Reinhard Zumkeller_, Jun 23 2009
%C A130747 A082447(n) = number of ones <= n. - _Reinhard Zumkeller_, Jul 01 2009
%C A130747 From _Benoit Cloitre_, Jul 17 2022: (Start)
%C A130747 Another way (less self-referent) to construct the sequence.
%C A130747 Step 1: Let's start from the integers separated by a hole:
%C A130747 1,_,2,_,3,_,4,_,5,_,6,_,7,_,8,_,9,_,10,_,11,_,12,_,...
%C A130747 Step 2: Put integers in the holes leaving 2 holes between each integer giving:
%C A130747 1,*1*,2,_,3,_,4,*2*,5,_,6,_,7,*3*,8,_,9,_,10,*4*,11,_,12,_,...
%C A130747 Step 3: Put integers in the holes leaving 3 holes between each integer giving:
%C A130747 1,1,2,*1*,3,_,4,2,5,_,6,_,7,3,8,*2*,9,_,10,4,11,_,12,_,...
%C A130747 Step 4: Put integers in the holes leaving 4 holes between each integer giving:
%C A130747 1,1,2,1,3,*1*,4,2,5,_,6,_,7,3,8,2,9,_,10,4,11,_,12,*2*,...
%C A130747 Iterating the process yields the sequence
%C A130747 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)
%D A130747 Benoit Cloitre, Pi in a hole, in preparation, 2007
%D A130747 Y. David, On a sequence generated by a sieving process, Riveon Lematematika, 11(1957), 26-31.
%D A130747 Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.4.7.
%H A130747 Reinhard Zumkeller, <a href="/A130747/b130747.txt">Table of n, a(n) for n = 1..10000</a>
%H A130747 Franklin T. Adams-Watters, <a href="/A002260/a002260.txt">Doubly Fractal Sequences and ordinal transform</a>
%H A130747 D. Betten, <a href="https://doi.org/10.1016/S0167-5060(08)70224-3">Kalahari and the Sequence "Sloane No. 377"</a>, Annals Discrete Math., 37, 51-58, 1988.
%H A130747 P. Erdős and E. Jabotinsky, <a href="https://users.renyi.hu/~p_erdos/1958-08.pdf">On sequences of integers generated by a sieving process, I</a>, Indagationes Math., 20, 115-123, 1958.
%H A130747 P. Erdős and E. Jabotinsky, <a href="https://users.renyi.hu/~p_erdos/1958-09.pdf">On sequences of integers generated by a sieving process, II</a>, Indagationes Math., 20, 124-128, 1958.
%t A130747 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 *)
%Y A130747 Cf. A002491.
%K A130747 nice,nonn
%O A130747 1,3
%A A130747 _Benoit Cloitre_, Jul 12 2007