A156077 -id:A156077 - cp's OEIS frontend

A289322 Number of 1s in the first 2^n entries of the Kolakoski sequence, A000002.

1, 1, 2, 4, 8, 17, 32, 64, 129, 256, 513, 1024, 2051, 4093, 8192, 16381, 32746, 65523, 131082, 262168, 524262, 1048547, 2097100, 4194345, 8388733, 16777351, 33554669, 67109796, 134219275, 268437750, 536872179

Offset: 0

Richard P. Brent, Jul 05 2017

nonn

			The first 32 entries of the Kolakoski sequence, A000002, are 12211212212211211221211212211211. From this we see that a(5)=17, since among the first 2^5 letters, 17 of them are 1s.

Richard P. Brent, Table of n, a(n) for n = 0..64
Richard P. Brent and Judy-anne H. Osborn, A fast algorithm for the Kolakoski sequence, Dec. 2016
J. Nilsson, A Space Efficient Algorithm for the Calculation of the Digit Distribution in the Kolakoski Sequence, arXiv preprint arXiv:1110.4228 [math.CO], 2011.
M. Rao, Trucs et bidules sur la séquence de Kolakoski, Oct. 2012.

Cf. A000002. Analogous for powers of ten is A195206. Equivalent but with smaller entries is A289323. Closely related are A054353, A074286, A088568, A156077.

a(n) = (2^n + A088568(2^n))/2 = (2^n - A289323(n))/2.

A289324 Number of twos minus number of ones in the first 10^n entries of the Kolakoski sequence, A000002.

Original entry on oeis.org

-1, 0, 2, -4, 8, 56, 28, -92, -1350, -2446, 4658, -3174, -101402, -16318, -632474, -1954842, 10724544, 45041304, 111069790, 548593100, 1818298480

Offset: 0

Richard P. Brent, Jul 07 2017

This is equivalent to A195206, since a(n) = (#twos)-(#ones) = 10^n-2*(#ones) in the first 10^n entries of A000002.
For example, a(2) = 51 - 49 = (100 - 49) - 49 = 100 - 2*49 = 2 because there are 49 ones and 51 twos in the first 100 = 10^2 entries of A000002.
The entries in this sequence appear to be of order 10^(n/2), whereas the entries in A195206 are larger (of order 10^n).
This sequence is analogous to A289323; the difference is that the indices are powers of ten instead of powers of two.

			The first 10 entries in the Kolakoski sequence, A000002, are 1221121221. There are 5 ones and 5 twos, so a(1) = 5 - 5 = 0.
The first 100=10^2 entries in the Kolakoski sequence A000002 include 49 ones and 51 twos, so a(2) = 51 - 49 = 2.

See the references and links for A195206, A289322.

Cf. A000002, A195206, A054353, A074286, A088568, A156077.

a(n) = 10^n - 2*A195206(n).

Additional (20th) term from Richard P. Brent, Mar 01 2018

cp's OEIS Frontend

A289322 Number of 1s in the first 2^n entries of the Kolakoski sequence, A000002.

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