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 1-5 of 5 results.

A008299 Triangle T(n,k) of associated Stirling numbers of second kind, n >= 2, 1 <= k <= floor(n/2).

Original entry on oeis.org

1, 1, 1, 3, 1, 10, 1, 25, 15, 1, 56, 105, 1, 119, 490, 105, 1, 246, 1918, 1260, 1, 501, 6825, 9450, 945, 1, 1012, 22935, 56980, 17325, 1, 2035, 74316, 302995, 190575, 10395, 1, 4082, 235092, 1487200, 1636635, 270270, 1, 8177, 731731, 6914908, 12122110
Offset: 2

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Author

Keywords

Comments

T(n,k) is the number of set partitions of [n] into k blocks of size at least 2. Compare with A008277 (blocks of size at least 1) and A059022 (blocks of size at least 3). See also A200091. Reading the table by diagonals gives A134991. The row generating polynomials are the Mahler polynomials s_n(-x). See [Roman, 4.9]. - Peter Bala, Dec 04 2011
Row n gives coefficients of moments of Poisson distribution about the mean expressed as polynomials in lambda [Haight]. The coefficients of the moments about the origin are the Stirling numbers of the second kind, A008277. - N. J. A. Sloane, Jan 24 2020
Rows are of lengths 1,1,2,2,3,3,..., a pattern typical of matrices whose diagonals are rows of another lower triangular matrix--in this instance those of A134991. - Tom Copeland, May 01 2017
For a relation to decomposition of spin correlators see Table 2 of the Delfino and Vito paper. - Tom Copeland, Nov 11 2012

Examples

			There are 3 ways of partitioning a set N of cardinality 4 into 2 blocks each of cardinality at least 2, so T(4,2)=3.
Table begins:
  1;
  1;
  1,    3;
  1,   10;
  1,   25,     15;
  1,   56,    105;
  1,  119,    490,     105;
  1,  246,   1918,    1260;
  1,  501,   6825,    9450,      945;
  1, 1012,  22935,   56980,    17325;
  1, 2035,  74316,  302995,   190575,   10395;
  1, 4082, 235092, 1487200,  1636635,  270270;
  1, 8177, 731731, 6914908, 12122110, 4099095, 135135;
  ...
Reading the table by diagonals produces the triangle A134991.
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.
  • Frank Avery Haight, "Handbook of the Poisson distribution," John Wiley, 1967. See pages 6,7, but beware of errors. [Haight on page 7 gives five different ways to generate these numbers (see link)].
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.
  • S. Roman, The Umbral Calculus, Dover Publications, New York (2005), pp. 129-130.

Crossrefs

Rows: A000247 (k=2), A000478 (k=3), A058844 (k=4).
Row sums: A000296, diagonal: A259877.

Programs

  • Maple
    A008299 := proc(n,k) local i,j,t1; if k<1 or k>floor(n/2) then t1 := 0; else
    t1 := add( (-1)^i*binomial(n, i)*add( (-1)^j*(k - i - j)^(n - i)/(j!*(k - i - j)!), j = 0..k - i), i = 0..k); fi; t1; end; # N. J. A. Sloane, Dec 06 2016
    G:= exp(lambda*(exp(x)-1-x)):
    S:= series(G,x,21):
    seq(seq(coeff(coeff(S,x,n)*n!,lambda,k),k=1..floor(n/2)),n=2..20); # Robert Israel, Jan 15 2020
    T := proc(n, k) option remember; if n < 0 then return 0 fi; if k = 0 then return k^n fi; k*T(n-1, k) + (n-1)*T(n-2, k-1) end:
    seq(seq(T(n,k), k=1..n/2), n=2..9); # Peter Luschny, Feb 11 2021
  • Mathematica
    t[n_, k_] := Sum[ (-1)^i*Binomial[n, i]*Sum[ (-1)^j*(k - i - j)^(n - i)/(j!*(k - i - j)!), {j, 0, k - i}], {i, 0, k}]; Flatten[ Table[ t[n, k], {n, 2, 14}, {k, 1, Floor[n/2]}]] (* Jean-François Alcover, Oct 13 2011, after David Wasserman *)
    Table[Sum[Binomial[n, k - j] StirlingS2[n - k + j, j] (-1)^(j + k), {j, 0, k}], {n, 15}, {k, n/2}] // Flatten (* Eric W. Weisstein, Nov 13 2018 *)
  • PARI
    {T(n, k) = if( n < 1 || 2*k > n, n==0 && k==0, sum(i=0, k, (-1)^i * binomial( n, i) * sum(j=0, k-i, (-1)^j * (k-i-j)^(n-i) / (j! * (k-i-j)!))))}; /* Michael Somos, Oct 19 2014 */
    
