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.

A062138 Coefficient triangle of generalized Laguerre polynomials n!*L(n,5,x)(rising powers of x).

Original entry on oeis.org

1, 6, -1, 42, -14, 1, 336, -168, 24, -1, 3024, -2016, 432, -36, 1, 30240, -25200, 7200, -900, 50, -1, 332640, -332640, 118800, -19800, 1650, -66, 1, 3991680, -4656960, 1995840, -415800, 46200, -2772, 84, -1, 51891840, -69189120
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

The row polynomials s(n,x) := n!*L(n,5,x)= sum(a(n,m)*x^m,m=0..n) have e.g.f. exp(-z*x/(1-z))/(1-z)^6. They are Sheffer polynomials satisfying the binomial convolution identity s(n,x+y) = sum(binomial(n,k)*s(k,x)*p(n-k,y),k=0..n), with polynomials sum(|A008297(n,m)|*(-x)^m, m=1..n), n >= 1 and p(0,x)=1 (for Sheffer polynomials see A048854 for S. Roman reference).
These polynomials appear in the radial part of the l=2 (d-wave) eigen functions for the discrete energy levels of the H-atom. See Messiah reference.
For m=0..5 the (unsigned) column sequences (without leading zeros) are: A001725(n+5), A062148-A062152. Row sums (signed) give A062191; row sums (unsigned) give A062192.
The unsigned version of this triangle is the triangle of unsigned 3-Lah numbers A143498. - Peter Bala, Aug 25 2008

Examples

			Triangle begins:
  {1};
  {6, -1};
  {42, -14, 1};
  {336, -168, 24, -1};
  ...
2!*L(2, 5, x) = 42-14*x+x^2.
		

References

  • A. Messiah, Quantum mechanics, vol. 1, p. 419, eq.(XI.18a), North Holland, 1969.

Crossrefs

For m=0..5 the (unsigned) column sequences (without leading zeros) are: A001725(n+5), A062148, A062149, A062150, A062151, A062152.
Row sums (signed) give A062191, row sums (unsigned) give A062192.
Cf. A143498.

Programs

  • Mathematica
    Flatten[Table[((-1)^m)*n!*Binomial[n+5,n-m]/m!,{n,0,8},{m,0,n}]] (* Indranil Ghosh, Feb 24 2017 *)
  • PARI
    tabl(nn) = {for (n=0, nn, for (m=0, n, print1(((-1)^m)*n!*binomial(n+5, n-m)/m!, ", "); ); print(); ); } \\ Indranil Ghosh, Feb 24 2017
    
  • PARI
    row(n) = Vecrev(n!*pollaguerre(n, 5)); \\ Michel Marcus, Feb 06 2021
    
  • Python
    import math
    f=math.factorial
    def C(n, r):return f(n)//f(r)//f(n-r)
    i=-1
    for n in range(26):
        for m in range(n+1):
            i += 1
            print(str(i)+" "+str(((-1)**m)*f(n)*C(n+5, n-m)//f(m))) # Indranil Ghosh, Feb 24 2017

Formula

T(n, m) = ((-1)^m)*n!*binomial(n+5, n-m)/m!.
E.g.f. for m-th column: ((-x/(1-x))^m)/(m!*(1-x)^6), m >= 0.

A086885 Lower triangular matrix, read by rows: T(i,j) = number of ways i seats can be occupied by any number k (0<=k<=j<=i) of persons.

Original entry on oeis.org

2, 3, 7, 4, 13, 34, 5, 21, 73, 209, 6, 31, 136, 501, 1546, 7, 43, 229, 1045, 4051, 13327, 8, 57, 358, 1961, 9276, 37633, 130922, 9, 73, 529, 3393, 19081, 93289, 394353, 1441729, 10, 91, 748, 5509, 36046, 207775, 1047376, 4596553, 17572114, 11, 111, 1021, 8501
Offset: 1

Views

Author

Hugo Pfoertner, Aug 22 2003

Keywords

Comments

Compare with A088699. - Peter Bala, Sep 17 2008
T(m, n) gives the number of matchings in the complete bipartite graph K_{m,n}. - Eric W. Weisstein, Apr 25 2017

Examples

			One person:
T(1,1)=a(1)=2: 0,1 (seat empty or occupied);
T(2,1)=a(2)=3: 00,10,01 (both seats empty, left seat occupied, right seat occupied).
Two persons:
T(2,2)=a(3)=7: 00,10,01,20,02,12,21;
T(3,2)=a(5)=13: 000,100,010,001,200,020,002,120,102,012,210,201,021.
Triangle starts:
  2;
  3  7;
  4 13  34;
  5 21  73 209;
  6 31 136 501 1546;
  ...
		

