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 1-3 of 3 results.

A076478 The binary Champernowne sequence: concatenate binary vectors of lengths 1, 2, 3, ... in numerical order.

Original entry on oeis.org

0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

N. J. A. Sloane, Nov 10 2002

Keywords

Comments

Can also be seen as triangle where row n contains all binary vectors of length n+1. - Reinhard Zumkeller, Aug 18 2015
From Clark Kimberling, Jul 18 2021: (Start)
In the following list, W represents the sequence of words w(n) represented by A076478. The list includes five partitions and two self-inverse permutations of the positive integers.
length of w(n): A000523
positions in W of words w(n) such that # 0's = # 1's: A258410;
positions in W of words w(n) such that # 0's < # 1's: A346299;
positions in W of words w(n) such that # 0's > # 1's: A346300;
positions in W of words w(n) that end with 0: A005498;
positions in W of words w(n) that end with 1: A005843;
positions in W of words w(n) such that first digit = last digit: A346301;
positions in W of words w(n) such that first digit != last digit: A346302;
positions in W of words w(n) such that 1st digit = 0 and last digit 0: A171757;
positions in W of words w(n) such that 1st digit = 0 and last digit 1: A346303;
positions in W of words w(n) such that 1st digit = 1 and last digit 0: A346304;
positions in W of words w(n) such that 1st digit = 1 and last digit 1: A346305;
position in W of n-th positive integer (base 2): A206332;
positions in W of binary complement of w(n): A346306;
sum of digits in w(n): A048881;
number of runs in w(n): A346307;
positions in W of palindromes: A346308;
positions in W of words such that #0's - #1's is odd: A346309;
positions in W of words such that #0's - #1's is even: A346310;
positions in W of the reversal of the n-th word in W: A081241. (End)

Examples

			0,
1,
0,0,
0,1,
1,0,
1,1,
0,0,0,
0,0,1,
0,1,0,
0,1,1,
1,0,0,
1,0,1,
...

References

  • Bodil Branner, Dynamics, Chap. IV.14 of The Princeton Companion to Mathematics, ed. T. Gowers, p. 499.
  • K. Dajani and C. Kraaikamp, Ergodic Theory of Numbers, Math. Assoc. America, 2002, p. 72.

Links

Crossrefs

Cf. A007931, A030308, A053645.
Cf. A341256, A341258, A341334, A342753, A342910.

