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.

Previous Showing 11-20 of 67 results. Next

A329313 Length of the Lyndon factorization of the reversed binary expansion of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 3, 1, 2, 2, 3, 1, 2, 1, 4, 1, 2, 2, 3, 1, 3, 2, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 2, 3, 2, 3, 2, 4, 1, 2, 3, 4, 1, 3, 2, 5, 1, 2, 2, 3, 1, 2, 2, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 2, 3, 2, 3, 2, 4, 1, 3, 3, 4, 2, 3, 2, 5, 1, 2, 2, 3, 1, 4, 3
Offset: 0

Views

Author

Gus Wiseman, Nov 11 2019

Keywords

Comments

We define the Lyndon product of two or more finite sequences to be the lexicographically maximal sequence obtainable by shuffling the sequences together. For example, the Lyndon product of (231) with (213) is (232131), the product of (221) with (213) is (222131), and the product of (122) with (2121) is (2122121). A Lyndon word is a finite sequence that is prime with respect to the Lyndon product. Every finite sequence has a unique (orderless) factorization into Lyndon words, and if these factors are arranged in lexicographically decreasing order, their concatenation is equal to their Lyndon product. For example, (1001) has sorted Lyndon factorization (001)(1).

Examples

			The sequence of reversed binary expansions of the nonnegative integers together with their Lyndon factorizations begins:
   0:      () = ()
   1:     (1) = (1)
   2:    (01) = (01)
   3:    (11) = (1)(1)
   4:   (001) = (001)
   5:   (101) = (1)(01)
   6:   (011) = (011)
   7:   (111) = (1)(1)(1)
   8:  (0001) = (0001)
   9:  (1001) = (1)(001)
  10:  (0101) = (01)(01)
  11:  (1101) = (1)(1)(01)
  12:  (0011) = (0011)
  13:  (1011) = (1)(011)
  14:  (0111) = (0111)
  15:  (1111) = (1)(1)(1)(1)
  16: (00001) = (00001)
  17: (10001) = (1)(0001)
  18: (01001) = (01)(001)
  19: (11001) = (1)(1)(001)
  20: (00101) = (00101)

Crossrefs

The non-reversed version is A211100.
Positions of 1's are A328596.
The "co" version is A329326.
Binary Lyndon words are counted by A001037 and ranked by A102659.
Numbers whose reversed binary expansion is a necklace are A328595.
Numbers whose reversed binary expansion is a aperiodic are A328594.
Length of the co-Lyndon factorization of the binary expansion is A329312.
Cf. A000031, A027375, A059966, A060223, A121016, A211097, A275692, A329131, A329314, A329317, A329325.

Programs

  • Mathematica
    lynQ[q_]:=Array[Union[{q,RotateRight[q,#]}]=={q,RotateRight[q,#]}&,Length[q]-1,1,And];
lynfac[q_]:=If[Length[q]==0,{},Function[i,Prepend[lynfac[Drop[q,i]],Take[q,i]]][Last[Select[Range[Length[q]],lynQ[Take[q,#1]]&]]]];
Table[If[n==0,0,Length[lynfac[Reverse[IntegerDigits[n,2]]]]],{n,0,30}]

A329398 Number of compositions of n with uniform Lyndon factorization and uniform co-Lyndon factorization.

Original entry on oeis.org

1, 2, 4, 7, 12, 18, 28, 40, 57, 80, 110, 148, 200, 266, 348, 457, 592, 764, 978, 1248, 1580, 2000, 2508, 3142, 3913
Offset: 1

Views

Author

Gus Wiseman, Nov 13 2019

Keywords

Comments

We define the Lyndon product of two or more finite sequences to be the lexicographically maximal sequence obtainable by shuffling the sequences together. For example, the Lyndon product of (231) with (213) is (232131), the product of (221) with (213) is (222131), and the product of (122) with (2121) is (2122121). A Lyndon word is a finite sequence that is prime with respect to the Lyndon product. Equivalently, a Lyndon word is a finite sequence that is lexicographically strictly less than all of its cyclic rotations. Every finite sequence has a unique (orderless) factorization into Lyndon words, and if these factors are arranged in lexicographically decreasing order, their concatenation is equal to their Lyndon product. For example, (1001) has sorted Lyndon factorization (001)(1).
Similarly, the co-Lyndon product is the lexicographically minimal sequence obtainable by shuffling the sequences together, and a co-Lyndon word is a finite sequence that is prime with respect to the co-Lyndon product, or, equivalently, a finite sequence that is lexicographically strictly greater than all of its cyclic rotations. For example, (1001) has sorted co-Lyndon factorization (1)(100).
A sequence of words is uniform if they all have the same length.
Conjecture: Also the number of compositions of n that are either weakly increasing or weakly decreasing. Hence a(n) = 2 * A000041(n) - A000005(n). - Gus Wiseman, Mar 05 2020

