A108267 -id:A108267 - cp's OEIS frontend

A108291 Triangle, read by rows, resulting from the matrix product of triangle A108267 with Pascal's triangle (A007318).

1, 2, 1, 9, 9, 1, 64, 96, 34, 1, 625, 1250, 750, 125, 1, 7776, 19440, 16470, 5265, 461, 1, 117649, 352947, 386561, 184877, 35329, 1715, 1, 2097152, 7340032, 9863168, 6307840, 1913408, 232288, 6434, 1, 43046721, 172186884, 274223556, 220016574

Offset: 0

Paul D. Hanna, May 31 2005

Row sums form A108292. Column 0 is A000169(n) = (n+1)^n. Triangle with rows reversed is A108290.

			Triangle begins:
1;
2,1;
9,9,1;
64,96,34,1;
625,1250,750,125,1;
7776,19440,16470,5265,461,1;
117649,352947,386561,184877,35329,1715,1;
2097152,7340032,9863168,6307840,1913408,232288,6434,1; ...

Cf. A108267, A060543, A108290, A000169.

{T(n,k)=local(X=x+x*O(x^(n-k))); polcoeff(sum(j=0,n,binomial(n+n*j+j,n*j+j)*(x/(1+X))^j)/(1+X),n-k)}

A108268 Column 2 of triangle A108267.

Original entry on oeis.org

1, 31, 381, 3431, 26769, 193705, 1343521, 9091270, 60632419, 401001030, 2639871326, 17339260251, 113792427233, 746807661549, 4903854042841, 32227106641988, 211992209767971, 1395903036647155, 9200826759772935

Offset: 2

Paul D. Hanna, May 29 2005

nonn

Cf. A108267.

a(n)=polcoeff((1-x)^(n+1)*sum(j=0,n,binomial(n+n*j+j,n*j+j)*x^j),2)

Offset corrected by Andrey Zabolotskiy, Nov 23 2021

A060543 Triangle, read by antidiagonals, where T(n,k) = C(n+nk+k, nk+k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 10, 5, 1, 1, 35, 28, 7, 1, 1, 126, 165, 55, 9, 1, 1, 462, 1001, 455, 91, 11, 1, 1, 1716, 6188, 3876, 969, 136, 13, 1, 1, 6435, 38760, 33649, 10626, 1771, 190, 15, 1, 1, 24310, 245157, 296010, 118755, 23751, 2925, 253, 17, 1, 1, 92378, 1562275

Offset: 0

Henry Bottomley, Apr 02 2001

Main diagonal is A108288. Antidiagonal sums is A108289. Inverse binomial transforms of each row give triangle A108290. G.f. of row n multiplied by (1-x)^(n+1) equals g.f. of row n of triangle A108267 (rows sums of A108267 equal (n+1)^n).

			row 1: (2*n+1)/1!
row 2: (3*n+1)*(3*n+2)/2!
row 3: (4*n+1)*(4*n+2)*(4*n+3)/3!
row 4: (5*n+1)*(5*n+2)*(5*n+3)*(5*n+4)/4!
row 5: (6*n+1)*(6*n+2)*(6*n+3)*(6*n+4)*(6*n+5)/5!.
Table begins:
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...
1,3,5,7,9,11,13,15,17,19,21,23,25,27,...
1,10,28,55,91,136,190,253,325,406,496,...
1,35,165,455,969,1771,2925,4495,6545,...
1,126,1001,3876,10626,23751,46376,82251,...
1,462,6188,33649,118755,324632,749398,...
1,1716,38760,296010,1344904,4496388,...

Harry J. Smith, Table of n, a(n) for n=0..252

Cf. A108290, A108267, A108288, A108289, A060544 (row 2), A015219 (row 3).

Rows include A000012, A001700, A025174. Columns include A000012, A005408, A060544. Main diagonal is A060545.