Programs

  • Haskell
    import Data.List (unfoldr)
a076478 n = a076478_list !! n
a076478_list = concat $ tail $ map (tail . reverse . unfoldr
   (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2 )) [1..]
-- Reinhard Zumkeller, Feb 08 2012
  • Haskell
    a076478_row n = a076478_tabf !! n :: [[Int]]
a076478_tabf = tail $ iterate (\bs -> map (0 :) bs ++ map (1 :) bs) [[]]
a076478_list' = concat $ concat a076478_tabf
-- Reinhard Zumkeller, Aug 18 2015
  • Mathematica
    d[n_] := Rest@IntegerDigits[n + 1, 2] + 1; -1 + Flatten[Array[d, 50]] (* Clark Kimberling, Feb 07 2012 *)
z = 1000;
t1 = Table[Tuples[{0, 1}, n], {n, 1, 10}];
"All binary words, lexicographic order:"
tt = Flatten[t1, 1]; (* all binary words, lexicographic order *)
"All binary words, flattened:"
Flatten[tt];
w[n_] := tt[[n]];
"List tt of all binary words:"
tt = Table[w[n], {n, 1, z}]; (*  all the binary words *)
u1 = Flatten[tt]; (* words, concatenated, A076478, binary Champernowne sequence *)
u2 = Map[Length, tt];
"Positions of 0^n:"
Flatten[Position[Map[Union, tt], {0}]]
"Positions of 1^n:"
Flatten[Position[Map[Union, tt], {1}]]
"Positions of words in which #0's = #1's:"  (* A258410 *)
"This and the next two sequences partition N."
u3 = Select[Range[Length[tt]], Count[tt[[#]], 0] == Count[tt[[#]], 1] &]
"Positions of words in which #0's < #1's:"  (* A346299 *)
u4 = Select[Range[Length[tt]], Count[tt[[#]], 0] < Count[tt[[#]], 1] &]
"Positions of words in which #0's > #1's:"  (* A346300 *)
u5 = Select[Range[Length[tt]], Count[tt[[#]], 0] > Count[tt[[#]], 1] &]
"Positions of words ending with 0:" (* A005498 *)
u6 = Select[Range[Length[tt]], Last[tt[[#]]] == 0 &]
"Positions of words ending with 1:" (* A005843 *)
u7 = Select[Range[Length[tt]], Last[tt[[#]]] == 1 &]
"Positions of words starting and ending with same digit:" (* A346301 *)
u8 = Select[Range[Length[tt]], First[tt[[#]]] == Last[tt[[#]]] &]
"Positions of words starting and ending with opposite digits:" (* A346302  *)
u9 = Select[Range[Length[tt]], First[tt[[#]]] != Last[tt[[#]]] &]
"Positions of words starting with 0 and ending with 0:" (* A346303 *)
"This and the next three sequences partition N."
u10 = Select[Range[Length[tt]], First[tt[[#]]] == 0 && Last[tt[[#]]] == 0 &]
"Positions of words starting with 0 and ending with 1:" (* A171757 *)
u11 = Select[Range[Length[tt]], First[tt[[#]]] == 0 && Last[tt[[#]]] == 1 &]
"Positions of words starting with 1 and ending with 0:" (* A346304 *)
u12 = Select[Range[Length[tt]], First[tt[[#]]] == 1 && Last[tt[[#]]] == 0 &]
"Positions of words starting with 1 and ending with 1:" (* A346305 *)
u13 = Select[Range[Length[tt]], First[tt[[#]]] == 1 && Last[tt[[#]]] == 1 &]
"Position of n-th positive integer (base 2) in tt:"
d[n_] := If[First[w[n]] == 1, FromDigits[w[n], 2]];
u14 = Flatten[Table[Position[Table[d[n], {n, 1, 200}], n], {n, 1, 200}]] (* A206332 *)
"Position of binary complement of w(n):"
u15 = comp = Flatten[Table[Position[tt, 1 - w[n]], {n, 1, 50}]] (* A346306 *)
"Sum of digits of w(n):"
u16 = Table[Total[w[n]], {n, 1, 100}] (* A048881 *)
"Number of runs in w(n):"
u17 = Map[Length, Table[Map[Length, Split[w[n]]], {n, 1, 100}]] (* A346307 *)
"Palindromes:"
Select[tt, # == Reverse[#] &]
"Positions of palindromes:"
u18 = Select[Range[Length[tt]], tt[[#]] == Reverse[tt[[#]]] &] (* A346308 *)
"Positions of words in which #0's - #1's is odd:"
u19 = Select[Range[Length[tt]], OddQ[Count[w[#], 0] - Count[w[#], 1]] &] (* A346309 *)
"Positions of words in which #0's - #1's is even:"
u20 = Select[Range[Length[tt]], EvenQ[Count[w[#], 0] - Count[w[#], 1]] &] (* A346310 *)
"Position of the reversal of the n-th word:"  (* A081241 *)
u21 = Flatten[Table[Position[tt, Reverse[w[n]]], {n, 1, 150}]]
(* Clark Kimberling, Jul 18 2011 *)
  • PARI
    {m=5; for(d=1,m, for(k=0,2^d-1,v=binary(k); while(matsize(v)[2]
  • PARI
    listn(n)= my(a=List(), i=0, s=0); while(s<=n, listput(~a, binary(i++)[^1]); s+=#a[#a]); concat(a)[1..n+1]; \\ Ruud H.G. van Tol, Mar 17 2025
  • Python
    from itertools import count, product
def agen():
    for digits in count(1):
        for b in product([0, 1], repeat=digits):
            yield from b
g = agen()
print([next(g) for n in range(105)]) # Michael S. Branicky, Jul 18 2021

Formula

To get the m-th binary vector, write m+1 in base 2 and remove the initial 1. - Clark Kimberling, Feb 07 2010

Extensions

Extended by Klaus Brockhaus, Nov 11 2002

A092754 a(1)=1, a(2n)=2a(n)+1, a(2n+1)=2a(n)+2.