  • PARI
    { T(n,k) = sum(i=0,min(n,k), (-1)^i * binomial(n,i) * stirling(n-i,k-i,2) ); } /* Max Alekseyev, Feb 27 2017 */

Formula

T(n,k) = abs(A137375(n,k)).
E.g.f. with additional constant 1: exp(t*(exp(x)-1-x)) = 1 + t*x^2/2! + t*x^3/3! + (t+3*t^2)*x^4/4! + ....
Recurrence relation: T(n+1,k) = k*T(n,k) + n*T(n-1,k-1).
T(n,k) = A134991(n-k,k); A134991(n,k) = T(n+k,k).
More generally, if S_r(n,k) gives the number of set partitions of [n] into k blocks of size at least r then we have the recurrence S_r(n+1,k) = k*S_r(n,k) + binomial(n,r-1)*S_r(n-r+1,k-1) (for this sequence, r=2), with associated e.g.f.: Sum_{n>=0, k>=0} S_r(n,k)*u^k*(t^n/n!) = exp(u*(e^t - Sum_{i=0..r-1} t^i/i!)).
T(n,k) = Sum_{i=0..k} (-1)^i*binomial(n, i)*Sum_{j=0..k-i} (-1)^j*(k-i-j)^(n-i)/(j!*(k-i-j)!). - David Wasserman, Jun 13 2007
G.f.: (R(0)-1)/(x^2*y), where R(k) = 1 - (k+1)*y*x^2/( (k+1)*y*x^2 - (1-k*x)*(1-x-k*x)/R(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
T(n,k) = Sum_{i=0..min(n,k)} (-1)^i * binomial(n,i) * Stirling2(n-i,k-i) = Sum_{i=0..min(n,k)} (-1)^i * A007318(n,i) * A008277(n-i,k-i). - Max Alekseyev, Feb 27 2017
T(n, k) = Sum_{j=0..n-k} binomial(j, n-2*k)*E2(n-k, n-k-j) where E2(n, k) are the second-order Eulerian numbers A340556. - Peter Luschny, Feb 11 2021

Extensions

Formula and cross-references from Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000
Edited by Peter Bala, Dec 04 2011
Edited by N. J. A. Sloane, Jan 24 2020

A057837 Number of partitions of a set of n elements where the partitions are of size > 3.

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 1, 1, 36, 127, 337, 793, 7525, 48764, 238954, 997790, 6401435, 49107697, 345482807, 2150694855, 14656830110, 116678887407, 978172378669, 7886661080873, 63905475745765, 553437891603452, 5122279358273976, 48331088541366296, 458771027309344261
Offset: 0

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Author

Steven C. Fairgrieve (fsteven(AT)math.wvu.edu), Nov 06 2000

Keywords

Crossrefs

Column k=3 of A293024.
Row sums of A059023.
Cf. A293039.

Programs

  • Maple
    G:={P=Set(Set(Atom,card>=4))}:combstruct[gfsolve](G,unlabeled,x):seq(combstruct[count]([P,G,labeled],size=i),i=0..26); # Zerinvary Lajos, Dec 16 2007
  • Mathematica
    With[{nn=30},CoefficientList[Series[Exp[Exp[x]-1-x-x^2/2-x^3/6],{x,0,nn}], x]Range[0,nn]!] (* Harvey P. Dale, Jun 28 2012 *)

Formula

E.g.f.: exp(exp(x)-1-x-x^2/2-x^3/6).
a(0) = 1; a(n) = Sum_{k=4..n} binomial(n-1,k-1) * a(n-k). - Ilya Gutkovskiy, Feb 09 2020

Extensions

Corrected and extended by Christian G. Bower and James Sellers, Nov 09 2000

A059022 Triangle of Stirling numbers of order 3.