Examples

			The a(1) = 1 through a(6) = 18 compositions:
  (1)  (2)   (3)    (4)     (5)      (6)
       (11)  (12)   (13)    (14)     (15)
             (21)   (22)    (23)     (24)
             (111)  (31)    (32)     (33)
                    (112)   (41)     (42)
                    (211)   (113)    (51)
                    (1111)  (122)    (114)
                            (221)    (123)
                            (311)    (222)
                            (1112)   (321)
                            (2111)   (411)
                            (11111)  (1113)
                                     (1122)
                                     (2211)
                                     (3111)
                                     (11112)
                                     (21111)
                                     (111111)

Crossrefs

Lyndon and co-Lyndon compositions are (both) counted by A059966.
Lyndon compositions that are not weakly increasing are A329141.
Lyndon compositions whose reverse is not co-Lyndon are A329324.
Cf. A000740, A001037, A001523, A008965, A059204, A060223, A211100, A328596, A329312, A329318, A329396, A329397, A329399, A332578, A332669, A332834.

Programs

  • Mathematica
    lynQ[q_]:=Array[Union[{q,RotateRight[q,#]}]=={q,RotateRight[q,#]}&,Length[q]-1,1,And];
lynfac[q_]:=If[Length[q]==0,{},Function[i,Prepend[lynfac[Drop[q,i]],Take[q,i]]][Last[Select[Range[Length[q]],lynQ[Take[q,#]]&]]]];
colynQ[q_]:=Array[Union[{RotateRight[q,#],q}]=={RotateRight[q,#],q}&,Length[q]-1,1,And];
colynfac[q_]:=If[Length[q]==0,{},Function[i,Prepend[colynfac[Drop[q,i]],Take[q,i]]]@Last[Select[Range[Length[q]],colynQ[Take[q,#]]&]]];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],SameQ@@Length/@lynfac[#]&&SameQ@@Length/@colynfac[#]&]],{n,10}]

Extensions

a(19)-a(25) from Robert Price, Jun 20 2021

A245563 Table read by rows: row n gives list of lengths of runs of 1's in binary expansion of n, starting with low-order bits.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 3, 4, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 2, 2, 3, 1, 3, 4, 5, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 1, 4, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 2, 2, 2, 3, 2, 3, 1, 3, 1, 3, 2, 3, 4, 1, 4, 5, 6, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

N. J. A. Sloane, Aug 10 2014

Keywords

Comments

A formula for A071053(n) depends on this table.

Examples

			Here are the run lengths for the numbers 0 through 21:
0, []
1, [1]
2, [1]
3, [2]
4, [1]
5, [1, 1]
6, [2]
7, [3]
8, [1]
9, [1, 1]
10, [1, 1]
11, [2, 1]
12, [2]
13, [1, 2]
14, [3]
15, [4]
16, [1]
17, [1, 1]
18, [1, 1]
19, [2, 1]
20, [1, 1]
21, [1, 1, 1]

Links

Crossrefs

Row sums = A000120 (the binary weight).
Cf. A245562, A071053.
Cf. A030308, A227736.
Row lengths are A069010.
The version for prime indices (instead of binary indices) is A124010.
Numbers with distinct run-lengths are A328592.
Numbers with equal run-lengths are A164707.
Cf. A003714, A014081, A328166, A328594, A328595, A328596.

Programs

  • Haskell
    import Data.List (group)
a245563 n k = a245563_tabf !! n !! k
a245563_row n = a245563_tabf !! n
a245563_tabf = [0] : map
   (map length . (filter ((== 1) . head)) . group) (tail a030308_tabf)
-- Reinhard Zumkeller, Aug 10 2014
  • Maple
    for n from 0 to 128 do
lis:=[]; t1:=convert(n,base,2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[op(lis),c]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[op(lis),c]; fi;
od:
lprint(n,lis);
od:
  • Mathematica
    Join@@Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,0,100}] (* Gus Wiseman, Nov 03 2019 *)
  • Python
    from re import split
A245563_list = [0]
for n in range(1,100):
    A245563_list.extend(len(d) for d in split('0+',bin(n)[:1:-1]) if d != '')
# Chai Wah Wu, Sep 07 2014

A121016 Numbers whose binary expansion is properly periodic.