Crossrefs

Diagonal: A002720, first subdiagonal: A000262, 2nd subdiagonal: A052852, 3rd subdiagonal: A062147, 4th subdiagonal: A062266, 5th subdiagonal: A062192, 2nd row/column: A002061. With column 0: A176120.

Programs

  • Magma
    [Factorial(k)*Evaluate(LaguerrePolynomial(k, n-k), -1): k in [1..n], n in [1..10]]; // G. C. Greubel, Feb 23 2021
    
  • Maple
    A086885 := proc(n,k)
        add( binomial(n,j)*binomial(k,j)*j!,j=0..min(n,k)) ;
    end proc: # R. J. Mathar, Dec 19 2014
  • Mathematica
    Table[Table[Sum[k! Binomial[n, k] Binomial[j, k], {k, 0, j}], {j, 1, n}], {n, 1, 10}] // Grid (* Geoffrey Critzer, Jul 09 2015 *)
    Table[m! LaguerreL[m, n - m, -1], {n, 10}, {m, n}] // Flatten (* Eric W. Weisstein, Apr 25 2017 *)
  • PARI
    T(i, j) = j!*pollaguerre(j, i-j, -1); \\ Michel Marcus, Feb 23 2021
  • Sage
    flatten([[factorial(k)*gen_laguerre(k, n-k, -1) for k in [1..n]] for n in (1..10)]) # G. C. Greubel, Feb 23 2021
    

Formula

a(n) = T(i, j) with n=(i*(i-1))/2+j; T(i, 1)=i+1, T(i, j)=T(i, j-1)+i*T(i-1, j-1) for j>1.
The role of seats and persons may be interchanged, so T(i, j)=T(j, i).
T(i, j) = j!*LaguerreL(j, i-j, -1). - Vladeta Jovovic, Aug 25 2003
T(i, j) = Sum_{k=0..j} k!*binomial(i, k)*binomial(j, k). - Vladeta Jovovic, Aug 25 2003

A143498 Triangle of unsigned 3-Lah numbers.

Original entry on oeis.org

1, 6, 1, 42, 14, 1, 336, 168, 24, 1, 3024, 2016, 432, 36, 1, 30240, 25200, 7200, 900, 50, 1, 332640, 332640, 118800, 19800, 1650, 66, 1, 3991680, 4656960, 1995840, 415800, 46200, 2772, 84, 1, 51891840, 69189120, 34594560, 8648640, 1201200, 96096, 4368
Offset: 3

Views

Author

Peter Bala, Aug 25 2008

Keywords

Comments

For a signed version of this triangle see A062138. This is the case r = 3 of the unsigned r-Lah numbers L(r;n,k). The unsigned 3-Lah numbers count the partitions of the set {1,2,...,n} into k ordered lists with the restriction that the elements 1, 2 and 3 belong to different lists. For other cases see A105278 (r = 1), A143497 (r = 2 and comments on the general case) and A143499 (r = 4).
The unsigned 3-Lah numbers are related to the 3-Stirling numbers: the lower triangular array of unsigned 3-Lah numbers may be expressed as the matrix product St1(3) * St2(3), where St1(3) = A143492 and St2(3) = A143495 are the arrays of 3-Stirling numbers of the first and second kind respectively. An alternative factorization for the array is as St1 * P^4 * St2, where P denotes Pascal's triangle, A007318, St1 is the triangle of unsigned Stirling numbers of the first kind, abs(A008275) and St2 denotes the triangle of Stirling numbers of the second kind, A008277.

