A081577 -id:A081577 - cp's OEIS frontend

A108350 Number triangle T(n,k) = Sum_{j=0..n-k} binomial(k,j)binomial(n-j,k)((j+1) mod 2).

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 4, 1, 1, 5, 13, 13, 5, 1, 1, 6, 21, 32, 21, 6, 1, 1, 7, 31, 65, 65, 31, 7, 1, 1, 8, 43, 116, 161, 116, 43, 8, 1, 1, 9, 57, 189, 341, 341, 189, 57, 9, 1, 1, 10, 73, 288, 645, 842, 645, 288, 73, 10, 1, 1, 11, 91, 417, 1121, 1827, 1827, 1121, 417, 91

Offset: 0

Paul Barry, May 31 2005

Or as a square array read by antidiagonals, T(n,k) = Sum_{j=0..n} binomial(k,j)*binomial(n+k-j,k)*((j+1) mod 2).
A symmetric number triangle based on 1/(1-x^2).
The construction of a symmetric triangle in this example is general. Let f(n) be a sequence, preferably with f(0)=1. Then T(n,k) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*f(j) yields a symmetric triangle. When f(n)=1^n, we get Pascal's triangle. When f(n)=2^n, we get the Delannoy triangle (see A008288). In general, f(n)=k^n yields a (1,k,1)-Pascal triangle (see A081577, A081578). Row sums of triangle are A100131. Diagonal sums of the triangle are A108351. Triangle mod 2 is A106465.

			Triangle rows begin
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3,  1;
  1,  4,  7,  4,  1;
  1,  5, 13, 13,  5,  1;
  1,  6, 21, 32, 21,  6,  1;
As a square array read by antidiagonals, rows begin
  1, 1,  1,   1,   1,    1,    1, ...
  1, 2,  3,   4,   5,    6,    7, ...
  1, 3,  7,  13,  21,   31,   43, ...
  1, 4, 13,  32,  65,  116,  189, ...
  1, 5, 21,  65, 161,  341,  645, ...
  1, 6, 31, 116, 341,  842, 1827, ...
  1, 7, 43, 189, 645, 1827, 4495, ...

trgn(nn) = {for (n= 0, nn, for (k = 0, n, print1(sum(j=0, n-k, binomial(k,j)*binomial(n-j,k)*((j+1) % 2)), ", ");); print(););} \\ Michel Marcus, Sep 11 2013

Row k (and column k) has g.f. (1+C(k,2)x^2)/(1-x)^(k+1).

A125171 Riordan array ((1-x)/(1-3*x+x^2),x/(1-x)) read by rows.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 13, 8, 4, 1, 34, 21, 12, 5, 1, 89, 55, 33, 17, 6, 1, 233, 144, 88, 50, 23, 7, 1, 610, 377, 232, 138, 73, 30, 8, 1, 1597, 987, 609, 370, 211, 103, 38, 9, 1, 4181, 2584, 1596, 979, 581, 314, 141, 47, 10, 1, 10946, 6765, 4180, 2575, 1560, 895, 455, 188, 57, 11, 1, 28657, 17711, 10945, 6755, 4135, 2455, 1350, 643

Offset: 0

Gary W. Adamson, Nov 22 2006

Partial column sums triangle of odd-indexed Fibonacci numbers.
Left border = odd-indexed Fibonacci numbers, next-to-left border = even-indexed Fibonacci numbers. Row sums = A061667: (1, 3, 9, 26, 73, 201, ...).
Diagonal sums are A027994(n). - Philippe Deléham, Jan 14 2014

			(6,3) = 33 = 12 + 21 = (5,3) + (5,2). First few rows of the triangle are:
   1;
   2,  1;
   5,  3,  1;
  13,  8,  4,  1;
  34, 21, 12,  5,  1;
  89, 55, 33, 17,  6,  1;
  ...

Vincenzo Librandi, Table of n, a(n) for n = 0..568
Peter Bala, A note on the diagonals of a proper Riordan Array

Cf. A027994, A061667 (row sums).

