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%I #7 Aug 22 2014 10:19:37
%S 0,2,7,10,15,23,31,36,44,49,57,70,78,91,104,112,125,138,159,193,214,
%T 248,282,303,337,371,392,426,447,481,515,536,570,591,625,659,680,714,
%U 748,803,892,981,1036,1125,1180,1269,1358,1413,1502,1557,1646,1735,1790
%N Index sequence for limit-block extending the (2,1)-version of the infinite Fibonacci word A014675 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 A014675 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^.
%C ...
%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 A246127.
%e S = the infinite Fibonacci word A014675, with B = (s(0)); that is, (m,k) = (0,0); S = (2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,...)
%e B'(0) = (2)
%e B'(1) = (2,2)
%e B'(2) = (2,2,1)
%e B'(3) = (2,2,1,2)
%e B'(4) = (2,2,1,2,1)
%e B'(5) = (2,2,1,2,1,2)
%e S^ = (2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2,...),
%e with index sequence (0,2,7,10,15,...)
%t seqPosition1[list_, seqtofind_] := If[Length[#] > Length[list], {}, Last[Last[ Position[Partition[list, Length[#], 1], Flatten[{___, #, ___}], 1, 1]]]] &[seqtofind]; s = Differences[Table[Floor[n*GoldenRatio], {n, 10000}]]; t = {{2}}; 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]; t1 = Last[t] (*A246127*)
%t q = -1 + Accumulate[Table[p[k], {k, 0, n - 1}]] (*A246128*)
%Y Cf. A245921, A246127, A246129, A014675.
%K nonn
%O 0,2
%A _Clark Kimberling_ and _Peter J. C. Moses_, Aug 15 2014
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