A245935 -id:A245935 - cp's OEIS frontend

A245933 Limit-reverse of A006337 (the difference sequence of Beatty sequence for sqrt(2)), with first term as initial block.

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, 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, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1

Offset: 1

Clark Kimberling and Peter J. C. Moses, Aug 07 2014

nonn

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-k), s(m-k+1),...,s(m)) be such a block, where m >= 0 and k >= 0.  Let m(1) be the least i > m such that (s(i-k), s(i-k+1),...,s(i)) = B(m,k), and put B(m(1),k+1) = (s(m(1)-k-1), s(m(1)-k),...,s(m(1))).  Let m(2) be the least i > m(1) such that (s(i-k-1), s(i-k),...,s(i)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)-k-2), s(m(2)-k-1),...,s(m(2))).  Continuing in this manner gives a sequence of blocks B(m(n),k+n).  Let B'(n) = reverse(B(m(n),k+n)), so that for n >= 1, B'(n) comes from B'(n-1) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limit-reverse of S with initial block B(m,k)", denoted by S*(m,k), or simply S*.  (Since Beatty sequences are usually written with offset 1, the above definition is adapted accordingly, so that s(n) = A006337(n+1) for n >= 0.)
...
The sequence (m(i)), where m(0) = 1, is the "index sequence for limit-reversing S with initial block B(m,k)" or simply the index sequence for S*, as in A245934.

			S = A006337 (re-indexed to start with s(0) = 1, with B = (s(0)); that is, (m,k) = (0,0)
S = (1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2,...)
B'(0) = (1)
B'(1) = (2,1)
B'(2) = (1,2,1)
B'(3) = (1, 2, 1, 1)
B'(4) = (1, 2, 1, 1, 2)
B'(5) = (1, 2, 1, 1, 2, 1)
S* = (1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1,...),
with index sequence (1, 3, 5, 8, 13, 20, 25, 32, 37,...)

Clark Kimberling, Table of n, a(n) for n = 1..300

Cf. A006337, A245934, A245935, A245920.

z = 100; seqPosition2[list_, seqtofind_] := Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 2]]] &[seqtofind]; x = Sqrt[2]; s = Differences[Table[Floor[n*x], {n, 1, z^2}]] ; ans = Join[{s[[p[0] = pos = seqPosition2[s, #] - 1]]}, #] &[{s[[1]]}]; cfs = Table[s = Drop[s, pos - 1]; ans = Join[{s[[p[n] = pos = seqPosition2[s, #] - 1]]}, #] &[ans], {n, z}]; rcf = Last[Map[Reverse, cfs]]

A245934 Index sequence for limit-reversing the A006337; see Comments.

Original entry on oeis.org

1, 3, 5, 8, 13, 20, 25, 32, 37, 42, 54, 66, 71, 83, 95, 107, 124, 136, 153, 165, 177, 194, 206, 223, 235, 252, 293, 322, 363, 392, 421, 462, 491, 532, 561, 602, 631, 660, 701, 730, 771, 800, 829, 870, 899, 940, 969, 1010, 1039, 1068, 1109, 1138, 1179, 1208

Offset: 1

Clark Kimberling and Peter J. C. Moses, Aug 07 2014

nonn

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-k), s(m-k+1),...,s(m)) be such a block, where m >= 0 and k >= 0.  Let m(1) be the least i > m such that (s(i-k), s(i-k+1),...,s(i)) = B(m,k), and put B(m(1),k+1) = (s(m(1)-k-1), s(m(1)-k),...,s(m(1))).  Let m(2) be the least i > m(1) such that (s(i-k-1), s(i-k),...,s(i)) = B(m(1),k+1), and put B(m(2),k+2) = (s(m(2)-k-2), s(m(2)-k-1),...,s(m(2))).  Continuing in this manner gives a sequence of blocks B(m(n),k+n).  Let B'(n) = reverse(B(m(n),k+n)), so that for n >= 1, B'(n) comes from B'(n-1) by suffixing a single term; thus the limit of B'(n) is defined; we call it the "limit-reverse of S with initial block B(m,k)", denoted by S*(m,k), or simply S*.  (Since Beatty sequences are usually written with offset 1, the above definition is adapted accordingly, so that s(n) = A006337(n+1) for n >= 0.)
...
The sequence (m(i)), where m(0) = 1, is the "index sequence for limit-reversing S with initial block B(m,k)" or simply the index sequence for S*, as in A245934.

			S = A006337 (re-indexed to start with s(0) = 1, with B = (s(0)); that is, (m,k) = (0,0); S = (1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2,...)
B'(0) = (1)
B'(1) = (2,1)
B'(2) = (1,2,1)
B'(3) = (1, 2, 1, 1)
B'(4) = (1, 2, 1, 1, 2)
B'(5) = (1, 2, 1, 1, 2, 1)
S* = (1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1,...),
with index sequence (1, 3, 5, 8, 13, 20, 25, 32, 37,...)

Cf. A006337, A245933, A245935, A245921.

z = 140; seqPosition2[list_, seqtofind_] := Last[Last[Position[Partition[list, Length[#], 1], Flatten[{_, #, _}], 1, 2]]] &[seqtofind]; x = Sqrt[2]; s =  Differences[Table[Floor[n*x], {n, 1, z^2}]];  ans = Join[{s[[p[0] = pos = seqPosition2[s, #] - 1]]}, #] &[{s[[1]]}]; cfs = Table[s = Drop[s, pos - 1]; ans = Join[{s[[p[n] = pos = seqPosition2[s, #] - 1]]}, #] &[ans], {n, z}]; q = Accumulate[Join[{1}, Table[p[n], {n, 0, z}]]]  (* A245934 *)
q1 = Differences[q]  (* A245935 *)

cp's OEIS Frontend

A245933 Limit-reverse of A006337 (the difference sequence of Beatty sequence for sqrt(2)), with first term as initial block.

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