Examples

			Triangle begins
n\k|......3......4......5......6......7......8
==============================================
3..|......1
4..|......6......1
5..|.....42.....14......1
6..|....336....168.....24......1
7..|...3024...2016....432.....36......1
8..|..30240..25200...7200....900.....50......1
...
T(4,3) = 6. The partitions of {1,2,3,4} into 3 ordered lists, such that the elements 1, 2 and 3 lie in different lists, are: {1}{2}{3,4} and {1}{2}{4,3}, {1}{3}{2,4} and {1}{3}{4,2}, {2}{3}{1,4} and {2}{3}{4,1}.
		

Crossrefs

Cf. A001725 (column 3), A007318, A008275, A008277, A062138, A062148 - A062152 (column 4 to column 8), A062191 (alt. row sums), A062192 (row sums), A105278 (unsigned Lah numbers), A143492, A143495, A143497, A143499.

Programs

  • Magma
    /* As triangle */ [[Factorial(n-3)/Factorial(k-3)*Binomial(n+2, k+2): k in [3..n]]: n in [3.. 15]]; // Vincenzo Librandi, Nov 27 2018
  • Maple
    with combinat: T := (n, k) -> (n-3)!/(k-3)!*binomial(n+2,k+2): for n from 3 to 12 do seq(T(n, k), k = 3..n) end do;
  • Mathematica
    T[n_, k_] := (n-3)!/(k-3)!*Binomial[n+2, k+2]; Table[T[n, k], {n, 3, 10}, {k, 3, n}] // Flatten (* Amiram Eldar, Nov 26 2018 *)

Formula

T(n,k) = (n-3)!/(k-3)!*binomial(n+2,k+2) for n,k >= 3.
Recurrence: T(n,k) = (n+k-1)*T(n-1,k) + T(n-1,k-1) for n,k >= 3, with the boundary conditions: T(n,k) = 0 if n < 3 or k < 3; T(3,3) = 1.
E.g.f. for column k: Sum_{n >= k} T(n,k)*t^n/(n-3)! = 1/(k-3)!*t^k/(1-t)^(k+3) for k >= 3.
E.g.f: Sum_{n = 3..inf} Sum_{k = 3..n} T(n,k)*x^k*t^n/(n-3)! = (x*t)^3/(1-t)^6*exp(x*t/(1-t)) = (x*t)^3*(1 + (6+x)*t +(42+14*x+x^2)*t^2/2! + ... ).
Generalized Lah identity: (x+5)*(x+6)*...*(x+n+1) = Sum_{k = 3..n} T(n,k)*(x-1)*(x-2)*...*(x-k+3).
The polynomials 1/n!*Sum_{k = 3..n+3} T(n+3,k)*(-x)^(k-3) for n >= 0 are the generalized Laguerre polynomials Laguerre(n,5,x). See A062138.
Array = A143492 * A143495 = abs(A008275) * ( A007318 )^4 * A008277 (apply Theorem 10 of [Neuwirth]). Array equals exp(D), where D is the array with the quadratic sequence (6,14,24,36, ... ) on the main subdiagonal and zeros everywhere else.

A216294 Triangular array read by rows: T(n,k) is the number of partial permutations of {1,2,...,n} that have exactly k cycles, 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 13, 14, 6, 1, 73, 84, 41, 10, 1, 501, 609, 325, 95, 15, 1, 4051, 5155, 2944, 965, 190, 21, 1, 37633, 49790, 30023, 10689, 2415, 343, 28, 1, 394353, 539616, 340402, 129220, 32179, 5348, 574, 36, 1, 4596553, 6478521, 4246842, 1698374, 455511, 84567, 10794, 906, 45, 1
Offset: 0

Views

Author

Geoffrey Critzer, Sep 04 2012

Keywords

Comments

A partial permutation on a set X is a bijection between two subsets of X.
Row sums are A002720.
First column (corresponding to k=0) is A000262.