Original entry on oeis.org

3, 7, 10, 15, 31, 36, 42, 45, 54, 63, 127, 136, 153, 170, 187, 204, 221, 238, 255, 292, 365, 438, 511, 528, 561, 594, 627, 660, 682, 693, 726, 759, 792, 825, 858, 891, 924, 957, 990, 1023, 2047, 2080, 2145, 2184, 2210, 2275, 2340, 2405, 2457, 2470, 2535
Offset: 1

Views

Author

Jacob A. Siehler, Sep 08 2006

Keywords

Comments

A finite sequence is aperiodic if its cyclic rotations are all different. - Gus Wiseman, Oct 31 2019

Examples

			For example, 204=(1100 1100)_2 and 292=(100 100 100)_2 belong to the sequence, but 30=(11110)_2 cannot be split into repeating periods.
From _Gus Wiseman_, Oct 31 2019: (Start)
The sequence of terms together with their binary expansions and binary indices begins:
   3:         11 ~ {1,2}
   7:        111 ~ {1,2,3}
   10:      1010 ~ {2,4}
   15:      1111 ~ {1,2,3,4}
   31:     11111 ~ {1,2,3,4,5}
   36:    100100 ~ {3,6}
   42:    101010 ~ {2,4,6}
   45:    101101 ~ {1,3,4,6}
   54:    110110 ~ {2,3,5,6}
   63:    111111 ~ {1,2,3,4,5,6}
  127:   1111111 ~ {1,2,3,4,5,6,7}
  136:  10001000 ~ {4,8}
  153:  10011001 ~ {1,4,5,8}
  170:  10101010 ~ {2,4,6,8}
  187:  10111011 ~ {1,2,4,5,6,8}
  204:  11001100 ~ {3,4,7,8}
  221:  11011101 ~ {1,3,4,5,7,8}
  238:  11101110 ~ {2,3,4,6,7,8}
  255:  11111111 ~ {1,2,3,4,5,6,7,8}
  292: 100100100 ~ {3,6,9}
(End)

Links

Crossrefs

A020330 is a subsequence.
Numbers whose binary expansion is aperiodic are A328594.
Numbers whose reversed binary expansion is Lyndon are A328596.
Numbers whose binary indices have equal run-lengths are A164707.
Cf. A000120, A003714, A014081, A065609, A069010, A275692, A328595.

Programs

  • Mathematica
    PeriodicQ[n_, base_] := Block[{l = IntegerDigits[n, base]}, MemberQ[ RotateLeft[l, # ] & /@ Most@ Divisors@ Length@l, l]]; Select[ Range@2599, PeriodicQ[ #, 2] &]
  • PARI
    is(n)=n=binary(n);fordiv(#n,d,for(i=1,#n/d-1, for(j=1,d, if(n[j]!=n[j+i*d], next(3)))); return(d<#n)) \\ Charles R Greathouse IV, Dec 10 2013

A326774 For any number m, let m* be the bi-infinite string obtained by repetition of the binary representation of m; this sequence lists the numbers n such that for any k < n, n* does not equal k* up to a shift.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 11, 16, 17, 18, 19, 21, 23, 32, 33, 34, 35, 37, 38, 39, 43, 47, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 77, 78, 79, 85, 87, 91, 95, 128, 129, 130, 131, 132, 133, 134, 135, 137, 138, 139, 140, 141, 142, 143, 146, 147, 149, 150, 151, 154
Offset: 0

Views

Author

Rémy Sigrist, Jul 27 2019

Keywords

Comments

This sequence contains every power of 2.
No term belongs to A121016.
Every terms belongs to A004761.
For any k > 0, there are A001037(k) terms with binary length k.
From Gus Wiseman, Apr 19 2020: (Start)
Also numbers k such that the k-th composition in standard order (row k of A066099) is a co-Lyndon word (regular Lyndon words being A275692). For example, the sequence of all co-Lyndon words begins:
0: () 37: (3,2,1) 79: (3,1,1,1,1)
1: (1) 38: (3,1,2) 85: (2,2,2,1)
2: (2) 39: (3,1,1,1) 87: (2,2,1,1,1)
4: (3) 43: (2,2,1,1) 91: (2,1,2,1,1)
5: (2,1) 47: (2,1,1,1,1) 95: (2,1,1,1,1,1)
8: (4) 64: (7) 128: (8)
9: (3,1) 65: (6,1) 129: (7,1)
11: (2,1,1) 66: (5,2) 130: (6,2)
16: (5) 67: (5,1,1) 131: (6,1,1)
17: (4,1) 68: (4,3) 132: (5,3)
18: (3,2) 69: (4,2,1) 133: (5,2,1)
19: (3,1,1) 70: (4,1,2) 134: (5,1,2)
21: (2,2,1) 71: (4,1,1,1) 135: (5,1,1,1)
23: (2,1,1,1) 73: (3,3,1) 137: (4,3,1)
32: (6) 74: (3,2,2) 138: (4,2,2)
33: (5,1) 75: (3,2,1,1) 139: (4,2,1,1)
34: (4,2) 77: (3,1,2,1) 140: (4,1,3)
35: (4,1,1) 78: (3,1,1,2) 141: (4,1,2,1)
(End)

