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A289323
Number of twos minus number of ones in the first 2^n entries of the Kolakoski sequence, A000002.
4
-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
0,6
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).
REFERENCES
See references for A289322.
LINKS
FORMULA
a(n) = 2^n - 2*A289322(n) = -A088568(2^n) = 2*A054353(2^n) - 3*2^n = 2^n - 2*A156077(2^n).
EXAMPLE
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.
CROSSREFS
KEYWORD
sign
AUTHOR
Richard P. Brent, Jul 05 2017
STATUS
approved