C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if;
end proc:
with(combinat):
for n from 0 to 10 do
    seq(C(n, n-k) + add(fibonacci(2*i)*C(n-i, n-k-i), i = 1..n), k = 0..n);
end do; # Peter Bala, Mar 21 2018

T(n,k)=if(k==n,1,if(k<=1,fibonacci(2*n-1),T(n-1,k)+T(n-1,k-1)));
for(n=1,15,for(k=1,n,print1(T(n,k),", "));print()); /* show triangle */
/* Joerg Arndt, Jun 17 2011 */

Let the left border = odd-indexed Fibonacci numbers, (1, 2, 5, 13, 34...); then for k>1, T(n,k) = T(n-1,k) + T(n-1,k-1).
G.f.: (1-x)^2/((1-3*x+x^2)*(1-x*(1+y))). - Paul Barry, Dec 05 2006
T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - 4*T(n-2,k) - 3*T(n-2,k-1) + T(n-3,k) + T(n-3,k-1), T(0,0)=1, T(1,0)=2, T(1,1)=1, T(2,0)=5, T(2,1)=3, T(2,2)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 14 2014
Exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(13 + 8*x + 4*x^2/2! + x^3/3!) = 13 + 21*x + 33*x^2/2! + 50*x^3/3! + 73*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014
T(n,k) = C(n, n-k) + Sum_{i = 1..n} Fibonacci(2*i)*C(n-i, n-k-i), where C(n,k) = n!/(k!*(n-k)!) for 0 <= k <= n, otherwise 0. - Peter Bala, Mar 21 2018

New description from Paul Barry, Dec 05 2006

Data error corrected by Johannes W. Meijer, Jun 16 2011

A129689 A007318 * A129688.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 7, 5, 3, 1, 15, 12, 8, 4, 1, 31, 27, 20, 12, 5, 1, 63, 58, 47, 32, 17, 6, 1, 127, 121, 105, 79, 49, 23, 7, 1, 255, 248, 226, 184, 128, 72, 30, 8, 1, 511, 503, 474, 410, 312, 200, 102, 38, 9, 1

Offset: 1

Gary W. Adamson, Apr 28 2007

Row sums = A057711: (1, 2, 6, 16, 40, 96, ...). A129690 = A129688 * A007318.

Riordan array ( (1-2*x+2*x^2)/((1-x)*(1-2*x)), x/(1-x) ). - Peter Bala, Mar 21 2018

			First few rows of the triangle are:
   1;
   1,  1;
   3,  2,  1;
   7,  5,  3,  1;
  15, 12,  8,  4,  1;
  31, 27, 20, 12,  5,  1;
  63, 58, 47, 32, 17,  6,  1;
  ...

Muniru A Asiru, Table of n, a(n) for n = 1..5151
Peter Bala, A note on the diagonals of a proper Riordan Array

Cf. A129688, A007318, A057711, A129690.

Flat(List([0..12],n->List([0..n],k->Binomial(n,k)+Sum([2..n],i->2^(i-1)*Binomial(n-i,n-k-i))))); # Muniru A Asiru, Mar 22 2018

C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if;
end proc:
for n from 0 to 12 do
   seq(C(n, n-k) + add(2^(i-1)*C(n-i, n-k-i), i = 2..n), k = 0..n)
end do; #  Peter Bala, Mar 21 2018

T[n_, k_] := Binomial[n, n-k] + Sum[2^(i-1) Binomial[n-i, n-k-i], {i, 2, n}]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 10 2019 *)

# uses[riordan_array from A256893]
# After Peter Bala.
riordan_array((1-2*x+2*x^2)/((1-x)*(1-2*x)), x/(1-x), 8) # Peter Luschny, Mar 21 2018

Binomial transform of A129688. A007318 * A129688 as infinite lower triangular matrices.
From Peter Bala, Mar 21 2018: (Start)
T(n,k) = C(n, n-k) + Sum_{i = 2..n} 2^(i-1)*C(n-i, n-k-i), where C(n,k) = n!/(k!*(n-k)!) for 0 <= k <= n, otherwise 0.
Exp(x) * the e.g.f. for row n = the e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(7 + 5*x + 3*x^2/2! + x^3/3!) = 7 + 12*x + 20*x^2/2! + 32*x^3/3! + 49*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1-x) ).
(End)

A210562 Triangle of coefficients of polynomials v(n,x) jointly generated with A210561; see the Formula section.