Examples

			3* = ...11... equals 1* = ...1..., so 3 is not a term.
6* = ...110... equals up to a shift 5* = ...101..., so 6 is not a term.
11* = ...1011... only equals up to a shift 13* = ...1101... and 14* = ...1110..., so 11 is a term.

Links

Crossrefs

Cf. A001037, A004761, A065609, A121016.
Necklace compositions are counted by A008965.
Lyndon compositions are counted by A059966.
Length of Lyndon factorization of binary expansion is A211100.
Numbers whose reversed binary expansion is a necklace are A328595.
Length of co-Lyndon factorization of binary expansion is A329312.
Length of Lyndon factorization of reversed binary expansion is A329313.
Length of co-Lyndon factorization of reversed binary expansion is A329326.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Necklaces are A065609.
- Sum is A070939.
- Runs are counted by A124767.
- Rotational symmetries are counted by A138904.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon compositions are A275692.
- Co-Lyndon compositions are A326774 (this sequence).
- Aperiodic compositions are A328594.
- Reversed co-necklaces are A328595.
- Rotational period is A333632.
- Co-necklaces are A333764.
- Co-Lyndon factorizations are counted by A333765.
- Lyndon factorizations are counted by A333940.
- Reversed necklaces are A333943.
- Length of co-Lyndon factorization is A334029.
Cf. A000740, A034691, A060223, A269134, A292884, A328596.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
colynQ[q_]:=Length[q]==0||Array[Union[{RotateRight[q,#],q}]=={RotateRight[q,#],q}&,Length[q]-1,1,And];
Select[Range[0,100],colynQ[stc[#]]&] (* Gus Wiseman, Apr 19 2020 *)
  • PARI
    See Links section.

A333764 Numbers k such that the k-th composition in standard order is a co-necklace.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 15, 16, 17, 18, 19, 21, 23, 31, 32, 33, 34, 35, 36, 37, 38, 39, 42, 43, 45, 47, 63, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 77, 78, 79, 85, 87, 91, 95, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140
Offset: 1

Views

Author

Gus Wiseman, Apr 12 2020

Keywords

Comments

A co-necklace is a finite sequence that is lexicographically greater than or equal to any cyclic rotation.
A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions

Examples

			The sequence together with the corresponding co-necklaces begins:
    1: (1)             32: (6)               69: (4,2,1)
    2: (2)             33: (5,1)             70: (4,1,2)
    3: (1,1)           34: (4,2)             71: (4,1,1,1)
    4: (3)             35: (4,1,1)           73: (3,3,1)
    5: (2,1)           36: (3,3)             74: (3,2,2)
    7: (1,1,1)         37: (3,2,1)           75: (3,2,1,1)
    8: (4)             38: (3,1,2)           77: (3,1,2,1)
    9: (3,1)           39: (3,1,1,1)         78: (3,1,1,2)
   10: (2,2)           42: (2,2,2)           79: (3,1,1,1,1)
   11: (2,1,1)         43: (2,2,1,1)         85: (2,2,2,1)
   15: (1,1,1,1)       45: (2,1,2,1)         87: (2,2,1,1,1)
   16: (5)             47: (2,1,1,1,1)       91: (2,1,2,1,1)
   17: (4,1)           63: (1,1,1,1,1,1)     95: (2,1,1,1,1,1)
   18: (3,2)           64: (7)              127: (1,1,1,1,1,1,1)
   19: (3,1,1)         65: (6,1)            128: (8)
   21: (2,2,1)         66: (5,2)            129: (7,1)
   23: (2,1,1,1)       67: (5,1,1)          130: (6,2)
   31: (1,1,1,1,1)     68: (4,3)            131: (6,1,1)