Examples

			1;
1,     1;
3,     3,   1;
13,   14,   6,  1;
73,   84,  41, 10,  1;
501, 609, 325, 95, 15,  1;
		

Crossrefs

Programs

  • Maple
    gf := exp(x / (1 - x)) / (1 - x)^y:
    serx := series(gf, x, 10): poly := n -> simplify(coeff(serx, x, n)):
    seq(print(seq(n!*coeff(poly(n), y, k), k = 0..n)), n = 0..9); # Peter Luschny, Feb 23 2023
  • Mathematica
    nn=10;t=Sum[n^(n-1)x^n/n!,{n,1,nn}];f[list_]:=Select[list,#>0&];Map[f,Range[0,nn]!CoefficientList[Series[Exp[ x/(1-x)]/(1-x)^y,{x,0,nn}],{x,y}]]//Flatten

Formula

E.g.f.: exp(x/(1-x))/(1-x)^y.
From Peter Bala, Aug 23 2013: (Start)
Exponential Riordan array [exp(x/(1-x)), log(1/(1-x))].
The row polynomials R(n,y), n > = 0, satisfy the 2nd order recurrence equation R(n,y) = (2*n + y - 1)*R(n-1,y) - (n - 1)*(n + y - 2)*R(n-2,y) with R(0,y) = 1 and R(1,y) = 1 + y.
Modulo variations in offset we have: R(n,0) = A000262, R(n,1) = A002720, R(n,2) = A000262, R(n,3) = A052852, R(n,4) = A062147, R(n,5) = A062266 and R(n,6) = A062192. In general, for fixed k, the sequence {R(n,k)}n>=1 gives the entries on a diagonal of the square array A088699. (End)

A293985 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f. exp(x/(1-x))/(1-x)^k.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 7, 13, 1, 4, 13, 34, 73, 1, 5, 21, 73, 209, 501, 1, 6, 31, 136, 501, 1546, 4051, 1, 7, 43, 229, 1045, 4051, 13327, 37633, 1, 8, 57, 358, 1961, 9276, 37633, 130922, 394353, 1, 9, 73, 529, 3393, 19081, 93289, 394353, 1441729, 4596553
Offset: 0

Views

Author

Seiichi Manyama, Oct 21 2017

Keywords

Examples

			Square array begins:
    1,    1,    1,    1,     1, ... A000012;
    1,    2,    3,    4,     5, ... A000027;
    3,    7,   13,   21,    31, ... A002061;
   13,   34,   73,  136,   229, ... A135859;
   73,  209,  501, 1045,  1961, ...
  501, 1546, 4051, 9276, 19081, ...
Antidiagonal rows begin as:
  1;
  1, 1;
  1, 2,  3;
  1, 3,  7, 13;
  1, 4, 13, 34,  73;
  1, 5, 21, 73, 209, 501; - _G. C. Greubel_, Mar 09 2021
		

Crossrefs

Columns k=0..6 give: A000262, A002720, A000262(n+1), A052852(n+1), A062147, A062266, A062192.
Main diagonal gives A152059.
Similar table: A086885, A088699, A176120.

Programs

  • Magma
    function t(n,k)
      if n eq 0 then return 1;
      else return Factorial(n-1)*(&+[(j+k)*t(n-j,k)/Factorial(n-j): j in [1..n]]);
      end if; return t;
    end function;
    [t(k,n-k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 09 2021
  • Mathematica
    t[n_, k_]:= t[n, k]= If[n==0, 1, (n-1)!*Sum[(j+k)*t[n-j,k]/(n-j)!, {j,n}]];
    T[n_,k_]:= t[k,n-k]; Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 09 2021 *)
  • Sage
    @CachedFunction
    def t(n,k): return 1 if n==0 else factorial(n-1)*sum( (j+k)*t(n-j,k)/factorial(n-j) for j in (1..n) )
    def T(n,k): return t(k,n-k)
    flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 09 2021
    

Formula

A(0,k) = 1 and A(n,k) = (n-1)! * Sum_{j=1..n} (j+k)*A(n-j,k)/(n-j)! for n > 0.
A(0,k) = 1, A(1,k) = k+1 and A(n,k) = (2*n-1+k)*A(n-1,k) - (n-1)*(n-2+k)*A(n-2,k) for n > 1.
From Seiichi Manyama, Jan 25 2025: (Start)
A(n,k) = n! * Sum_{j=0..n} binomial(n+k-1,j)/(n-j)!.
A(n,k) = n! * LaguerreL(n, k-1, -1). (End)
Showing 1-5 of 5 results.