A157129 -id:A157129 - cp's OEIS frontend

A229785 Partial sums of A157129.

1, 2, 4, 6, 7, 8, 9, 10, 12, 14, 16, 18, 19, 20, 22, 24, 25, 26, 28, 30, 31, 32, 33, 34, 36, 38, 40, 42, 43, 44, 45, 46, 48, 50, 52, 54, 55, 56, 58, 60, 61, 62, 63, 64, 66, 68, 70, 72, 73, 74, 76, 78, 79, 80, 81, 82, 84, 86, 88, 90, 91, 92, 94, 96, 97, 98, 100, 102, 103, 104

Offset: 1

Benoit Cloitre, Sep 29 2013

nonn

Although the behavior of the partial sums of the Kolakoski sequence (A054353) is mysterious, this sequence is much easier to handle.

Cf. A054353, A157129.

a(n)=(3/2)n+O(1). More precisely, let b(n)=3*n-2*a(n); then b(n) satisfies the following recurrence modulo 12: b(n)=1,2,1,0,1,2,3,4,3,2,1 for n=1,2,3,4,5,6,7,8,9,10,11. Then for k>=1 we have b(12k)=b(4k), b(12k+1)=b(4k+1), b(12k+2)=b(4k+2), b(12k+2)=b(4k+2), b(12k+3)=b(4k+2)-1, b(12k+4)=b(4k+2)-2, b(12k+5)=b(4k+2)-1, b(12k+6)=b(4k+2), b(12k+7)=4-b(4k+3), b(12k+8)=4-b(4k+4), b(12k+9)=4-b(4k+3), b(12k+10)=4-b(4k+2), b(12k+11)=b(4k+3).

A157196 a(n) = (1/2)*(sum of elements of n-th run) using 1 and 2 starting with 1,1.

Original entry on oeis.org

1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1

Offset: 1

Benoit Cloitre, Feb 24 2009

nonn

We conjecture that the density of 1's in the sequence approaches 2/3 as n -> infinity. This conjecture is proved in the paper of Shallit.

			Write the sums of elements in each run, you obtain: 2,2,4,2,2,2,2,4,2,2,4,2,2,4,4,... dividing by 2 you get: 1,1,2,1,1,1,1,2,1,1,2,1,1,2,2,... the sequence itself.

Daniel Hoyt, Visual representation of the sequence in the style of the Kolakoski sequence.
Daniel Hoyt, Representing the sequence visually.
Jeffrey Shallit, Cloitre's Self-Generating Sequence, arXiv:2501.00784 [math.CO], 2025.

Cf. A000002, A157129, A336810.

mx:= 1000: l:= [1$2]: a:= n-> l[n]:
for h from 2 while nops(l)Alois P. Heinz, May 31 2012

More terms from Alvin Hoover Belt, May 31 2012

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A229785 Partial sums of A157129.

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A157196 a(n) = (1/2)*(sum of elements of n-th run) using 1 and 2 starting with 1,1.

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