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Limiting block extension of A000002 (Kolakoski sequence) with first term as initial block.
6

%I #5 Nov 22 2017 01:15:03

%S 1,1,2,2,1,2,1,1,2,1,2,2,1,1,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2,1,

%T 1,2,1,1,2,2,1,2,2,1,2,1,1,2,2,1,2,2,1,1,2,1,1,2,1,2,2,1,2,2,1,1,2,1,

%U 2,2,1,2,1,1,2,2,1,2,2,1,2,1,1,2,1,2

%N Limiting block extension of A000002 (Kolakoski sequence) with first term as initial block.

%C Suppose S = (s(0), s(1), s(2), ...) is an infinite sequence such that every finite block of consecutive terms occurs infinitely many times in S. (It is assumed that A000002 is such a sequence.) Let B = B(m,k) = (s(m), s(m+1),...s(m+k)) be such a block, where m >= 0 and k >= 0. Let m(1) be the least i > m such that (s(i), s(i+1),...,s(i+k)) = B(m,k), and put B(m(1),k+1) = (s(m(1)), s(m(1)+1),...s(m(1)+k+1)). Let m(2) be the least i > m(1) such that (s(i), s(i+1),...,s(i+k)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)), s(m(2)+1),...s(m(2)+k+2)). Continuing in this manner gives a sequence of blocks B'(n) = B(m(n),k+n), so that for n >= 0, B'(n+1) comes from B'(n) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limiting block extension of S with initial block B(m,k)", denoted by S^ in case the initial block is s(0).

%C The sequence (m(i)), where m(0) = 0, is the "index sequence for limit-block extending S with initial block B(m,k)", as in A246128. If the sequence S is given with offset 1, then the role played by s(0) in the above definitions is played by s(1) instead, as in the case of A246144 and A246145.

%C Limiting block extensions are analogous to limit-reverse sequences, S*, defined at A245920. The essential difference is that S^ is formed by extending each new block one term to the right, whereas S* is formed by extending each new block one term to the left (and then reversing).

%e S = A000002, with B = (s(1)); that is, (m,k) = (1,0)

%e S = (1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2,1,...)

%e B'(0) = (1)

%e B'(1) = (1,1)

%e B'(2) = (1,1,2)

%e B'(3) = (1,1,2,2)

%e B'(4) = (1,1,2,2,1)

%e B'(5) = (1,1,2,2,1,2)

%e S^ = (1,1,2,2,1,2,1,1,2,1,2,2,1,1,2,1,1,...),

%e with index sequence (1,4,13,16,51,78,97,124,178,247,322,...)

%t seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[Position[Partition[list, Length[#], 1], Flatten[{___, #, ___}], 1, 1]]]] &[seqtofind]; n = 30; s = Prepend[Nest[Flatten[Partition[#, 2] /. {{2, 2} -> {2, 2, 1, 1}, {2, 1} -> {2, 2, 1}, {1, 2} -> {2, 1, 1}, {1, 1} -> {2, 1}}] &, {2, 2}, n], 1]; (* A246144 *)

%t Take[s, 30]

%t t = {{1}}; p[0] = seqPosition1[s, Last[t]]; s = Drop[s, p[0]]; Off[Last::nolast]; n = 1; While[(p[n] = seqPosition1[s, Last[t]]) > 0, (AppendTo[t, Take[s, {#, # + Length[Last[t]]}]]; s = Drop[s, #]) &[p[n]]; n++]; On[Last::nolast]; Last[t] (* A246144*)

%t Accumulate[Table[p[k], {k, 0, n - 1}]] (*A246145*)

%Y Cf. A246145, A246127, A246140, A246142, A000002.

%K nonn

%O 1,3

%A _Clark Kimberling_ and _Peter J. C. Moses_, Aug 17 2014