PARI
```
T(n,k)=binomial(n+n*k+k,n*k+k)
```

{ i=0; write("b060543.txt", "0 1"); for (m=0, 20, for (k=0, m + 1, n=m - k + 1; write("b060543.txt", i++, " ", binomial(n + n*k + k, n*k + k))); ) } \\ Harry J. Smith, Jul 06 2009

a(n) = A060539(n, k)/n = A007318(nk, k)/n = A060540(n, k)/A060540(n-1, k).

Entry revised by Paul D. Hanna, May 31 2005

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 17 2007

A108290 Triangle, read by rows, such that row n equals the inverse binomial transform of row n of table A060543, where A060543(n,k) = C(n+nk+k, nk+k).

Original entry on oeis.org

1, 1, 2, 1, 9, 9, 1, 34, 96, 64, 1, 125, 750, 1250, 625, 1, 461, 5265, 16470, 19440, 7776, 1, 1715, 35329, 184877, 386561, 352947, 117649, 1, 6434, 232288, 1913408, 6307840, 9863168, 7340032, 2097152, 1, 24309, 1513656, 18921924, 92365758, 220016574

Offset: 0

Paul D. Hanna, May 31 2005

Row sums form A108292. Main diagonal is A000169(n) = (n+1)^n. Triangle with rows reversed is A108291.

The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial (1+n*x)*(2+n*x)*...*(n-1+n*x)/(n-1)! in the basis made of the binomial(x+i,i). - F. Chapoton, Nov 01 2022

			BINOMIAL[1, 9, 9] = {1, 10, 28, 55, 91, 136, 190, 253, ...}.
BINOMIAL[1, 34, 96, 64] = {1, 35, 165, 455, 969, 1771, 2925, ...}.
BINOMIAL[1, 125, 750, 1250, 625] = {1, 126, 1001, 3876, 10626, ...}.
Triangle begins:
  1;
  1,    2;
  1,    9,      9;
  1,   34,     96,      64;
  1,  125,    750,    1250,     625;
  1,  461,   5265,   16470,   19440,    7776;
  1, 1715,  35329,  184877,  386561,  352947,  117649;
  1, 6434, 232288, 1913408, 6307840, 9863168, 7340032, 2097152; ...

Cf. A108267, A060543, A108290, A000169.

{T(n,k)=local(X=x+x*O(x^k)); polcoeff(sum(j=0,n,binomial(n+n*j+j,n*j+j)*(x/(1+X))^j)/(1+X),k)}

A108292 Row sums of triangle A108290.

Original entry on oeis.org

1, 3, 19, 195, 2751, 49413, 1079079, 27760323, 822299383, 27565191753, 1031671508495, 42643092165765, 1929325374428791, 94835735736471369, 5032700868665421519, 286770182910733076163, 17463186681730290301671

Offset: 0

Paul D. Hanna, May 31 2005

nonn

Cf. A108267, A060543, A108290.

a(n)=local(X=x+x*O(x^n));sum(k=0,n, polcoeff(sum(j=0,n,binomial(n+n*j+j,n*j+j)*(x/(1+X))^j)/(1+X),k))

a(n)=sum(k=0,n,2^k*polcoeff( (1-x)^(n+1)*sum(j=0,n,binomial(n+n*j+j,n*j+j)*x^j),k))

a(n) = Sum_{k=0..n} A108267(n, k)*2^k.

A108288 Main diagonal of table A060543; a(n) = C((n+1)^2-1, n*(n+1)).

Original entry on oeis.org

1, 3, 28, 455, 10626, 324632, 12271512, 553270671, 28987537150, 1731030945644, 116068178638776, 8634941152058949, 705873715441872264, 62895036884524942320, 6067037854078498539696, 629921975126394617164575, 70043473196734767582082230

Offset: 0

Paul D. Hanna, May 31 2005

nonn

Cf. A060545, A108267, A060543, A108289.

PARI
```
a(n)=binomial((n+1)^2-1,n*(n+1))
```

a(n) = A060545(n+1). - R. J. Mathar, Aug 24 2008

A108289 Antidiagonal sums of table A060543.

