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.

A051168 Triangular array T(h,k) for 0 <= k <= h read by rows: T(h,k) = number of binary Lyndon words with k ones and h-k zeros.

Original entry on oeis.org

1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 3, 2, 1, 0, 0, 1, 3, 5, 5, 3, 1, 0, 0, 1, 3, 7, 8, 7, 3, 1, 0, 0, 1, 4, 9, 14, 14, 9, 4, 1, 0, 0, 1, 4, 12, 20, 25, 20, 12, 4, 1, 0, 0, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 0, 0, 1, 5, 18, 40, 66, 75, 66, 40, 18, 5, 1, 0, 0, 1, 6
Offset: 0

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Author

Keywords

Comments

T(h,k) is the number of classes of aperiodic binary words of k ones and h-k zeros; words u,v are in the same class if v is a cyclic permutation of u (e.g., u=111000, v=110001) and a word is aperiodic if not a juxtaposition of 2 or more identical subwords.
T(2n, n), T(2n+1, n), T(n, 3) match A022553, A000108, A001840, respectively. Row sums match A001037.
From R. J. Mathar, Jul 31 2008: (Start)
This triangle may also be regarded as the square array A(r,n), the n-th term of the r-th Witt transform of the all-1 sequence, r >= 1, n >= 0, read by antidiagonals:
This array begins as follows:
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9
0 1 2 3 5 7 9 12 15 18 22 26 30 35 40 45 51 57 63
0 1 2 5 8 14 20 30 40 55 70 91 112 140 168 204 240 285 330
0 1 3 7 14 25 42 66 99 143 200 273 364 476 612 775 969 1197 1463
0 1 3 9 20 42 75 132 212 333 497 728 1026 1428 1932 2583 3384 4389 5598
0 1 4 12 30 66 132 245 429 715 1144 1768 2652 3876 5537 7752 10659 14421 19228
0 1 4 15 40 99 212 429 800 1430 2424 3978 6288 9690 14520 21318 30624 43263 60060
0 1 5 18 55 143 333 715 1430 2700 4862 8398 13995 22610 35530 54477 81719 120175
0 1 5 22 70 200 497 1144 2424 4862 9225 16796 29372 49742 81686 130750 204248
0 1 6 26 91 273 728 1768 3978 8398 16796 32065 58786 104006 178296 297160 482885
0 1 6 30 112 364 1026 2652 6288 13995 29372 58786 112632 208012 371384 643842
0 1 7 35 140 476 1428 3876 9690 22610 49742 104006 208012 400023 742900 1337220
0 1 7 40 168 612 1932 5537 14520 35530 81686 178296 371384 742900 1432613 2674440
...
It is essentially symmetric: A(r,r+i) = A(r,r-i+1).
Some of the diagonals are:
A(r,r+1): A000108
A(r,r): A022553
A(r,r-1): A000108
A(r,r+2): A000150
A(r,r+3): A050181
A(r,r+4): A050182
A(r,r+5): A050183
A(r,r-2): A000150 (End)
Fredman (1975) proved that the number S(n, k, v) of vectors (a_0, ..., a_{n-1}) of nonnegative integer components that satisfy a_0 + ... + a_{n-1} = k and Sum_{i=0..n-1} i*a_i = v (mod n) is given by S(n, k, v) = (1/(n + k)) * Sum_{d | gcd(n, k)} A054533(d, v) * binomial((n + k)/d, k/d) = S(k, n, v). This was also proved by Elashvili et al. (1999), who also proved that S(n, k, v) = Sum_{d | gcd(n, k, v)} S(n/d, k/d, 1). Here, S(n, k, 1) = T(n + k, k). - Petros Hadjicostas, Jul 09 2019

Examples

			Triangle begins with:
h=0: 1
h=1: 1, 1
h=2: 0, 1, 0
h=3: 0, 1, 1, 0
h=4: 0, 1, 1, 1,  0
h=5: 0, 1, 2, 2,  1,  0
h=6: 0, 1, 2, 3,  2,  1, 0
h=7: 0, 1, 3, 5,  5,  3, 1, 0
h=8: 0, 1, 3, 7,  8,  7, 3, 1, 0
h=9: 0, 1, 4, 9, 14, 14, 9, 4, 1, 0
...
T(6,3) counts classes {111000},{110100},{110010}, each of 6 aperiodic. The class {100100} contains 3 periodic words, counted by T(3,1) as {100}, consisting of 3 aperiodic words 100,010,001.
		

