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.

A090238 Triangle T(n, k) read by rows. T(n, k) is the number of lists of k unlabeled permutations whose total length is n.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 6, 4, 1, 0, 24, 16, 6, 1, 0, 120, 72, 30, 8, 1, 0, 720, 372, 152, 48, 10, 1, 0, 5040, 2208, 828, 272, 70, 12, 1, 0, 40320, 14976, 4968, 1576, 440, 96, 14, 1, 0, 362880, 115200, 33192, 9696, 2720, 664, 126, 16, 1, 0, 3628800, 996480, 247968, 64704, 17312, 4380, 952, 160, 18, 1
Offset: 0

Views

Author

Philippe Deléham, Jan 23 2004, Jun 14 2007

Keywords

Comments

T(n,k) is the number of lists of k unlabeled permutations whose total length is n. Unlabeled means each permutation is on an initial segment of the positive integers. Example: with dashes separating permutations, T(3,2) = 4 counts 1-12, 1-21, 12-1, 21-1. - David Callan, Nov 29 2007
For n > 0, -Sum_{i=0..n} (-1)^i*T(n,i) is the number of indecomposable permutations A003319. - Peter Luschny, Mar 13 2009
Also the convolution triangle of the factorial numbers for n >= 1. - Peter Luschny, Oct 09 2022

Examples

			Triangle begins:
  1;
  0,       1;
  0,       2,      1;
  0,       6,      4,      1;
  0,      24,     16,      6,     1;
  0,     120,     72,     30,     8,     1;
  0,     720,    372,    152,    48,    10,     1;
  0,    5040,   2208,    828,   272,    70,    12,    1;
  0,   40320,  14976,   4968,  1576,   440,    96,   14,   1;
  0,  366880, 115200,  33192,  9696,  2720,   664,  126,  16,   1;
  0, 3628800, 996480, 247968, 64704, 17312,  4380,  952, 160,  18,  1;
  ...

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 171, #34.

Crossrefs

Another version: A059369.
Row sums: A051296, A003319 (n>0).
Diagonals: A000007, A000142, A059371, A000012, A005843, A054000.
Cf. A084938.

Programs

  • Maple
    T := proc(n,k) option remember; if n=0 and k=0 then return 1 fi;
if n>0 and k=0 or k>0 and n=0 then return 0 fi;
T(n-1,k-1)+(n+k-1)*T(n-1,k)/k end:
for n from 0 to 10 do seq(T(n,k),k=0..n) od; # Peter Luschny, Mar 03 2016
# Uses function PMatrix from A357368.
PMatrix(10, factorial); # Peter Luschny, Oct 09 2022
  • Mathematica
    T[n_, k_] := T[n, k] = T[n-1, k-1] + ((n+k-1)/k)*T[n-1, k]; T[0, 0] = 1; T[, 0] = T[0, ] = 0;
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 20 2018 *)

Formula

T(n, k) is given by [0, 2, 1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.
T(n, k) = T(n-1, k-1) + ((n+k-1)/k)*T(n-1, k); T(0, 0)=1, T(n, 0)=0 if n > 0, T(0, k)=0 if k > 0.
G.f. for the k-th column: (Sum_{i>=1} i!*t^i)^k = Sum_{n>=k} T(n, k)*t^n.
Sum_{k=0..n} T(n, k)*binomial(m, k) = A084938(m+n, m). - Philippe Deléham, Jan 31 2004
T(n, k) = Sum_{j>=0} A090753(j)*T(n-1, k+j-1). - Philippe Deléham, Feb 18 2004
From Peter Bala, May 27 2017: (Start)
Conjectural o.g.f.: 1/(1 + t - t*F(x)) = 1 + t*x + (2*t + t^2)*x^2 + (6*t + 4*t^2 + t^3)*x^3 + ..., where F(x) = Sum_{n >= 0} n!*x^n.
If true then a continued fraction representation of the o.g.f. is 1 - t + t/(1 - x/(1 - t*x - x/(1 - 2*x/(1 - 2*x/(1 - 3*x/(1 - 3*x/(1 - 4*x/(1 - 4*x/(1 - ... ))))))))). (End)

Extensions

New name using a comment from David Callan by Peter Luschny, Sep 01 2022

A136633 G.f.: A(x) = Series_Reversion( x / Sum_{n>=0} (n+1)!*x^n ).