Original entry on oeis.org

1, 2, 2, 2, 5, 4, 2, 6, 12, 8, 2, 6, 17, 28, 16, 2, 6, 18, 46, 64, 32, 2, 6, 18, 53, 120, 144, 64, 2, 6, 18, 54, 152, 304, 320, 128, 2, 6, 18, 54, 161, 424, 752, 704, 256, 2, 6, 18, 54, 162, 474, 1152, 1824, 1536, 512, 2, 6, 18, 54, 162, 485, 1372, 3056, 4352

Offset: 1

Clark Kimberling, Mar 22 2012

Last term in row n:  2^(n-1)
Limiting row:  2*3^(n-1)
Alternating row sums: 1,0,1,0,1,0,1,0,...
For a discussion and guide to related arrays, see A208510.

			First five rows:
  1
  2   2
  2   5   4
  2   6   12   8
  2   6   17   28   16
First three polynomials v(n,x): 1, 2 + 2*x, 2 + 5*x + 4*x^2.

P. Bala, A note on the diagonals of a proper Riordan Array

Row sums A005409. Cf. A208510, A210561.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*u[n - 1, x] + x*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A210561 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A210562 *)

u(n,x) = x*u(n-1,x)+x*v(n-1,x)+1,
v(n,x) = (x+1)*u(n-1,x)+x*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
From Peter Bala, Mar 06 2017: (Start)
T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1).
E.g.f. for the n-th subdiagonal: exp(2*x)*(2 + 2*x + 2*x^2/2! + 2*x^3/3! + ... + 2*x^(n-1)/(n-1)! + x^n/n!).
Riordan array ((1 + x)/(1 - x), x*(2 + x)).
Row sums A005409 (except for the initial term). (End)

A283150 Riordan array (1/(1-9x)^(1/3), x/(9x-1)).

Original entry on oeis.org

1, 3, -1, 18, -12, 1, 126, -126, 21, -1, 945, -1260, 315, -30, 1, 7371, -12285, 4095, -585, 39, -1, 58968, -117936, 49140, -9360, 936, -48, 1, 480168, -1120392, 560196, -133380, 17784, -1368, 57, -1, 3961386, -10563696, 6162156, -1760616, 293436, -30096, 1881, -66, 1, 33011550, -99034650, 66023100

Offset: 0

Tom Richardson, Mar 01 2017

Triangle read by rows. This is an example of a Riordan group involution. Dual Riordan array of A283151. With A283151 and A248324, forms doubly infinite Riordan array. For b and c the sequences A283151 and A248324, respectively, and i,j >= 0, the doubly infinite array with d(i,j) = a(i,j), d(-j,-i) = b(i,j), d(i,-j) = c(j,i), and d(-i,j) = 0 (except d(0,0)=1) is a doubly infinite Riordan array.

Matrix inverse of a(m,n) is a(m,n). - Werner Schulte, Aug 05 2017

			The triangle begins
        1;
        3,        -1;
       18,       -12,       1;
      126,      -126,      21,       -1;
      945,     -1260,     315,      -30,      1;
     7371,    -12285,    4095,     -585,     39,     -1;
    58968,   -117936,   49140,    -9360,    936,    -48,    1;
   480168,  -1120392,  560196,  -133380,  17784,  -1368,   57,  -1;
  3961386, -10563696, 6162156, -1760616, 293436, -30096, 1881, -66, 1;

Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
H. Prodinger, Some information about the binomial transform, The Fibonacci Quarterly, 32, 1994, 412-415.
Thomas M. Richardson, The three 'R's and Dual Riordan Arrays, arXiv:1609.01193 [math.CO], 2016.

Cf. A004987, A248324, A283151, A046521, A155579.

T := (n, k) -> (-1)^k*binomial(n - 2/3, n - k)*9^(n - k):
for n from 0 to 6 do seq(T(n, k), k = 0..n) od; # Peter Luschny, Sep 03 2021

a(m,n) = binomial(-n-1/3, m-n)*(-1)^m*9^(m-n);
tabl(nn) = for(n=0, nn, for (k=0, n, print1(a(n, k), ", ")); print); \\ Michel Marcus, Aug 07 2017