Original entry on oeis.org

1, 3, 4, 7, 8, 9, 10, 15, 16, 17, 18, 19, 20, 21, 22, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 127, 128, 129, 130, 131, 132
Offset: 1

Views

Author

Benoit Cloitre, Apr 13 2004

Keywords

Comments

More generally the sequence b(1)=1, b(2n)=2b(n)+x, b(2n+1)=2b(n)+y is given by the formula b(n)=A053644(n)+x*(n-A053644(n))+y*(A053644(n)-1).

Links

Crossrefs

Cf. A053644 (x=y=0), A054429(x=-1, y=+1), A062050(x=+1, y=-1).
Cf. A206332 (complement).
Cf. A004754.

Programs

  • Haskell
    a092754 n = if n < 2 then n else 2 * a092754 n' + m + 1
            where (n',m) = divMod n 2
a092754_list = map a092754 [1..]
-- Reinhard Zumkeller, May 07 2012
  • PARI
    a(n)=if(n<2,1,if(n%2,a(n-1)+1,a(n/2)*2+1))
  • PARI
    a(n) = n + 1<Kevin Ryde, Jun 19 2021

Formula

a(n) = 2^floor(log(n)/log(2)) + n - 1.
a(n) = A004754(n) - 1. - Rémy Sigrist, May 05 2019

A307433 A special version of Pascal's triangle where only powers of 2 are permitted.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 4, 4, 1, 1, 1, 2, 1, 8, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 1, 8, 8, 8, 8, 8, 1, 2, 1, 1, 1, 1, 1, 16, 16, 16, 16, 1, 1, 1, 1, 1, 2, 2, 2, 1, 32, 32, 32, 1, 2, 2, 2, 1
Offset: 0

Views

Author

Rémy Sigrist, May 05 2019

Keywords

Comments

If the sum of the two numbers above in the triangular array is not a power of 2 (A000079), then a 1 is put in its place.
The ones in the table form a Sierpinski gasket (A047999).
Apparently, for any k > 0, the value 2^k first occurs on row A206332(k).
From Bernard Schott, May 05 2019: (Start)
For any m, at row 2^m - 1, there is only a string of 2^m times the number 1, then at row 2^(m+1) - 2, comes out for the first time and only once, the power of 2 equals to 2^(2^m-1). At row 2^(m+1) - 1, there are again 2^(m+1) times the number 1. This cycle can go on. Under, a part of this triangle between row 2^3 -1 and 2^4 - 2 that visualizes the explanations.
1 1 1 1 1 1 1 1
2 2 2 2 2 2 2
4 4 4 4 4 4
8 8 8 8 8
16 16 16 16
32 32 32
64 64
128
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (End)

Examples

			The triangle begins:
                                1
                              1   1
                            1   2   1
                          1   1   1   1
                        1   2   2   2   1
                      1   1   4   4   1   1
                    1   2   1   8   1   2   1
                  1   1   1   1   1   1   1   1
                1   2   2   2   2   2   2   2   1
              1   1   4   4   4   4   4   4   1   1
            1   2   1   8   8   8   8   8   1   2   1
          1   1   1   1  16  16  16  16   1   1   1   1
        1   2   2   2   1  32  32  32   1   2   2   2   1
      1   1   4   4   1   1  64  64   1   1   4   4   1   1
    1   2   1   8   1   2   1  128  1   2   1   8   1   2   1
  1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1

Links

Crossrefs

Cf. A000079, A007318, A047999, A206332, A307116 (analog with Fibonacci numbers).

Programs

  • PARI
    for (r=1, 13, apply(v -> print1 (v", "), row=vector(r, k, if (k==1 || k==r, 1, hammingweight(s=row[k-1]+row[k])==1, s, 1))))
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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