A289323 Number of twos minus number of ones in the first 2^n entries of the Kolakoski sequence, A000002.

-1, 0, 0, 0, 0, -2, 0, 0, -2, 0, -2, 0, -6, 6, 0, 6, 44, 26, -20, -48, 52, 58, 104, -82, -250, -270, -474, -1864, -3094, -4588, -2534, -7574, -1522, 1818, 9264, 18082, 8898, -30500, -20586, -3232, -90522, -127446, -231384, -83574, -87364, 267886

Offset: 0

Author

Richard P. Brent, Jul 05 2017

Keywords

• sign

Comments

This is equivalent to A289322, since a(n) = (#twos)-(#ones) = 2^n-2*(#ones) in the first 2^n entries of A000002.

For example, a(5)=15-17=(32-17)-17=32-2*17=-2 because there are 15 twos and 17 ones in the first 32=2^5 entries of A000002.

The entries in this sequence appear to be of order 2^(n/2), whereas the entries in A289322 are larger (of order 2^n).

Examples

			The first 32 entries of the Kolakoski sequence, A000002, are 12211212212211211221211212211211. From this we see that a(5)=15-17=-2, since among the first 2^5 letters, 15 of them are twos and 17 of them are ones.
		

References

• See references for A289322.

Links

• Richard P. Brent, Table of n, a(n) for n = 0..64

Crossrefs

Cf. A000002, A054353, A074286, A088568, A156077, A289322.

Formula

a(n) = 2^n - 2*A289322(n) = -A088568(2^n) = 2*A054353(2^n) - 3*2^n = 2^n - 2*A156077(2^n).