a(m,n) = binomial(-n-1/3, m-n)*(-1)^m*9^(m-n).
G.f.: (1-9x)^(2/3)/(xy-9x+1).
Recurrence: a(m,n) = a(m, n-1)*(n-1-m)/(9*n-6) for 0 < n <= m. - Werner Schulte, Aug 05 2017
From Peter Bala, Mar 05 2018 (Start):
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then (-1)^n*P(n,x) is the n-th degree Taylor polynomial of (1 - 9*x)^(n-2/3) about 0. For example, for n = 4 we have (1 - 9*x)^(10/3) = 945*x^4 - 1260*x^3 + 315*x^2 - 30*x + 1 + O(x^5).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,9*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(9*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(-x) * the e.g.f. for the polynomial R(n,9*x). For example, when n = 3 we have exp(-x)*(126 - 126*(9*x) + 21*(9*x)^2/2! - (9*x)^3/3!) = 126 - 1260*x + 4095*x^2/2! - 9360*x^3/3! + 17784*x^4/4! - ....
Let F(x) = (1 - ( 1 - 9*x)^(2/3))/(3*x) denote the o.g.f. of A155579. The derivatives of F(x) are related to the row polynomials P(n,x) by the identity x^n/n! * (d/dx)^n(F(x)) = 1/(3*x)*( (-1)^n - P(n,x)/(1 - 9*x)^(n-2/3) ), n = 0,1,2,.... Cf. A283151 and A046521. (End)
From Peter Bala, Aug 18 2021: (Start)
T(n,k) = (-1)^k*binomial(n-2/3, n-k)*9^(n-k).
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 9*b, c = -1 and d = 1/3.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of formal power series then
G(x) = (1/(1 - 9*b*x)^(1/3)) * F(x/(1 - 9*b*x)) iff F(x) = (1/(1 + 9*b*x)^(1/3)) * G(x/(1 + 9*b*x)).
The infinitesimal generator of the unsigned array has the sequence (9*n+3) n>=0 on the main subdiagonal and zeros elsewhere. The m-th power of the unsigned array has entries m^(n-k)*|T(n,k)|. (End)

Offset corrected by Werner Schulte, Aug 05 2017

A283151 Triangle read by rows: Riordan array (1/(1-9x)^(2/3), x/(9x-1)).

Original entry on oeis.org

1, 6, -1, 45, -15, 1, 360, -180, 24, -1, 2970, -1980, 396, -33, 1, 24948, -20790, 5544, -693, 42, -1, 212058, -212058, 70686, -11781, 1071, -51, 1, 1817640, -2120580, 848232, -176715, 21420, -1530, 60, -1, 15677145, -20902860, 9754668, -2438667, 369495, -35190, 2070, -69, 1, 135868590, -203802885

Offset: 0

Tom Richardson, Mar 01 2017

This is an example of a Riordan group involution.
Dual Riordan array of A283150.
With A283150 and A248324, forms doubly infinite Riordan array. For b and c the sequences A283150 and A248324, respectively, and i,j >= 0, the doubly infinite array with d(i,j) = a(i,j), d(-j,-i) = b(i,j), d(i,-j) = c(i,j), and d(-i,j) = 0 (except d(0,0)=1) is a doubly infinite Riordan array.

			Triangle begins
         1;
         6,        -1;
        45,       -15,       1;
       360,      -180,      24,       -1;
      2970,     -1980,     396,      -33,      1;
     24948,    -20790,    5544,     -693,     42,     -1;
    212058,   -212058,   70686,   -11781,   1071,    -51,    1;
   1817640,  -2120580,  848232,  -176715,  21420,  -1530,   60,  -1;
  15677145, -20902860, 9754668, -2438667, 369495, -35190, 2070, -69, 1;

Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
H. Prodinger, Some information about the binomial transform, The Fibonacci Quarterly, 32, 1994, 412-415.
Thomas M. Richardson, The three 'R's and Dual Riordan Arrays, arXiv:1609.01193 [math.CO], 2016.

Cf. A004988, A248324, A283150, A025748, A046521.

