cp's OEIS Frontend

From Tilman Piesk, Oct 28 2014: (Start)

Number of permutations of a multiset that contains m different elements n times. These multisets have the signatures A249543(m,n-1) for m>=1 and n>=2.

In an m-dimensional Pascal tensor (the generalization of a symmetric Pascal matrix) P(x1,...,xn) = (x1+...+xn)!/(x1!*...*xn!), so the main diagonal of an m-dimensional Pascal tensor is D(n) = (m*n)!/(n!^m). These diagonals are the rows of this array (with m>0), which begins like this:

m\n:0 1 2 3 4 5

0: 1 1 1 1 1 1 ... A000012;

1: 1 1 1 1 1 1 ... A000012;

2: 1 2 6 20 70 252 ... A000984;

3: 1 6 90 1680 34650 756756 ... A006480;

4: 1 24 2520 369600 63063000 11732745024 ... A008977;

5: 1 120 113400 168168000 305540235000 623360743125120 ... A008978;

6: 1 720 7484400 137225088000 3246670537110000 88832646059788350720 ... A008979;

with columns: A000142 (n=1), A000680 (n=2), A014606 (n=3), A014608 (n=4), A014609 (n=5).

A089759 is the transpose of this matrix. A034841 is its diagonal. A141906 is its lower triangle. A120666 is the upper triangle of this matrix with indices starting from 1. A248827 are the diagonal sums (or the row sums of the triangle).

(End)

T(n,k) is the number of lattice paths from {n}^k to {0}^k using steps that decrement one component by 1. - Alois P. Heinz, May 06 2013

In general, define b(k,e,p) = Sum_{i=0..k} (-1)^i*binomial(e*p+1,i)*binomial(k+e-i,e)^p. Then T(n,k) = b(k,4,n).

Using these coefficients we can obtain formulas for binomial(n,e)^p and for Sum_{i=1..n} binomial(e-1+i,e)^p.

In particular:

binomial(n, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+k, e*p).

Sum_{i=1..n} binomial(e-1+i, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+e+k, e*p+1).

T(n,k) is the number of permutations of 4 indistinguishable copies of 1..n with exactly k descents. A descent is a pair of adjacent elements with the second element less than the first. - Andrew Howroyd, May 08 2020

a(n) is also the constant term in product 1 <= i,j <= n, i different from j (1 - x_i/x_j)^5. - Sharon Sela (sharonsela(AT)hotmail.com), Feb 18 2002

The generalized Eulerian polynomials F_{m}(x) are defined F_{m; 0}(x) = 1 for all m >= 0 and for n > 0:

F_{0; n}(x) = Sum_{k=0..n} A097805(n, k)*(x-1)^(n-k) with coeffs. in A129186.

F_{1; n}(x) = Sum_{k=0..n} A131689(n, k)*(x-1)^(n-k) with coeffs. in A173018.

F_{2; n}(x) = Sum_{k=0..n} A241171(n, k)*(x-1)^(n-k) with coeffs. in A292604.

F_{3; n}(x) = Sum_{k=0..n} A278073(n, k)*(x-1)^(n-k) with coeffs. in A292605.

F_{4; n}(x) = Sum_{k=0..n} A278074(n, k)*(x-1)^(n-k) with coeffs. in A292606.

The case m = 1 are the Eulerian polynomials whose coefficients are the Eulerian numbers which are displayed in Euler's triangle A173018.

Evaluated at x in {-1, 1, 0} these families of polynomials give for the first few m:

F_{m} : F_{0} F_{1} F_{2} F_{3} F_{4}

x = -1: A165326 A155585 A002105 A292609 A292607

x = 1: A000012 A000142 A000680 A014606 A014608 ... (m*n)!/m!^n

x = 0: -- A000012 A000364 A002115 A211212 ... m-alternating permutations of length m*n.

Note that the constant terms of the polynomials are the generalized Euler numbers as defined in A181985. In this sense generalized Euler numbers are also generalized Eulerian numbers.

See the comments in A292604.

T_{m}(n, k) gives the number of ordered set partitions of the set {1, 2, ..., m*n} into sized blocks of shape m*P(n, k), where P(n, k) is the k-th integer partition of n in the 'canonical' order A080577. Here we assume the rows of A080577 to be 0-based and m*[a, b, c,..., h] = [m*a, m*b, m*c,..., m*h]. Here is case m = 4. For instance 4*P(4, .) = [[16], [12, 4], [8, 8], [8, 4, 4], [4, 4, 4, 4]].