Original entry on oeis.org

1, 2, 5, 17, 72, 357, 2022, 12900, 91448, 711180, 6004981, 54619489, 531854438, 5515551251, 60642234815, 704106298738, 8603658260904, 110306422692488, 1479905106340895, 20727595895871297, 302423908621734606

Offset: 0

Paul D. Hanna, May 31 2005

nonn

Cf. A108267, A060543, A108288.

a(n)=sum(k=0,n,binomial(n+(n-k)*k,(n-k)*k+k))

a(n)=Sum_{k=0..n} C(n+(n-k)*k, (n-k)*k+k).

A333829 Triangle read by rows: T(n,k) is the number of parking functions of length n with k strict descents. T(n,k) for n >= 1 and 0 <= k <= n-1.

Original entry on oeis.org

1, 2, 1, 5, 10, 1, 14, 73, 37, 1, 42, 476, 651, 126, 1, 132, 2952, 8530, 4770, 422, 1, 429, 17886, 95943, 114612, 31851, 1422, 1, 1430, 107305, 987261, 2162033, 1317133, 202953, 4853, 1, 4862, 642070, 9613054, 35196634, 39471355, 13792438, 1262800, 16786, 1

Offset: 1

Matthieu Josuat-Vergès, Apr 06 2020

In a parking function w(1), ..., w(n), a strict descent is an index i such that w(i) > w(i+1).

Define an n-dimensional polytope as the convex hull of length n+1 nondecreasing parking functions. Then, the Ehrhart h*-polynomial of this polytope is Sum_{k=0..n-1} T(n,k) * z^(n-1-k).

			The triangle T(n,k) begins:
n/k  0    1    2    3    4    5
1    1
2    2    1
3    5   10    1
4   14   73   37    1
5   42  476  651  126    1
6  132 2952 8530 4770  422    1
...
The 10 parking functions of length 3 with 1 strict descent are: [[1, 2, 1], [2, 1, 1], [1, 3, 1], [3, 1, 1], [2, 1, 2], [2, 2, 1], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2]].

Ari Cruz, Pamela E. Harris, Kimberly J. Harry, Jan Kretschmann, Matt McClinton, Alex Moon, John O. Museus, and Eric Redmon, On some discrete statistics of parking functions, arXiv:2312.16786 [math.CO], 2023.
Paul R. F. Schumacher, Descents in Parking Functions, J. Int. Seq. 21 (2018), #18.2.3.

Row sums give A000272(n+1).
Column k=0 gives A000108.
Similar to A108267.

SageMath
```
var('z,t')
assume(0
				
```

# using[latte_int from LattE]
# Install with "sage -i latte_int".
# Another method is to compute the Ehrhart h^*-polynomial of a polytope.
var('z,t')
def Tpol(n):
    p = Polyhedron( NonDecreasingParkingFunctions(n+1) ).ehrhart_polynomial()
    return expand(factor( (1-z)**(n+1) * sum( p * z**t , t , 0 , oo ) ))
def T(n,k):
    return Tpol(n).list()[n-1-k]

cp's OEIS Frontend

A108291 Triangle, read by rows, resulting from the matrix product of triangle A108267 with Pascal's triangle (A007318).

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A108268 Column 2 of triangle A108267.

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A060543 Triangle, read by antidiagonals, where T(n,k) = C(n+n*k+k, n*k+k).

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A108290 Triangle, read by rows, such that row n equals the inverse binomial transform of row n of table A060543, where A060543(n,k) = C(n+n*k+k, n*k+k).

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A108292 Row sums of triangle A108290.

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Formula

A108288 Main diagonal of table A060543; a(n) = C((n+1)^2-1, n*(n+1)).

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Formula

A108289 Antidiagonal sums of table A060543.

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Formula

A333829 Triangle read by rows: T(n,k) is the number of parking functions of length n with k strict descents. T(n,k) for n >= 1 and 0 <= k <= n-1.

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SageMath

SageMath

A060543 Triangle, read by antidiagonals, where T(n,k) = C(n+nk+k, nk+k).

A108290 Triangle, read by rows, such that row n equals the inverse binomial transform of row n of table A060543, where A060543(n,k) = C(n+nk+k, nk+k).