Original entry on oeis.org

1, 1, 1, 1, 10, 1, 35, 1, 91, 1, 210, 280, 1, 456, 2100, 1, 957, 10395, 1, 1969, 42735, 15400, 1, 4004, 158301, 200200, 1, 8086, 549549, 1611610, 1, 16263, 1827826, 10335325, 1401400, 1, 32631, 5903898, 57962905, 28028000, 1, 65382, 18682014, 297797500
Offset: 3

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Author

Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000

Keywords

Comments

The number of partitions of the set N, |N|=n, into k blocks, all of cardinality greater than or equal to 3. This is the 3-associated Stirling number of the second kind (Comtet) or the Stirling number of order 3 (Fekete).
This is entered as a triangular array. The entries S_3(n,k) are zero for 3k>n, so these values are omitted. The initial entry in the sequence is S_3(3,1).
Rows are of lengths 1,1,1,2,2,2,3,3,3,...

Examples

			There are 10 ways of partitioning a set N of cardinality 6 into 2 blocks each of cardinality at least 3, so S_3(6,2) = 10.
From _Wesley Ivan Hurt_, Feb 24 2022: (Start)
Triangle starts:
  1;
  1;
  1;
  1,   10;
  1,   35;
  1,   91;
  1,  210,   280;
  1,  456,  2100;
  1,  957, 10395;
  1, 1969, 42735, 15400;
  ...
(End)
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.

Crossrefs

Row sums give A006505.
Cf. A008299, A059023, A059024, A059025, A100861, A272352 (column 2), A272982 (column 3), A261724 (column 4), A352611 (column 5).

Programs

  • Maple
    b:= proc(n) option remember; `if`(n=0, 1, add(
          expand(x*b(n-j))*binomial(n-1, j-1), j=3..n))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n)):
    seq(T(n), n=3..20);  # Alois P. Heinz, Feb 21 2022
    # alternative
    A059022 := proc(n, k)
        option remember;
        if n<3 then
            0;
        elif n < 6 and k=1 then
            1 ;
        else
            k*procname(n-1, k)+binomial(n-1, 2)*procname(n-3, k-1) ;
        end if;
    end proc:  # R. J. Mathar, Apr 15 2022
  • Mathematica
    S3[3, 1] = S3[4, 1] = S3[5, 1] = 1; S3[n_, k_] /; 1 <= k <= Floor[n/3] := S3[n, k] = k*S3[n-1, k] + Binomial[n-1, 2]*S3[n-3, k-1]; S3[, ] = 0; Flatten[ Table[ S3[n, k], {n, 3, 20}, {k, 1, Floor[n/3]}]] (* Jean-François Alcover, Feb 21 2012 *)

Formula

S_r(n+1,k) = k*S_r(n,k) + binomial(n,r-1)*S_r(n-r+1,k-1); for this sequence, r=3.
G.f.: Sum_{n>=0, k>=0} S_r(n,k)*u^k*t^n/n! = exp(u(e^t - Sum_{i=0..r-1} t^i/i!)).
T(n,k) = Sum_{j=0..min(n/2,k)} (-1)^j*B(n,j)*S_2(n-2j,k-j), where B are the Bessel numbers A100861 and S_2 are the 2-associated Stirling numbers of the second kind A008299. - Fabián Pereyra, Feb 20 2022

A059024 Triangle of Stirling numbers of order 5.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 126, 1, 462, 1, 1254, 1, 3003, 1, 6721, 1, 14443, 126126, 1, 30251, 1009008, 1, 62322, 5309304, 1, 127024, 23075052, 1, 257108, 89791416, 1, 518092, 325355316, 488864376, 1, 1041029, 1122632043, 6844101264, 1, 2088043
Offset: 5

Views

Author

Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000

Keywords

Comments

The number of partitions of the set N, |N|=n, into k blocks, all of cardinality greater than or equal to 5. This is the 5-associated Stirling number of the second kind.
This is entered as a triangular array. The entries S_5(n,k) are zero for 5k>n, so these values are omitted. Initial entry in sequence is S_5(5,1).
Rows are of lengths 1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,...

