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A246140 Limiting block extension of A006337 (difference sequence of the Beatty sequence for sqrt(2)) with first term as initial block.

%I A246140 #15 Jul 27 2018 09:36:28
%S A246140 1,2,1,1,2,1,2,1,2,1,1,2,1,2,1,1,2,1,2,1,2,1,1,2,1,2,1,1,2,1,2,1,1,2,
%T A246140 1,2,1,2,1,1,2,1,2,1,1,2,1,2,1,2,1,1,2,1,2,1,1,2,1,2,1,1,2,1,2,1,2,1,
%U A246140 1,2,1,2,1,1,2,1,2,1,2,1,1,2,1,2,1,1
%N A246140 Limiting block extension of  A006337 (difference sequence of the Beatty sequence for sqrt(2)) with first term as initial block.
%C A246140 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 A006337 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 A246140 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 A246140 and A246141.
%C A246140 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).
%H A246140 G. C. Greubel, <a href="/A246140/b246140.txt">Table of n, a(n) for n = 1..550</a>
%e A246140 S = A006337, with B = (s(1)); that is, (m,k) = (1,0)
%e A246140 S = (1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2,...)
%e A246140 B'(0) = (1)
%e A246140 B'(1) = (1,2)
%e A246140 B'(2) = (1,2,1)
%e A246140 B'(3) = (1,2,1,1)
%e A246140 B'(4) = (1,2,1,1,2)
%e A246140 B'(5) = (1,2,1,1,2,1)
%e A246140 S^ = (1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1,...),
%e A246140 with index sequence (1,3,6,8,15,...)
%t A246140 seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[      Position[Partition[list, Length[#], 1], Flatten[{___, #, ___}], 1, 1]]]] &[seqtofind]; s =  Differences[Table[Floor[n*Sqrt[2]], {n, 10000}]]; Take[s, 60]
%t A246140 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] (* A246140 *)
%t A246140 Accumulate[Table[p[k], {k, 0, n - 1}]] (* A246141 *)
%Y A246140 Cf. A246141, A246127, A246142, A246144, A246146, A006337.
%K A246140 nonn
%O A246140 1,2
%A A246140 _Clark Kimberling_ and _Peter J. C. Moses_, Aug 17 2014