Crossrefs

The non-"co" version is A065609.
The reversed version is A328595.
Binary necklaces are A000031.
Necklace compositions are A008965.
Necklaces covering an initial interval are A019536.
Numbers whose prime signature is a necklace are A329138.
Length of co-Lyndon factorization of binary expansion is A329312.
Length of Lyndon factorization of reversed binary expansion is A329313.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Runs are counted by A124767.
- Rotational symmetries are counted by A138904.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon compositions are A275692.
- Co-Lyndon compositions are A326774.
- Aperiodic compositions are A328594.
- Length of Lyndon factorization is A329312.
- Rotational period is A333632.
- Reversed necklaces are A333943.
Cf. A000740, A001037, A027375, A059966, A211100, A302291, A328596, A329142, A333765, A333939, A333941.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
coneckQ[q_]:=Array[OrderedQ[{RotateRight[q,#],q}]&,Length[q]-1,1,And];
Select[Range[100],coneckQ[stc[#]]&]

A329326 Length of the co-Lyndon factorization of the reversed binary expansion of n.

Original entry on oeis.org

1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 2, 4, 3, 4, 4, 5, 2, 3, 2, 4, 3, 3, 2, 5, 3, 4, 3, 5, 4, 5, 5, 6, 2, 3, 2, 4, 2, 3, 2, 5, 3, 4, 2, 4, 3, 3, 2, 6, 3, 4, 3, 5, 4, 4, 3, 6, 4, 5, 4, 6, 5, 6, 6, 7, 2, 3, 2, 4, 2, 3, 2, 5, 3, 3, 2, 4, 3, 3, 2, 6, 3, 4, 2, 5, 4, 3, 2
Offset: 1

Views

Author

Gus Wiseman, Nov 11 2019

Keywords

Comments

First differs from A211100 at a(77) = 3, A211100(77) = 2. The reversed binary expansion of 77 is (1011001), with co-Lyndon factorization (10)(1100)(1), while the binary expansion is (1001101), with Lyndon factorization of (1)(001101).
The co-Lyndon product of two or more finite sequences is defined to be the lexicographically minimal sequence obtainable by shuffling the sequences together. For example, the co-Lyndon product of (231) and (213) is (212313), the product of (221) and (213) is (212213), and the product of (122) and (2121) is (1212122). A co-Lyndon word is a finite sequence that is prime with respect to the co-Lyndon product. Equivalently, a co-Lyndon word is a finite sequence that is lexicographically strictly greater than all of its cyclic rotations. Every finite sequence has a unique (orderless) factorization into co-Lyndon words, and if these factors are arranged in certain order, their concatenation is equal to their co-Lyndon product. For example, (1001) has sorted co-Lyndon factorization (1)(100).

Examples

			The reversed binary expansion of each positive integer together with their co-Lyndon factorizations begins:
   1:     (1) = (1)
   2:    (01) = (0)(1)
   3:    (11) = (1)(1)
   4:   (001) = (0)(0)(1)
   5:   (101) = (10)(1)
   6:   (011) = (0)(1)(1)
   7:   (111) = (1)(1)(1)
   8:  (0001) = (0)(0)(0)(1)
   9:  (1001) = (100)(1)
  10:  (0101) = (0)(10)(1)
  11:  (1101) = (110)(1)
  12:  (0011) = (0)(0)(1)(1)
  13:  (1011) = (10)(1)(1)
  14:  (0111) = (0)(1)(1)(1)
  15:  (1111) = (1)(1)(1)(1)
  16: (00001) = (0)(0)(0)(0)(1)
  17: (10001) = (1000)(1)
  18: (01001) = (0)(100)(1)
  19: (11001) = (1100)(1)
  20: (00101) = (0)(0)(10)(1)

Crossrefs

The non-"co" version is A211100.
Positions of 2's are A329357.
Numbers whose binary expansion is co-Lyndon are A275692.
Length of the co-Lyndon factorization of the binary expansion is A329312.
Cf. A000031, A001037, A059966, A060223, A211097, A296372, A296658, A328596, A329131, A329314, A329318, A329324, A329325.

Programs

  • Mathematica
    colynQ[q_]:=Array[Union[{RotateRight[q,#],q}]=={RotateRight[q,#],q}&,Length[q]-1,1,And];
colynfac[q_]:=If[Length[q]==0,{},Function[i,Prepend[colynfac[Drop[q,i]],Take[q,i]]]@Last[Select[Range[Length[q]],colynQ[Take[q,#]]&]]];
Table[Length[colynfac[Reverse[IntegerDigits[n,2]]]],{n,100}]

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219
Offset: 1

Views

Author

Olivier Gérard

Keywords

Comments

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Links

Crossrefs

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Programs

  • Fortran
    c See link in A139351.
  • Maple
    N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015
  • Mathematica
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)
  • PARI
    for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Formula

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002
That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A329131 Numbers whose prime signature is a Lyndon word.