Examples

			There are 126 ways of partitioning a set N of cardinality 10 into 2 blocks each of cardinality at least 5, so S_5(10,2) = 126.
Triangle begins:
1;
1;
1;
1;
1;
1,    126;
1,    462;
1,   1254;
1,   3003;
1,   6721;
1,  14443,    126126;
1,  30251,   1009008;
1,  62322,   5309304;
1, 127024,  23075052;
1, 257108,  89791416;
1, 518092, 325355316, 488864376;
...
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.

Crossrefs

Row sums give A057814.

Programs

  • Maple
    T:= proc(n,k) option remember; `if`(k<1 or k>n/5, 0,
          `if`(k=1, 1, k*T(n-1, k)+binomial(n-1, 4)*T(n-5, k-1)))
        end:
    seq(seq(T(n, k), k=1..n/5), n=5..25);  # Alois P. Heinz, Aug 18 2017
  • Mathematica
    S5[n_ /; 5 <= n <= 9, 1] = 1; S5[n_, k_] /; 1 <= k <= Floor[n/5] := S5[n, k] = k*S5[n-1, k] + Binomial[n-1, 4]*S5[n-5, k-1]; S5[, ] = 0; Flatten[ Table[ S5[n, k], {n, 5, 25}, {k, 1, Floor[n/5]}]] (* Jean-François Alcover, Feb 21 2012 *)

Formula

S_r(n+1, k) = k*S_r(n, k) + binomial(n, r-1)*S_r(n-r+1, k-1); for this sequence, r=5.
G.f.: Sum_{n>=0, k>=0} S_r(n,k)*u^k*t^n/n! = exp(u(e^t-sum(t^i/i!, i=0..r-1))).
T(n,k) = Sum_{j=0..min(n/4,k)} (-1)^j*n!/(24^j*j!*(n-4j)!)*S_4(n-4j,k-j), where S_4 are the 4-associated Stirling numbers of the second kind A059023. - Fabián Pereyra, Feb 21 2022

A059025 Triangle of Stirling numbers of order 6.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 462, 1, 1716, 1, 4719, 1, 11440, 1, 25883, 1, 56134, 1, 118456, 2858856, 1, 245480, 23279256, 1, 502588, 124710300, 1, 1020680, 551496660, 1, 2061709, 2181183147, 1, 4149752, 8021782197, 1, 8333153, 28051272535
Offset: 6

Views

Author

Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Dec 14 2000

Keywords

Comments

The number of partitions of the set N, |N|=n, into k blocks, all of cardinality greater than or equal to 6. This is the 6-associated Stirling number of the second kind.
This is entered as a triangular array. The entries S_6(n,k) are zero for 6k>n, so these values are omitted. Initial entry in sequence is S_6(6,1).
Rows are of lengths 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, ...

Examples

			There are 462 ways of partitioning a set N of cardinality 12 into 2 blocks each of cardinality at least 6, so S_6(12,2)=462.
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 222.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 76.

Crossrefs

Programs

  • Mathematica
    S6[n_ /; 6 <= n <= 11, 1] = 1; S6[n_, k_] /; 1 <= k <= Floor[n/6] := S6[n, k] = k*S6[n-1, k] + Binomial[n-1, 5]*S6[n-6, k-1]; S6[, ] = 0; Flatten[ Table[ S6[n, k], {n, 6, 24}, {k, 1, Floor[n/6]}]] (* Jean-François Alcover, Feb 21 2012 *)

Formula

S_r(n+1, k)=k S_r(n, k)+binomial(n, r-1)S_r(n-r+1, k-1) for this sequence, r=6.
G.f.: Sum_{n>=0, k>=0} S_r(n,k)*u^k*t^n/n! = exp(u(e^t - Sum_{i=0..r-1} t^i/i!)).
Showing 1-5 of 5 results.