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Search: a245933 -id:a245933
Displaying 1-2 of 2 results found. page 1
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A245934 Index sequence for limit-reversing the A006337; see Comments. +10
3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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.
LINKS
EXAMPLE
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,...)
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved
A245935 First differences of A245934; see Comments. +10
3
2, 2, 3, 5, 7, 5, 7, 5, 5, 12, 12, 5, 12, 12, 12, 17, 12, 17, 12, 12, 17, 12, 17, 12, 17, 41, 29, 41, 29, 29, 41, 29, 41, 29, 41, 29, 29, 41, 29, 41, 29, 29, 41, 29, 41, 29, 41, 29, 29, 41, 29, 41, 29, 29, 70, 70, 29, 70, 70, 70, 29, 70, 70, 29, 70, 70, 70 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
It appears that every term is a term of A002965. The sequence A245935 arises from A245933 and A245934, in which the limit-reverse of certain sequences is defined, as follows. Suppose that 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.
LINKS
EXAMPLE
S = A006337 (the Beatty sequence of sqrt(2)), re-indexed to start with s(0) = 1, with B = (s(0)); that is, (m,k) = (0,0)
S = (1,2,1,2,1,1,2,1,2,1,1,2,1,2,1,2,1,1,2,1,2,1,1,2,...)
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,...), with difference sequence (2,2,3,5,7,5,7,5,5,12,12,...).
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved
page 1

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Last modified August 29 23:09 EDT 2024. Contains 375519 sequences. (Running on oeis4.)