cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A208569 Duplicate of A131271.

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 3, 1, 8, 4, 5, 2, 7, 3, 6, 1, 16, 8, 9, 4, 13, 5, 12, 2, 15, 7, 10, 3, 14, 6, 11, 1, 32, 16, 17, 8, 25, 9, 24, 4, 29, 13, 20, 5, 28, 12, 21, 2, 31, 15, 18, 7, 26, 10, 23, 3, 30, 14, 19, 6, 27, 11, 22, 1, 64, 32, 33, 16, 49, 17, 48, 8, 57
Offset: 1

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Author

Keywords

A049773 Triangular array T read by rows: if row n is r(1),...,r(m), then row n+1 is 2r(1)-1,...,2r(m)-1,2r(1),...,2r(m).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 4, 1, 5, 3, 7, 2, 6, 4, 8, 1, 9, 5, 13, 3, 11, 7, 15, 2, 10, 6, 14, 4, 12, 8, 16, 1, 17, 9, 25, 5, 21, 13, 29, 3, 19, 11, 27, 7, 23, 15, 31, 2, 18, 10, 26, 6, 22, 14, 30, 4, 20, 12, 28, 8, 24, 16, 32, 1, 33, 17, 49, 9, 41, 25, 57, 5, 37, 21, 53, 13, 45, 29, 61, 3, 35, 19
Offset: 1

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Author

Keywords

Comments

n-th row = (r(1),r(2),...,r(m)), where m=2^(n-1), satisfies r(r(k))=k for k=1,2,...,m and has exactly 2^[ n/2 ] solutions of r(k)=k. (The function r(k) reverses bits. Or rather, r(k)=revbits(k-1)+1.)
In a knockout competition with m players, arranging the competition brackets (see links) in r(k) order, where k is the rank of the player, ensures that highest ranked players cannot meet until the later stages of the competition. None of the top 2^p ranked players can meet earlier than the p-th from last round of the competition. At the same time the top ranked players in each match meet the highest ranked player possible consistent with this rule. The sequence for the top ranked players meeting the lowest ranked player possible is A131271. - Colin Hall, Jul 31 2011, Feb 29 2012
Row n contains one of A003407(2^(n-1)) non-averaging permutations of [2^(n-1)], i.e., a permutation of [2^(n-1)] without 3-term arithmetic progressions. - Alois P. Heinz, Dec 05 2017

Examples

			Triangle begins:
1;
1,  2;
1,  3, 2,  4;
1,  5, 3,  7, 2,  6,  4,  8;
1,  9, 5, 13, 3, 11,  7, 15, 2, 10,  6, 14, 4, 12,  8, 16;
1, 17, 9, 25, 5, 21, 13, 29, 3, 19, 11, 27, 7, 23, 15, 31, 2, 18, 10, 26, ...
		

Crossrefs

Sum of odd-indexed terms of n-th row gives A007582. Sum of even-indexed terms gives A049775.
A030109 is another version.
Cf. A131271.
Cf. A088370.

Programs

  • Haskell
    a049773 n k = a049773_tabf !! (n-1) !! (k-1)
    a049773_row n = a049773_tabf !! (n-1)
    a049773_tabf = iterate f [1] where
       f vs = (map (subtract 1) ws) ++ ws where ws = map (* 2) vs
    -- Reinhard Zumkeller, Mar 14 2015
  • Maple
    T:= proc(n) option remember; `if`(n=1, 1,
          [map(x->2*x-1, [T(n-1)])[], map(x->2*x, [T(n-1)])[]][])
        end:
    seq(T(n), n=1..7);  # Alois P. Heinz, Oct 28 2011
  • Mathematica
    row[1] = {1}; row[n_] := row[n] = Join[ 2*row[n-1] - 1, 2*row[n-1] ]; Flatten[ Table[ row[n], {n, 1, 7}]] (* Jean-François Alcover, May 03 2012 *)
  • PARI
    (a(n, k) = if( k<=0 || k>=n, 0, if( k%2, n\2) + a(n\2, k\2))); {T(n, k) = if( k<=0 || k>2^n/2, 0, 1 + a(2^n/2, k-1))}; /* Michael Somos, Oct 13 1999 */
    

A088208 Table read by rows where T(0,0)=1; n-th row has 2^n terms T(n,j),j=0 to 2^n-1. For j==0 mod 2, T(n+1,2j)=T(n,j) and T(n+1,2j+1)=T(n,j)+2^n. For j==1 mod 2, T(n+1,2j+1)=T(n,j) and T(n+1,2j)=T(n,j)+2^n.

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 2, 1, 5, 7, 3, 4, 8, 6, 2, 1, 9, 13, 5, 7, 15, 11, 3, 4, 12, 16, 8, 6, 14, 10, 2, 1, 17, 25, 9, 13, 29, 21, 5, 7, 23, 31, 15, 11, 27, 19, 3, 4, 20, 28, 12, 16, 32, 24, 8, 6, 22, 30, 14, 10, 26, 18, 2, 1, 33, 49, 17, 25, 57, 41, 9, 13, 45, 61, 29, 21, 53, 37, 5, 7, 39, 55, 23
Offset: 1

