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 1-10 of 16 results. Next

A059966 a(n) = (1/n) * Sum_{ d divides n } mu(n/d) * (2^d - 1).

Original entry on oeis.org

1, 1, 2, 3, 6, 9, 18, 30, 56, 99, 186, 335, 630, 1161, 2182, 4080, 7710, 14532, 27594, 52377, 99858, 190557, 364722, 698870, 1342176, 2580795, 4971008, 9586395, 18512790, 35790267, 69273666, 134215680, 260300986, 505286415, 981706806
Offset: 1

Views

Author

Roland Bacher, Mar 05 2001

Keywords

Comments

Dimensions of the homogeneous parts of the free Lie algebra with one generator in 1,2,3, etc. (Lie analog of the partition numbers).
This sequence is the Lie analog of the partition sequence (which gives the dimensions of the homogeneous polynomials with one generator in each degree) or similarly, of the partitions into distinct (or odd numbers) (which gives the dimensions of the homogeneous parts of the exterior algebra with one generator in each dimension).
The number of cycles of length n of rectangle shapes in the process of repeatedly cutting a square off the end of the rectangle. For example, the one cycle of length 1 is the golden rectangle. - David Pasino (davepasino(AT)yahoo.com), Jan 29 2009
In music, the number of distinct rhythms, at a given tempo, produced by a continuous repetition of measures with identical patterns of 1's and 0's (where 0 means no beat, and 1 means one beat), where each measure allows for n possible beats of uniform character, and when counted under these two conditions: (i) the starting and ending times for the measure are unknown or irrelevant and (ii) identical rhythms that can be produced by using a measure with fewer than n possible beats are excluded from the count. - Richard R. Forberg, Apr 22 2013
Richard R. Forberg's comment does not hold for n=1 because a(1)=1 but there are the two possible rhythms: "0" and "1". - Herbert Kociemba, Oct 24 2016
The comment does hold for n=1 as the rhythm "0" can be produced by using a measure of 0 beats and so is properly excluded from a(1)=1 by condition (ii) of the comment. - Travis Scott, May 28 2022
a(n) is also the number of Lyndon compositions (aperiodic necklaces of positive integers) with sum n. - Gus Wiseman, Dec 19 2017
Mobius transform of A008965. - Jianing Song, Nov 13 2021
a(n) is the number of cycles of length n for the map x->1 - abs(2*x-1) applied on rationals 0Michel Marcus, Jul 16 2025

Examples

			a(4)=3: the 3 elements [a,c], [a[a,b]] and d form a basis of all homogeneous elements of degree 4 in the free Lie algebra with generators a of degree 1, b of degree 2, c of degree 3 and d of degree 4.
From _Gus Wiseman_, Dec 19 2017: (Start)
The sequence of Lyndon compositions organized by sum begins:
  (1),
  (2),
  (3),(12),
  (4),(13),(112),
  (5),(14),(23),(113),(122),(1112),
  (6),(15),(24),(114),(132),(123),(1113),(1122),(11112),
  (7),(16),(25),(115),(34),(142),(124),(1114),(133),(223),(1213),(1132),(1123),(11113),(1222),(11212),(11122),(111112). (End)
		

References

  • C. Reutenauer, Free Lie algebras, Clarendon press, Oxford (1993).

Crossrefs

Apart from initial terms, same as A001037.

Programs

  • Haskell
    a059966 n = sum (map (\x -> a008683 (n `div` x) * a000225 x)
                         [d | d <- [1..n], mod n d == 0]) `div` n
    -- Reinhard Zumkeller, Nov 18 2011
    
  • Mathematica
    Table[1/n Apply[Plus, Map[(MoebiusMu[n/# ](2^# - 1)) &, Divisors[n]]], {n, 20}]
    (* Second program: *)
    Table[(1/n) DivisorSum[n, MoebiusMu[n/#] (2^# - 1) &], {n, 35}] (* Michael De Vlieger, Jul 22 2019 *)
  • Python
    from sympy import mobius, divisors
    def A059966(n): return sum(mobius(n//d)*(2**d-1) for d in divisors(n,generator=True))//n # Chai Wah Wu, Feb 03 2022

