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A131987
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Representation of a dense para-sequence.
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5
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1, 2, 1, 3, 4, 2, 5, 1, 6, 3, 7, 8, 4, 9, 2, 10, 5, 11, 1, 12, 6, 13, 3, 14, 7, 15, 16, 8, 17, 4, 18, 9, 19, 2, 20, 10, 21, 5, 22, 11, 23, 1, 24, 12, 25, 6, 26, 13, 27, 3, 28, 14, 29, 7, 30, 15, 31, 32, 16, 33, 8, 34, 17, 35, 4, 36, 18, 37, 9, 38, 19, 39, 2, 40, 20, 41, 10, 42, 21, 43
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OFFSET
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1,2
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COMMENTS
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A fractal sequence. The para-sequence may be regarded as a sort of "limit" of the concatenated segments. The para-sequence (itself not a sequence) is dense in the sense that every pair of terms i and j are separated by another term (and hence separated by infinitely many terms).
The para-sequence accounts for positions of dyadic rational numbers in the following way: Label 1/2 as 1; label 1/4, 3/4 as 2 and 3; label 1/8, 3/8, 5/8, 7/8 as 4,5,6,7, etc. Then, for example, the ordering 1/8 < 1/4 < 3/8 < 1/2 < 5/8 < 3/4 < 7/8 matches the labels 4,2,5,1,6,3,7, which is the 3rd segment of A131987. The n-th segment consists of labels for rationals having denominators 2, 4, 8, ..., 2^n.
Could be seen as a "fuzzy table" with row lengths 2^n-1. In row n one has the numbers, read from the leftmost to the rightmost, as they appear in a perfect binary tree of 2^n-1 nodes when inserted in "storage order" into the tree, cf. illustration in A101279 and stackexchange link. These rows are obviously permutations of [1..2^n-1], their inverse is given in A269752. - M. F. Hasler, Mar 04 2016
The sequence obtained by adding 1 to every term of this sequence is the same as A003602 with all 1's removed. - David James Sycamore, Jul 25 2022
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REFERENCES
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C. Kimberling, Proper self-containing sequences, fractal sequences and para-sequences, preprint, 2007.
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LINKS
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FORMULA
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When viewed as a table, T(h,p), related to the in order traversal of a full binary tree, T(h,p) = 2^h+(p-1)/2, p odd, 2^(h-m(p)) + (p-2^m(p)) / 2^(m(p)+1), where m(p) is the greatest value of n such that p mod 2^n == 0. m(p) = p-adic[ordp](2*p,2)-1. - Gary Detlefs, Sep 28 2018
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EXAMPLE
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Start with 1 and isolate it using 2,3, like this: 2,1,3. Then isolate those using 4,5,6,7, like this: 4,2,5,1,6,3,7. The next segment, to be concatenated after 4,2,5,1,6,3,7, is 8,4,9,2,10,5,11,1,12,6,13,3,14,7,15.
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MAPLE
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m:=p->padic[ordp](2*p, 2)-1:podd:=(h, p)->2^h+(p-2)/2:peven:=(h, p)->2^(h-m(p))+(p-2^m(p))/2^(m(p)+1):for i from 0 to 5 do for j from 1 to 2^(i+1)-1 do if j mod 2 =1 then print(podd(i, j)) else print(peven(i, j)) fi od od # Gary Detlefs, Sep 28 2018
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MATHEMATICA
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Flatten@NestList[Riffle[Range[Length[#] + 1, 2 Length[#] + 1], #] &, {1}, 4] (* Birkas Gyorgy, Mar 11 2011 *)
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PROG
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(PARI) A131987_row(n, r=[1])={for(k=2, n, r=vector(2^k-1, j, if(bittest(j, 0), j\2+2^(k-1), r[j\2]))); r}
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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