a(m,n) = binomial(-n-2/3, m-n)*(-1)^m*9^(m-n).
G.f.: (1-9x)^(1/3)/(xy-9x+1).
Recurrence: a(m,n) = a(m,n-1)*(n-1-m)/(9*n-3) for 0 < n <= m; matrix inverse of a(m,n) is a(m,n). - Werner Schulte, Aug 05 2017
From Peter Bala, Mar 05 2018 (Start):
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then (-1)^n*P(n,x) is the n-th degree Taylor polynomial of (1 - 9*x)^(n-1/3) about 0. For example, for n = 4 we have (1 - 9*x)^(11/3) = 2970*x^4 - 1980*x^3 + 396*x^2 - 33*x + 1 + O(x^5).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,9*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(9*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(-x) * the e.g.f. for the polynomial R(n,9*x). For example, when n = 3 we have exp(-x)*(360 - 180*(9*x) + 24*(9*x)^2/2! - (9*x)^3/3!) = 360 - 1980*x + 5544*x^2/2! - 11781*x^3/3! + 21420*x^4/4! - ....
Let F(x) = (1 - ( 1 - 9*x)^(1/3))/(3*x). See A025748. The derivatives of F(x) are related to the row polynomials P(n,x) by the identity x^n/n! * (d/dx)^n(F(x)) = 1/(3*x)*( (-1)^n - P(n,x)/(1 - 9*x)^(n-1/3) ), n = 0,1,2,.... Cf. A283151 and A046521. (End)
From Peter Bala, Aug 18 2021: (Start)
T(n,k) = (-1)^k*binomial(n-1/3, n-k)*9^(n-k).
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 9*b, c = -1 and d = 2/3.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of formal power series then
G(x) = (1/(1 - 9*b*x)^(2/3)) * F(x/(1 - 9*b*x)) iff F(x) = (1/(1 + 9*b*x)^(2/3)) * G(x/(1 + 9*b*x)).
The infinitesimal generator of the unsigned array has the sequence (9*n+6) n>=0 on the main subdiagonal and zeros elsewhere. The m-th power of the unsigned array has entries m^(n-k)*|T(n,k)|. (End)

Offset corrected by Werner Schulte, Aug 05 2017

A159853 Riordan array ((1-2x+2x^2)/(1-x), x/(1-x)).

Original entry on oeis.org

1, -1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 3, 4, 4, 3, 1, 1, 4, 7, 8, 7, 4, 1, 1, 5, 11, 15, 15, 11, 5, 1, 1, 6, 16, 26, 30, 26, 16, 6, 1, 1, 7, 22, 42, 56, 56, 42, 22, 7, 1, 1, 8, 29, 64, 98, 112, 98, 64, 29, 8, 1, 1, 9, 37, 93, 162, 210, 210, 162, 93, 37, 9, 1, 1, 10, 46, 130, 255

Offset: 0

Philippe Deléham, Apr 24 2009

Essentially the same as A087698.

			Triangle begins:
   1;
  -1, 1;
   1, 0, 1;
   1, 1, 1, 1;
   1, 2, 2, 2, 1;
   1, 3, 4, 4, 3, 1;
   ...

Muniru A Asiru, Table of n, a(n) for n = 0..5151
Peter Bala, A note on the diagonals of a proper Riordan Array

Cf. A087698.

Flat(List([0..12],n->List([0..n],k->Binomial(n,k)-2*Binomial(n-1,n-k-1)+2*Binomial(n-2,n-k-2)))); # Muniru A Asiru, Mar 22 2018

/* As triangle */ [[Binomial(n, n-k)-2*Binomial(n-1, n-k-1)+2*Binomial(n-2, n-k-2): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Mar 22 2018

C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if;
end proc:
for n from 0 to 10 do
  seq(C(n, n-k) - 2*C(n-1, n-k-1) + 2*C(n-2, n-k-2), k = 0..n);
end do; # Peter Bala, Mar 20 2018

Join[{1, -1}, Rest[T[0, 0]=1; T[n_, k_]:=Binomial[n, n - k] - 2 Binomial[n - 1, n - k - 1] + 2 Binomial[n - 2, n - k - 2]; Table[T[n, k], {n, 1, 15}, {k, 0, n}]//Flatten]] (* Vincenzo Librandi, Mar 22 2018 *)

# uses[riordan_array from A256893]
riordan_array((1-2*x+2*x^2)/(1-x), x/(1-x), 8) # Peter Luschny, Mar 21 2018

From Peter Bala, Mar 20 2018: (Start)
T(n,k) = C(n,k) - 2*C(n-1,n-k-1) + 2*C(n-2,n-k-2), where C(n,k) = n!/(k!*(n-k)!) for 0 <= k <= n, otherwise 0.
Exp(x) * the e.g.f. for row n = the e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + x + x^2/2! + x^3/3!) = 1 + 2*x + 2*x^2/2! + 4*x^3/3! + 8*x^4/4! + 15*x^5/5! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1-x) ). (End)

A210561 Triangle of coefficients of polynomials u(n,x) jointly generated with A210562; see the Formula section.