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Author

Gary W. Adamson, Sep 23 2003

Keywords

Comments

Schroeder, p. 281 states "The ordering with which the iterates x_n fall into the 2^m different chaos bands [order as to magnitude] is also the same as the ordering of the iterates in a stable orbit of period length P = 2^m. For example, for both the period-4 orbit and the four chaos bands, the iterates, starting with the largest iterate x_1, are ordered as follows: x_1 > x_3 > x_4 > x_2."
From Andrey Zabolotskiy, Dec 06 2024: (Start)
For n>0, row n-1 is the permutation relating row n of the left half of Stern-Brocot tree with row n of Kepler's tree of fractions. Specifically, if K_n(k) [resp. SB_n(k)] is the k-th fraction in the n-th row of A294442 [resp. A057432], where 1/2 is in row 1 and k=1..2^(n-1), then SB_n(k) = K_n(T(n-1, k)).
The inverse permutation is row n of A131271.
Equals A362160+1. (End)

Examples

			1
1 2
1 3 4 2
1 5 7 3 4 8 6 2
1 9 13 5 7 15 11 3 4 12 16 8 6 14 10 2
		

References

  • Manfred R. Schroeder, "Fractals, Chaos, Power Laws", W.H. Freeman, 1991, p. 282.

Crossrefs

Programs

  • Haskell
    a088208 n k = a088208_tabf !! (n-1) !! (k-1)
    a088208_row n = a088208_tabf !! (n-1)
    a088208_tabf = iterate f [1] where
       f vs = (map (subtract 1) ws) ++ reverse ws where ws = map (* 2) vs
    -- Reinhard Zumkeller, Mar 14 2015
  • Mathematica
    nmax = 6;
    T[, 0] = 1; T[n, j_] /; j == 2^n = n;
    Do[Which[
      EvenQ[j], T[n+1, 2j] = T[n, j]; T[n+1, 2j+1] = T[n, j] + 2^n,
      OddQ[j], T[n+1, 2j+1] = T[n, j]; T[n+1, 2j] = T[n, j] + 2^n],
    {n, 0, nmax}, {j, 0, 2^n-1}];
    Table[T[n, j], {n, 0, nmax}, {j, 0, 2^n-1}] // Flatten (* Jean-François Alcover, Aug 03 2018 *)

Extensions

Edited by Ray Chandler and N. J. A. Sloane, Oct 08 2003

A240909 The sequency numbers of the 16 rows of a Hadamard-Walsh matrix of order 16.

Original entry on oeis.org

0, 15, 7, 8, 3, 12, 4, 11, 1, 14, 6, 9, 2, 13, 5, 10
Offset: 1

Views

Author

Ross Drewe, Apr 14 2014

Keywords

Comments

See A240908 for context. This sequence is the natural sequency ordering for an order 16 matrix.

Examples

			This is a fixed length sequence of only 16 values, as given.
		

Crossrefs

Cf. A240908 "natural order" sequencies for Hadamard-Walsh matrix, order 8.
Cf. A240910 "natural order" sequencies for Hadamard-Walsh matrix, order 32.
Cf. A153141 "dyadic order" sequencies for Hadamard-Walsh matrix, all orders.
Cf. A000975(n) is sequency of last row of H(n). - William P. Orrick, Jun 28 2015
Cf. A131271 (row 4, minus 1).

Formula

Recursion: H(2)=[1 1; 1 -1]; H(n) = H(n-1)*H(2), where * is Kronecker matrix product.
a(n) = A131271(4,n) - 1. - Mathew Englander, Jun 19 2021

Extensions

Definition of H(n) corrected by William P. Orrick, Jun 28 2015

A057432 Obtained by reading first the numerator then the denominator of fractions in left-hand half of Stern-Brocot tree (A007305/A007306).

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 2, 3, 1, 4, 2, 5, 3, 5, 3, 4, 1, 5, 2, 7, 3, 8, 3, 7, 4, 7, 5, 8, 5, 7, 4, 5, 1, 6, 2, 9, 3, 11, 3, 10, 4, 11, 5, 13, 5, 12, 4, 9, 5, 9, 7, 12, 8, 13, 7, 11, 7, 10, 8, 11, 7, 9, 5, 6, 1, 7, 2, 11, 3, 14, 3, 13, 4, 15, 5, 18, 5, 17, 4, 13, 5, 14, 7, 19, 8, 21, 7, 18, 7, 17, 8, 19, 7
Offset: 0

Views

Author

N. J. A. Sloane, Sep 08 2000

Keywords

Examples

			The tree begins:
                                     1/1
                                     1/2
                  1/3                                   2/3
        1/4                 2/5               3/5                 3/4
    1/5      2/7       3/8       3/7     4/7       5/8       5/7      4/5
  1/6 2/9 3/11 3/10 4/11 5/13 5/12 4/9 5/9 7/12 8/13 7/11 7/10 8/11 7/9 5/6
		

Crossrefs

Related to the Kepler tree A294442 via row permutations given by A088208 or A131271.

Programs

  • Mathematica
    sbt[n_]:=Module[{P,L,Y},P={{1,0},{1,1}};L={{1,1},{0,1}};Y={{1,0},{0,1}}; w[b_]:=Fold[ #1.If[ #2==0,L,P]&,Y,b]; u[a_]:={a[[2,1]]+a[[2,2]],a[[1,1]]+a[[1,2]]}; s[l_]:={l,{Last[l],First[l]}}; Map[s,Map[u,Map[w,Part[Partition[Tuples[{0,1},n],2^(n-1)],1]]]]]
    Flatten[Append[{1,1},Table[Map[First,sbt[i]],{i,1,6}]]] (* Peter Luschny, Apr 27 2009 *)

Extensions

More terms from Alford Arnold, Sep 11 2000
More terms from Joshua Zucker, May 11 2006
Showing 1-5 of 5 results.