Original entry on oeis.org

1, 2, 10, 68, 544, 4832, 46312, 471536, 5055328, 56795840, 667286656, 8197599104, 105446118784, 1423627264256, 20234885027968, 303737480337152, 4827671316780544, 81385455480335360, 1455806861755411456
Offset: 0

Views

Author

Paul D. Hanna, Jan 14 2008

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 10*x^2 + 68*x^3 + 544*x^4 + 4832*x^5 + 46312*x^6 +...
Let F(x) = 1 + 2x + 6x^2 + 24x^3 + 120x^4 + 720x^5 +...+ (n+1)!*x^n +...
then A(x) = F(x*A(x)) and A(x/F(x)) = F(x);
also, a(n) = coefficient of x^n in F(x)^n divided by (n+1).
The g.f. A(x) also satisfies:
A(x) = 1 + 2*x*A(x)/(1+x*A(x)) + 2*2^2*x^2*A(x)^2/(1+2*x*A(x))^2 + 2*3^3*x^3*A(x)^3/(1+3*x*A(x))^3 + 2*4^4*x^4*A(x)^4/(1+4*x*A(x))^4 +...

Links

Crossrefs

Cf. A090753, A075834, A211207.

Programs

  • PARI
    a(n)=polcoeff(serreverse(x/sum(k=0,n,(k+1)!*x^k +x*O(x^n))),n)
  • PARI
    a(n)=local(A=1+x); for(i=1, n, A=1+2*sum(m=1, n, m^m*x^m*A^m/(1+m*x*A+x*O(x^n))^m)); polcoeff(A, n)
for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Feb 04 2013

Formula

G.f. A(x) satisfies:
(1) A(x) = F(x*A(x)) and A(x/F(x)) = F(x);
(2) a(n) = [x^n] F(x)^n / (n+1);
where F(x) = Sum_{n>=0} (n+1)!*x^n.
G.f. satisfies: A(x) = 1 + 2*Sum_{n>=1} n^n * x^n * A(x)^n / (1 + n*x*A(x))^n. - Paul D. Hanna, Feb 04 2013
a(n) ~ exp(2) * n! * n. - Vaclav Kotesovec, Nov 23 2024

A230253 Coefficients of power series A(x) such that coefficient of x^n in A(x)^(n+1) equals (n+1)*(n+2)!/2 for n>=0.

Original entry on oeis.org

1, 3, 3, 6, 27, 162, 1206, 10476, 103059, 1125738, 13473378, 174997908, 2448791838, 36706645908, 586646510796, 9957100024152, 178868488496643, 3390603439026618, 67639341903290730, 1416612563019545220, 31079692422132040170, 712855563504590236860, 17061654943814209044660
Offset: 0

Views

Author

Paul D. Hanna, Oct 13 2013

Keywords

Examples

			G.f. A(x) = 1 + 3*x + 3*x^2 + 6*x^3 + 27*x^4 + 162*x^5 + 1206*x^6 +...
The coefficients in A(x)^n begin:
n=1: [(1),3,   3,    6,    27,    162,   1206,   10476,   103059, ...];
n=2: [1, (6), 15,   30,    99,    522,   3582,   29484,   278883, ...];
n=3: [1,  9, (36),  99,   297,   1323,   8208,   63342,   572751, ...];
n=4: [1, 12,  66, (240),  783,   3132,  17298,  123552,  1060155, ...];
n=5: [1, 15, 105,  480, (1800),  7083,  35415,  231660,  1869885, ...];
n=6: [1, 18, 153,  846,  3672, (15120), 71415,  428490,  3226797, ...];
n=7: [1, 21, 210, 1365,  6804,  30240,(141120), 789939,  5529573, ...];
n=8: [1, 24, 276, 2064, 11682,  56736, 270720,(1451520), 9485343, ...];
n=9: [1, 27, 351, 2970, 18873, 100440, 500904, 2643840,(16329600), ...]; ...
where the coefficient of x^n in A(x)^(n+1) equals (n+1)*(n+2)!/2 for n>=0.
Notice that the diagonal above the main diagonal forms A111546.

Crossrefs

Cf. A075834, A090753, A111546.

Programs

  • PARI
    {a(n)=polcoeff(x/serreverse(sum(m=1, n+1, (m+1)!/2*x^m)+x^2*O(x^n)), n)}

Formula

[x^n] A(x)^n = A111546(n) for n>=0.
Showing 1-3 of 3 results.

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