Crossrefs

Columns 1-11: A000012, A004526(n-1), A001840(n-4), A006918(n-4), A011795(n-1), A011796(n-6), A011797(n-1), A031164(n-9), A011845, A032168, A032169. See also A000150.

Programs

  • Maple
    A := proc(r,n) local gf,d,genf; genf := 1/(1-x) ; gf := 0 ; for d in numtheory[divisors](r) do gf := gf + numtheory[mobius](d)*(subs(x= x^d,genf))^(r/d) ; od: gf := expand(gf/r) ; coeftayl(gf,x=0,n) ; end proc:
    A051168 := proc(n,k) if n<=1 then 1; elif n=0 or n=k then 0; else A(n-k,k) ; end if;
    end proc:
    seq(seq(A051168(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Mar 29 2011
  • Mathematica
    Table[If[n===0,1,1/n Plus@@(MoebiusMu[ # ]Binomial[n/#,k/# ]&/@ Divisors[GCD[n,k]])],{n,0,12},{k,0,n}] (* Wouter Meeussen, Jul 20 2008 *)
  • PARI
    {T(n, k) = local(A, ps, c); if( k<0 || k>n, 0, if( n==0 && k==0, 1, A = x * O(x^n) + y * O(y^n); ps = 1 - x - y + A; for( m=1, n, for( i=0, m, c = polcoeff( polcoeff(ps, i, x), m-i, y); if( m==n && i==k, break(2), ps *= (1 -y^(m-i) * x^i + A)^c))); -c))} /* Michael Somos, Jul 03 2004 */
    
  • PARI
    T(n,k) = if (n==0, 1, (1/n) * sumdiv(gcd(n,k), d, moebius(d) * binomial(n/d,k/d)));
    tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, May 16 2018

Formula

T(h, k) = 1 for (h, k) in {(0, 0), (1, 0), (1, 1)}; T(h, k) = 0 if h >= 2 and k = 0 or k = h. Otherwise, T(h, k) = (1/h)*(C(h, k)-S(h, k)), where S(h, k) = Sum_{d <= 2, d|h, d|k} (h/d)*T(h/d, k/d).
1 - x - y = Product_{i,j} (1 - x^i * y^j)^T(i+j, j) where i >= 0, j >= 0 are not both zero. - Michael Somos, Jul 03 2004
The prime rows are given by (1+x)^p/p with noninteger coefficients rounded to zero. E.g., for h = 2 below, (1 + x)^2/2 = (1 + 2*x + x^2)/2 = 0.5 + x + 0.5*x^2 gives (0,1,0). - Tom Copeland, Oct 21 2014
T(n,k) = (1/n) * Sum_{d | gcd(n,k)} mu(d) * binomial(n/d, k/d), for n > 0. - Andrew Howroyd, Mar 26 2017
From Petros Hadjicostas, Jun 16 2019: (Start)
O.g.f. for column k >= 1: (x^k/k) * Sum_{d|k} mu(d)/(1 - x^d)^(k/d).
Bivariate o.g.f.: Sum_{n,k >= 0} T(n, k)*x^n*y^k = 1 - Sum_{d >= 1} (mu(d)/d) *log(1 - x^d * (1 + y^d)).
(End)

A092964 Numbers > 1 in A051168, with a(0) = 1.