Formula

G.f.: Product_{n>0} (1-q^n)^a(n) = 1-q-q^2-q^3-q^4-... = 2-1/(1-q).
Inverse Euler transform of A011782. - Alois P. Heinz, Jun 23 2018
G.f.: Sum_{k>=1} mu(k)*log((1 - x^k)/(1 - 2*x^k))/k. - Ilya Gutkovskiy, May 19 2019
a(n) ~ 2^n / n. - Vaclav Kotesovec, Aug 10 2019
Dirichlet g.f.: f(s+1)/zeta(s+1) - 1, where f(s) = Sum_{n>=1} 2^n/n^s. - Jianing Song, Nov 13 2021

Extensions

Explicit formula from Paul D. Hanna, Apr 15 2002
Description corrected by Axel Kleinschmidt, Sep 15 2002

A102659 List of Lyndon words on {1,2} sorted first by length and then lexicographically.

Original entry on oeis.org

1, 2, 12, 112, 122, 1112, 1122, 1222, 11112, 11122, 11212, 11222, 12122, 12222, 111112, 111122, 111212, 111222, 112122, 112212, 112222, 121222, 122222, 1111112, 1111122, 1111212, 1111222, 1112112, 1112122, 1112212, 1112222, 1121122
Offset: 1

Views

Author

N. J. A. Sloane, Feb 03 2005

Keywords

Comments

A Lyndon word is primitive (not a power of another word) and is earlier in lexicographic order than any of its cyclic shifts.

Crossrefs

The "co" version is A329318.
A triangular version is A296657.
A sequence listing all Lyndon compositions is A294859.
Numbers whose binary expansion is Lyndon are A328596.
Length of the Lyndon factorization of the binary expansion is A211100.

Programs

  • Haskell
    cf. link.
    
  • Mathematica
    lynQ[q_]:=Array[Union[{q,RotateRight[q,#]}]=={q,RotateRight[q,#]}&,Length[q]-1,1,And];
    Join@@Table[FromDigits/@Select[Tuples[{1,2},n],lynQ],{n,5}] (* Gus Wiseman, Nov 14 2019 *)
  • PARI
    is_A102659(n)={ vecsort(d=digits(n))!=d&&for(i=1,#d-1, n>[1,10^(#d-i)]*divrem(n,10^i)&&return); fordiv(#d,L,L<#d && d==concat(Col(vector(#d/L,i,1)~*vecextract(d,2^L-1))~)&&return); !setminus(Set(d),[1,2])} \\ The last check is the least expensive one, but not useful if we test only numbers with digits {1,2}.
    for(n=1,6,p=vector(n,i,10^(n-i))~;forvec(d=vector(n,i,[1,2]),is_A102659(m=d*p)&&print1(m","))) \\ One could use is_A102660 instead of is_A102659 here. - M. F. Hasler, Mar 08 2014

Formula

A102659 = A102660 intersect A007931 = A213969 intersect A239016. - M. F. Hasler, Mar 10 2014

Extensions

More terms from Franklin T. Adams-Watters, Dec 14 2006
Definition improved by Reinhard Zumkeller, Mar 23 2012

A228369 Triangle read by rows in which row n lists the compositions (ordered partitions) of n in lexicographic order.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Aug 28 2013

Keywords

Comments

The representation of the compositions (for fixed n) is as lists of parts, the order between individual compositions (for the same n) is lexicographic. - Joerg Arndt, Sep 02 2013
The equivalent sequence for partitions is A026791.
Row n has length A001792(n-1).
Row sums give A001787, n >= 1.
The m-th composition has length A008687(m+1), m >= 1. - Andrey Zabolotskiy, Jul 19 2017

