A289323 - OEIS

%I #21 Jul 07 2017 03:29:25

%S -1,0,0,0,0,-2,0,0,-2,0,-2,0,-6,6,0,6,44,26,-20,-48,52,58,104,-82,

%T -250,-270,-474,-1864,-3094,-4588,-2534,-7574,-1522,1818,9264,18082,

%U 8898,-30500,-20586,-3232,-90522,-127446,-231384,-83574,-87364,267886

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

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

%C 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.

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

%D See references for A289322.

%H Richard P. Brent, <a href="/A289323/b289323.txt">Table of n, a(n) for n = 0..64</a>

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

%e 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.

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

%K sign

%O 0,6

%A _Richard P. Brent_, Jul 05 2017