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 1, 3, 8, 8, 1, 3, 9, 20, 16, 1, 3, 9, 26, 48, 32, 1, 3, 9, 27, 72, 112, 64, 1, 3, 9, 27, 80, 192, 256, 128, 1, 3, 9, 27, 81, 232, 496, 576, 256, 1, 3, 9, 27, 81, 242, 656, 1248, 1280, 512, 1, 3, 9, 27, 81, 243, 716, 1808, 3072, 2816, 1024, 1, 3, 9

Offset: 1

Clark Kimberling, Mar 22 2012

Last term in row n:  2^(n-1)
Limiting row:  3^(k-1)
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
1...2
1...3...4
1...3...8...8
1...3...9...20...16
First three polynomials u(n,x): 1, 1 + 2x, 1 + 3x + 4x^2.

P. Bala, A note on the diagonals of a proper Riordan Array

Cf. A210562, A208510, A004070, A048739.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*u[n - 1, x] + v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A210559 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A210560 *)

u(n,x)=x*u(n-1,x)+x*v(n-1,x)+1,
v(n,x)=(x+1)*u(n-1,x)+x*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
From Peter Bala, Mar 06 2017: (Start)
T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1).
E.g.f. for n-th subdiagonal: exp(2*x)*(1 + x + x^2/2! + x^3/3! + ... + x^n/n!). Cf. A004070.
Riordan array (1/(1 - x), x*(2 + x)).
Row sums A048739.
(End)

A143685 Pascal-(1,9,1) array.

Original entry on oeis.org

1, 1, 1, 1, 11, 1, 1, 21, 21, 1, 1, 31, 141, 31, 1, 1, 41, 361, 361, 41, 1, 1, 51, 681, 1991, 681, 51, 1, 1, 61, 1101, 5921, 5921, 1101, 61, 1, 1, 71, 1621, 13151, 29761, 13151, 1621, 71, 1, 1, 81, 2241, 24681, 96201, 96201, 24681, 2241, 81, 1, 1, 91, 2961, 41511, 239241, 460251, 239241, 41511, 2961, 91, 1

Offset: 0

Paul Barry, Aug 28 2008

			Square array begins as:
  1,  1,    1,     1,      1,       1,        1, ... A000012;
  1, 11,   21,    31,     41,      51,       61, ... A017281;
  1, 21,  141,   361,    681,    1101,     1621, ...
  1, 31,  361,  1991,   5921,   13151,    24681, ...
  1, 41,  681,  5921,  29761,   96201,   239241, ...
  1, 51, 1101, 13151,  96201,  460251,  1565301, ...
  1, 61, 1621, 24681, 239241, 1565301,  7272861, ...
Antidiagonal triangle begins as:
  1;
  1,  1;
  1, 11,    1;
  1, 21,   21,     1;
  1, 31,  141,    31,     1;
  1, 41,  361,   361,    41,     1;
  1, 51,  681,  1991,   681,    51,    1;
  1, 61, 1101,  5921,  5921,  1101,   61,  1;
  1, 71, 1621, 13151, 29761, 13151, 1621, 71, 1;

G. C. Greubel, Antidiagonal rows n = 0..50, flattened

Cf. A002534, A143680, A143682.

Pascal (1,m,1) array: A123562 (m = -3), A098593 (m = -2), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081578 (m = 3), A081579 (m = 4), A081580 (m = 5), A081581 (m = 6), A081582 (m = 7), A143683 (m = 8), this sequence (m = 9).

A143685:= func< n,k,q | (&+[Binomial(k, j)*Binomial(n-j, k)*q^j: j in [0..n-k]]) >;
[A143685(n,k,9): k in [0..n], n in [0..12]]; // G. C. Greubel, May 29 2021

Table[Hypergeometric2F1[-k, k-n, 1, 10], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, May 24 2013 *)

flatten([[hypergeometric([-k, k-n], [1], 10).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 29 2021

Square array: T(n, k) = T(n, k-1) + 9*T(n-1, k-1) + T(n-1, k) with T(n, 0) = T(0, k) = 1.
Number triangle: T(n,k) = Sum_{j=0..n-k} binomial(n-k,j)*binomial(k,j)*10^j.
Riordan array (1/(1-x), x*(1+9*x)/(1-x)).
T(n, k) = Hypergeometric2F1([-k, k-n], [1], 10). - Jean-François Alcover, May 24 2013
Sum_{k=0..n} T(n, k) = A002534(n+1). - G. C. Greubel, May 29 2021

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