Examples

			Illustration of initial terms:
-----------------------------------
n  j       Diagram   Composition j
-----------------------------------
.               _
1  1           |_|   1;
.             _ _
2  1         | |_|   1, 1,
2  2         |_ _|   2;
.           _ _ _
3  1       | | |_|   1, 1, 1,
3  2       | |_ _|   1, 2,
3  3       |   |_|   2, 1,
3  4       |_ _ _|   3;
.         _ _ _ _
4  1     | | | |_|   1, 1, 1, 1,
4  2     | | |_ _|   1, 1, 2,
4  3     | |   |_|   1, 2, 1,
4  4     | |_ _ _|   1, 3,
4  5     |   | |_|   2, 1, 1,
4  6     |   |_ _|   2, 2,
4  7     |     |_|   3, 1,
4  8     |_ _ _ _|   4;
.
Triangle begins:
[1];
[1,1],[2];
[1,1,1],[1,2],[2,1],[3];
[1,1,1,1],[1,1,2],[1,2,1],[1,3],[2,1,1],[2,2],[3,1],[4];
[1,1,1,1,1],[1,1,1,2],[1,1,2,1],[1,1,3],[1,2,1,1],[1,2,2],[1,3,1],[1,4],[2,1,1,1],[2,1,2],[2,2,1],[2,3],[3,1,1],[3,2],[4,1],[5];
...
		

Crossrefs

Programs

  • Haskell
    a228369 n = a228369_list !! (n - 1)
    a228369_list = concatMap a228369_row [1..]
    a228369_row 0 = []
    a228369_row n
      | 2^k == 2 * n + 2 = [k - 1]
      | otherwise        = a228369_row (n `div` 2^k) ++ [k] where
        k = a007814 (n + 1) + 1
    -- Peter Kagey, Jun 27 2016
    
  • Mathematica
    Table[Sort[Join@@Permutations/@IntegerPartitions[n],OrderedQ[PadRight[{#1,#2}]]&],{n,5}] (* Gus Wiseman, Dec 14 2017 *)
  • PARI
    gen_comp(n)=
    {  /* Generate compositions of n as lists of parts (order is lex): */
        my(ct = 0);
        my(m, z, pt);
        \\ init:
        my( a = vector(n, j, 1) );
        m = n;
        while ( 1,
            ct += 1;
            pt = vector(m, j, a[j]);
            /* for A228369  print composition: */
            for (j=1, m, print1(pt[j],", ") );
    \\        /* for A228525 print reversed (order is colex): */
    \\        forstep (j=m, 1, -1, print1(pt[j],", ") );
            if ( m<=1,  return(ct) );  \\ current is last
            a[m-1] += 1;
            z = a[m] - 2;
            a[m] = 1;
            m += z;
        );
        return(ct);
    }
    for(n=1, 12, gen_comp(n) );
    \\ Joerg Arndt, Sep 02 2013
    
  • Python
    a = [[[]], [[1]]]
    for s in range(2, 9):
        a.append([])
        for k in range(1, s+1):
            for ss in a[s-k]:
                a[-1].append([k]+ss)
    print(a)
    # Andrey Zabolotskiy, Jul 19 2017

A296774 Triangle read by rows in which row n lists the compositions of n ordered first by length and then reverse-lexicographically.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Dec 20 2017

Keywords

Examples

			Triangle of compositions begins:
(1),
(2),(11),
(3),(21),(12),(111),
(4),(31),(22),(13),(211),(121),(112),(1111),
(5),(41),(32),(23),(14),(311),(221),(212),(131),(122),(113),(2111),(1211),(1121),(1112),(11111).
		

Crossrefs

Programs

  • Mathematica
    Table[Sort[Join@@Permutations/@IntegerPartitions[n],Or[Length[#1]
    				

A296302 Number of aperiodic compositions of n with relatively prime parts. Number of compositions of n with relatively prime parts and relatively prime run-lengths.

Original entry on oeis.org

1, 0, 2, 5, 14, 24, 62, 114, 249, 480, 1022, 1978, 4094, 8064, 16348, 32520, 65534, 130512, 262142, 523270, 1048444, 2095104, 4194302, 8384316, 16777185, 33546240, 67108356, 134201398, 268435454, 536837136, 1073741822, 2147418240, 4294965244, 8589803520
Offset: 1

Views

Author

Gus Wiseman, Dec 11 2017

Keywords

Examples

			The a(6) = 24 aperiodic compositions with relatively prime parts are:
(15), (51),
(114), (123), (132), (141), (213), (231), (312), (321), (411),
(1113), (1122), (1131), (1221), (1311), (2112), (2211), (3111),
(11112), (11121), (11211), (12111), (21111).
		

Crossrefs

Programs

  • Mathematica
    Table[DivisorSum[n,Function[d,MoebiusMu[n/d]*DivisorSum[d,MoebiusMu[#]*2^(d/#-1)&]]],{n,20}]

Formula

a = mu * mu * c, where * is Dirichlet convolution and c(n) = 2^(n-1).