Original entry on oeis.org

1, 2, 2, 2, 3, 2, 3, 5, 5, 3, 3, 7, 8, 7, 3, 4, 9, 14, 14, 9, 4, 4, 12, 20, 25, 20, 12, 4, 5, 15, 30, 42, 42, 30, 15, 5, 5, 18, 40, 66, 75, 66, 40, 18, 5, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 6, 26, 70, 143, 212, 245, 212, 143, 70, 26, 6, 7, 30, 91, 200, 333, 429, 429, 333, 200
Offset: 0

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Author

Thomas O. Hoffbauer (Thomas.Hoffbauer(AT)cibamberg.de), Apr 20 2004

Keywords

Comments

Original definition: Triangle read by rows in which row n gives the number of equal configurations under cyclic shift.
T(n,k) = A051168(n+3,k+1), if 0Michael Somos, Jul 17 2004
From Paul Weisenhorn, Dec 21 2010 (Start):
T(n,k) is the number of classes that have (k+1) ordered sums of (n+4) with (k+1) positive integers, that can be transformed into each other by a cyclic permutation.
m has 2^(m-1) ordered sums; for each sum one remove the first part z(1) and add 1 to the next z(1) parts to get a new ordered sum until a period is reached. T(n,k)=a(m) with m=(n+4)*(n+3)/2+k+1 gives for m the number of periods with length (n+4).
The numbers m=n*(n+3)/2 with 1<=n have one period with length (n+1).
The numbers m=n*(n+3)/2+2 with 1<=n have one period with length (n+2).
The triangular numbers n*(n+1)/2 with 1<=n have one period [(n+(n-1)+...+2+1)] with length 1. (End)

Examples

			As triangle, starts:
  1;
  2,2;
  2,3,2;
  3,5,5,3;
  3,7,8,7,3;
  4,9,14,14,9,4;
  4,12,20,25,20,12,4; ...
From _Paul Weisenhorn_, Dec 21 2010: (Start)
T(2,2)=3 classes with 3 ordered sums of 6; [(1+1+4),(1+4+1),(4+1+1)]; [(1+2+3),(2+3+1),(3+1+2)]; [(1+3+2),(3+2+1),(2+1+3)].
T(2,2)=a(m)=3 periods with length 6 for m=6*5/2+3=18 [(5+5+4+3+1),(6+5+4+2+1),(6+5+3+2+1+1),(6+4+3+2+2+1),(5+4+3+3+2+1),(5+4+4+3+2)]; [(5+5+3+3+2),(6+4+4+3+1),(5+5+4+2+1+1),(6+5+3+2+2),(6+4+3+3+1+1),(5+4+4+2+2+1)]; [(5+5+3+2+2+1),(6+4+3+3+2),(5+4+4+3+1+1),(5+5+4+2+2),(6+5+3+3+1),(6+4+4+2+1+1)]. (End)
		

Crossrefs

Row sums give A093210. Essentially the same as A051168. See A185158 for another version.

Programs

  • Mathematica
    T[n_, k_] := DivisorSum[GCD[k + 1, n + 4], Binomial[(n + 4)/#, (k + 1)/#] * MoebiusMu[#] & ]/(n + 4); Table[T[n, k], {n, 0, 12}, {k, 1, n + 1}] // Flatten (* Jean-François Alcover, Dec 02 2015 *)
  • PARI
    T(n,k)=local(A,ps,c); n+=3; k++; if(k<1||k>=n-1, 0, A=x*O(x^n) + y*O(y^n); ps=1-x-y+A; for(m=1,n,for(i=0,m,c=polcoeff(polcoeff(ps,i,x),m-i, y); if(m==n&i==k,break(2),ps*=(1-y^(m-i)*x^i+A)^c)));-c) /* Michael Somos, Jul 17 2004 */

Extensions

Edited with better definition by Omar E. Pol, Jan 05 2009

A111492 Triangle read by rows: a(n,k) = (k-1)! * C(n,k).