Original entry on oeis.org

2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 18, 19, 23, 25, 27, 29, 31, 32, 37, 41, 43, 47, 49, 50, 53, 54, 59, 61, 64, 67, 71, 73, 75, 79, 81, 83, 89, 97, 98, 101, 103, 107, 108, 109, 113, 121, 125, 127, 128, 131, 137, 139, 147, 149, 150, 151, 157, 162, 163, 167
Offset: 1

Views

Author

Gus Wiseman, Nov 06 2019

Keywords

Comments

First differs from A133811 in having 50.
A Lyndon word is a finite sequence that is lexicographically strictly less than all of its cyclic rotations.
A number's prime signature is the sequence of positive exponents in its prime factorization.

Examples

			The prime signature of 30870 is (1,2,1,3), which is a Lyndon word, so 30870 is in the sequence.
The sequence of terms together with their prime indices begins:
    2: {1}
    3: {2}
    4: {1,1}
    5: {3}
    7: {4}
    8: {1,1,1}
    9: {2,2}
   11: {5}
   13: {6}
   16: {1,1,1,1}
   17: {7}
   18: {1,2,2}
   19: {8}
   23: {9}
   25: {3,3}
   27: {2,2,2}
   29: {10}
   31: {11}
   32: {1,1,1,1,1}

Crossrefs

Numbers whose reversed binary expansion is Lyndon are A328596.
Numbers whose prime signature is a necklace are A329138.
Numbers whose prime signature is aperiodic are A329139.
Lyndon compositions are A059966.
Prime signature is A124010.
Cf. A000031, A000740, A000961, A001037, A025487, A027375, A097318, A112798, A118914, A304678, A318731, A329140, A329142.

Programs

  • Mathematica
    lynQ[q_]:=Array[Union[{q,RotateRight[q,#]}]=={q,RotateRight[q,#]}&,Length[q]-1,1,And];
Select[Range[2,100],lynQ[Last/@FactorInteger[#]]&]

Formula

Intersection of A329138 and A329139.

A333632 Rotational period of the k-th composition in standard order; a(0) = 0.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 3, 2, 3, 3, 1, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 1, 1, 2, 2, 3, 1, 3, 3, 4, 2, 3, 1, 4, 3, 2, 4, 5, 2, 3, 3, 4, 3, 4, 2, 5, 3, 4, 4, 5, 4, 5, 5, 1, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4
Offset: 0

Views

Author

Gus Wiseman, Apr 12 2020

Keywords

Comments

A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The a(299) = 5 rotations:
  (1,1,3,2,2)
  (1,3,2,2,1)
  (3,2,2,1,1)
  (2,2,1,1,3)
  (2,1,1,3,2)
The a(9933) = 4 rotations:
  (1,2,1,3,1,2,1,3)
  (1,3,1,2,1,3,1,2)
  (2,1,3,1,2,1,3,1)
  (3,1,2,1,3,1,2,1)

Crossrefs

Aperiodic compositions are counted by A000740.
Aperiodic binary words are counted by A027375.
The orderless period of prime indices is A052409.
Numbers whose binary expansion is periodic are A121016.
Periodic compositions are counted by A178472.
The version for binary expansion is A302291.
Numbers whose prime signature is aperiodic are A329139.
Compositions by number of distinct rotations are A333941.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Necklaces are A065609.
- Sum is A070939.
- Equal runs are counted by A124767.
- Rotational symmetries are counted by A138904.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon compositions are A275692.
- Co-Lyndon compositions are A326774.
- Aperiodic compositions are A328594.
- Rotational period is A333632 (this sequence).
- Co-necklaces are A333764.
- Reversed necklaces are A333943.
Cf. A000031, A001037, A008965, A019536, A211100, A328595, A328596, A329312, A329313, A329326.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Length[Union[Array[RotateRight[stc[n],#]&,DigitCount[n,2,1]]]],{n,0,100}]

Formula

a(n) = A000120(n)/A138904(n) = A302291(n) - A023416(n)/A138904(n).
Previous Showing 11-20 of 67 results. Next

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

Content is available under The OEIS End-User License Agreement.