A296372 Triangle read by rows: T(n,k) is the number of normal sequences of length n whose standard factorization into Lyndon words (aperiodic necklaces) has k factors.

Original entry on oeis.org

1, 1, 2, 4, 5, 4, 18, 31, 18, 8, 108, 208, 153, 56, 16, 778, 1700, 1397, 616, 160, 32, 6756, 15980, 14668, 7197, 2196, 432, 64, 68220, 172326, 171976, 93293, 31564, 7208, 1120, 128
Offset: 1

Views

Author

Gus Wiseman, Dec 11 2017

Keywords

Comments

A finite sequence is normal if its union is an initial interval of positive integers.

Examples

			The T(3,2) = 5 normal sequences are {2,1,2}, {1,2,1}, {2,1,3}, {2,3,1}, {3,1,2}.
Triangle begins:
     1;
     1,     2;
     4,     5,     4;
    18,    31,    18,     8;
   108,   208,   153,    56,    16;
   778,  1700,  1397,   616,   160,    32;
  6756, 15980, 14668,  7197,  2196,   432,    64;
		

Crossrefs

Programs

  • Mathematica
    neckQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And];
    aperQ[q_]:=UnsameQ@@Table[RotateRight[q,k],{k,Length[q]}];
    qit[q_]:=If[#===Length[q],{q},Prepend[qit[Drop[q,#]],Take[q,#]]]&[Max@@Select[Range[Length[q]],neckQ[Take[q,#]]&&aperQ[Take[q,#]]&]];
    allnorm[n_]:=Function[s,Array[Count[s,y_/;y<=#]+1&,n]]/@Subsets[Range[n-1]+1];
    Table[Length[Select[Join@@Permutations/@allnorm[n],Length[qit[#]]===k&]],{n,5},{k,n}]
  • PARI
    \\ here U(n,k) is A074650(n,k).
    EulerMT(u)={my(n=#u, p=x*Ser(u), vars=variables(p)); Vec(exp( sum(i=1, n, substvec(p + O(x*x^(n\i)), vars, apply(v->v^i,vars))/i ))-1)}
    U(n,k)={sumdiv(n, d, moebius(n/d) * k^d)/n}
    A(n)={[Vecrev(p/y) | p<-sum(k=1, n, EulerMT(vector(n, n, y*U(n,k)))*sum(j=k, n, (-1)^(k-j)*binomial(j,k)))]}
    { my(T=A(10)); for(n=1, #T, print(T[n])) } \\ Andrew Howroyd, Dec 08 2018

Extensions

Example and program corrected by Gus Wiseman, Dec 08 2018

A296373 Triangle T(n,k) = number of compositions of n whose factorization into Lyndon words (aperiodic necklaces) is of length k.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 6, 5, 3, 1, 1, 9, 12, 6, 3, 1, 1, 18, 21, 14, 6, 3, 1, 1, 30, 45, 27, 15, 6, 3, 1, 1, 56, 84, 61, 29, 15, 6, 3, 1, 1, 99, 170, 120, 67, 30, 15, 6, 3, 1, 1, 186, 323, 254, 136, 69, 30, 15, 6, 3, 1, 1, 335, 640, 510, 295, 142, 70, 30, 15, 6, 3, 1, 1
Offset: 1

Views

Author

Gus Wiseman, Dec 11 2017

Keywords

Examples

			Triangle begins:
    1;
    1,   1;
    2,   1,   1;
    3,   3,   1,   1;
    6,   5,   3,   1,   1;
    9,  12,   6,   3,   1,   1;
   18,  21,  14,   6,   3,   1,   1;
   30,  45,  27,  15,   6,   3,   1,   1;
   56,  84,  61,  29,  15,   6,   3,   1,   1;
   99, 170, 120,  67,  30,  15,   6,   3,   1,   1;
  186, 323, 254, 136,  69,  30,  15,   6,   3,   1,   1;
  335, 640, 510, 295, 142,  70,  30,  15,   6,   3,   1,   1;
		