Original entry on oeis.org

1, 2, 1, 3, 3, 2, 4, 6, 8, 6, 5, 10, 20, 30, 24, 6, 15, 40, 90, 144, 120, 7, 21, 70, 210, 504, 840, 720, 8, 28, 112, 420, 1344, 3360, 5760, 5040, 9, 36, 168, 756, 3024, 10080, 25920, 45360, 40320, 10, 45, 240, 1260, 6048, 25200, 86400, 226800, 403200, 362880
Offset: 1

Views

Author

Ross La Haye, Nov 15 2005

Keywords

Comments

For k > 1, a(n,k) = the number of permutations of the symmetric group S_n that are pure k-cycles.
Reverse signed array is A238363. For a relation to (Cauchy-Euler) derivatives of the Vandermonde determinant, see Chervov link. - Tom Copeland, Apr 10 2014
Dividing the k-th column of T by (k-1)! for each column generates A135278 (the f-vectors, or face-vectors for the n-simplices). Then ignoring the first column gives A104712, so T acting on the column vector (-0,d,-d^2/2!,d^3/3!,...) gives the Euler classes for hypersurfaces of degree d in CP^n. Cf. A104712 and Dugger link therein. - Tom Copeland, Apr 11 2014
With initial i,j,n=1, given the n X n Vandermonde matrix V_n(x_1,...,x_n) with elements a(i=row,j=column)=(x_j)^(i-1), its determinant |V_n|, and the column vector of n ones C=(1,1,...,1), the n-th row of the lower triangular matrix T is given by the column vector determined by (1/|V_n|) * V_n(:x_1*d/dx_1:,...,:x_n*d/dx_n:)|V_n| * C, where :x_j*d/dx_j:^n = (x_j)^n*(d/dx_j)^n. - Tom Copeland, May 20 2014
For some other combinatorial interpretations of the first three columns of T, see A208535 and the link to necklace polynomials therein. Because of the simple relation of the array to the Pascal triangle, it can easily be related to many other arrays, e.g., T(p,k)/(p*(k-1)!) with p prime gives the prime rows of A185158 and A051168 when the non-integers are rounded to 0. - Tom Copeland, Oct 23 2014

Examples

			a(3,3) = 2 because (3-1)!C(3,3) = 2.
1;
2 1;
3 3 2;
4 6 8 6;
5 10 20 30 24;
6 15 40 90 144 120;
7 21 70 210 504 840 720;
8 28 112 420 1344 3360 5760 5040;
9 36 168 756 3024 10080 25920 45360 40320;
		

Programs

  • Magma
    /* As triangle: */ [[Factorial(k-1)*Binomial(n,k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 21 2014
  • Mathematica
    Flatten[Table[(k - 1)!Binomial[n, k], {n, 10}, {k, n}]]

Formula

a(n, k) = (k-1)!C(n, k) = P(n, k)/k.
E.g.f. (by columns) = exp(x)((x^k)/k).
a(n, 1) = A000027(n);
a(n, 2) = A000217(n-1);
a(n, 3) = A007290(n);
a(n, 4) = A033487(n-3).
a(n, n) = A000142(n-1);
a(n, n-1) = A001048(n-1) for n > 1.
Sum[a(n, k), {k, 1, n}] = A002104(n);
Sum[a(n, k), {k, 2, n}] = A006231(n).
a(n,k) = sum(j=k..n-1, j!/(j-k)!) (cf. Chervov link). - Tom Copeland, Apr 10 2014
From Tom Copeland, Apr 28 2014: (Start)
E.g.f. by row: [(1+t)^n-1]/t.
E.g.f. of row e.g.f.s: {exp[(1+t)*x]-exp(x)}/t.
O.g.f. of row e.g.f.s: {1/[1-(1+t)*x] - 1/(1-x)}/t.
E.g.f. of row o.g.f.s: -exp(x) * log(1-t*x). (End)
Showing 1-3 of 3 results.