Crossrefs

Programs

  • Mathematica
    neckQ[q_]:=Array[OrderedQ[{RotateRight[q,#],q}]&,Length[q]-1,1,And];
    aperQ[q_]:=UnsameQ@@Table[RotateRight[q,k],{k,Length[q]}];
    qit[q_]:=If[#===Length[q],{q},Prepend[qit[Drop[q,#]],Take[q,#]]]&[Max@@Select[Range[Length[q]],neckQ[Take[q,#]]&&aperQ[Take[q,#]]&]];
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[qit[#]]===k&]],{n,12},{k,n}]
  • PARI
    EulerMT(u)={my(n=#u, p=x*Ser(u), vars=variables(p)); Vec(exp( sum(i=1, n, substvec(p + O(x*x^(n\i)), vars, apply(v->v^i,vars))/i ))-1)}
    A(n)=[Vecrev(p/y) | p<-EulerMT(y*vector(n, n, sumdiv(n, d, moebius(n/d) * (2^d-1))/n))]
    { my(T=A(12)); for(n=1, #T, print(T[n])) } \\ Andrew Howroyd, Dec 01 2018

Formula

First column is A059966.

A296658 Length of the standard Lyndon word factorization of the first n terms of A000002.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Dec 18 2017

Keywords

Examples

			The standard Lyndon word factorization of (12211212212211211) is (122)(112122122)(112)(1)(1), so a(17) = 5.
The standard factorizations of initial terms of A000002:
1
12
122
122,1
122,1,1
122,112
122,112,1
122,11212
122,112122
122,112122,1
122,11212212
122,112122122
122,112122122,1
122,112122122,1,1
122,112122122,112
122,112122122,112,1
122,112122122,112,1,1
122,112122122,112,112
122,112122122,1121122
122,112122122,1121122,1
		

Crossrefs

Programs

  • Mathematica
    LyndonQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And]&&Array[RotateRight[q,#]&,Length[q],1,UnsameQ];
    qit[q_]:=If[#===Length[q],{q},Prepend[qit[Drop[q,#]],Take[q,#]]]&[Max@@Select[Range[Length[q]],LyndonQ[Take[q,#]]&]];
    kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],Part[q,-2],Last[q]},{1,1,1},0,{1,1,2},1,{1,2,1},2,{1,2,2},0,{2,1,1},2,{2,1,2},2,{2,2,1},1,{2,2,2},1]]];
    Table[Length[qit[Nest[kolagrow,1,n]]],{n,150}]

A294859 Triangle whose n-th row is the concatenated sequence of all Lyndon compositions of n in lexicographic order.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Dec 18 2017

Keywords

Examples

			Triangle of Lyndon compositions begins:
(1),
(2),
(12),(3),
(112),(13),(4),
(1112),(113),(122),(14),(23),(5),
(11112),(1113),(1122),(114),(123),(132),(15),(24),(6),
(111112),(11113),(11122),(1114),(11212),(1123),(1132),(115),(1213),(1222),(124),(133),(142),(16),(223),(25),(34),(7).
		

Crossrefs

Programs

  • Mathematica
    LyndonQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And]&&Array[RotateRight[q,#]&,Length[q],1,UnsameQ];
    Table[Sort[Select[Join@@Permutations/@IntegerPartitions[n],LyndonQ],OrderedQ[PadRight[{#1,#2}]]&],{n,7}]

Formula

Row n is a concatenation of A059966(n) Lyndon words with total length A000740(n).

A296772 Triangle read by rows in which row n lists the compositions of n ordered first by decreasing length and then reverse-lexicographically.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Dec 20 2017

Keywords

Comments

The ordering of compositions in each row is consistent with the reverse-Mathematica ordering of expressions (cf. A124734).
Length of k-th composition is A124748(k-1)+1. - Andrey Zabolotskiy, Dec 20 2017

Examples

			Triangle of compositions begins:
(1),
(11),(2),
(111),(21),(12),(3),
(1111),(211),(121),(112),(31),(22),(13),(4),
(11111),(2111),(1211),(1121),(1112),(311),(221),(212),(131),(122),(113),(41),(32),(23),(14),(5).
		

Crossrefs

Programs

  • Mathematica
    Table[Reverse[Sort[Join@@Permutations/@IntegerPartitions[n]]],{n,6}]
Showing 1-10 of 16 results. Next