A029635 -id:A029635 - cp's OEIS frontend

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024

Offset: 1

Boris Putievskiy, Aug 15 2013

The third row is (n^4 - n^2 + 24*n + 24)/12.

For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Rely Pellicer and David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4
Index entries for triangles and arrays related to Pascal's triangle

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).

Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

T:= function(n,k)
    if k=0 then return n^2;
    elif k=n then return 2^n;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019

T:= proc(n, k) option remember;
      if k=0 then n^2
    elif k=n then 2^k
    else T(n-1, k-1) + T(n-1, k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019

T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)

T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019

def funcL(n):
   q = n**2
   return q
def funcR(n):
   q = 2**n
   return q
for n in range (1,9871):
   t=int((math.sqrt(8*n-7) - 1)/ 2)
   i=n-t*(t+1)/2-1
   j=(t*t+3*t+4)/2-n-1
   sum1=0
   sum2=0
   for m1 in range (1,i+1):
      sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
   for m2 in range (1,j+1):
      sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
   sum=sum1+sum2

@CachedFunction
def T(n, k):
    if (k==0): return n^2
    elif (k==n): return 2^n
    else: return T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A029600 Numbers in the (2,3)-Pascal triangle (by row).

Original entry on oeis.org

1, 2, 3, 2, 5, 3, 2, 7, 8, 3, 2, 9, 15, 11, 3, 2, 11, 24, 26, 14, 3, 2, 13, 35, 50, 40, 17, 3, 2, 15, 48, 85, 90, 57, 20, 3, 2, 17, 63, 133, 175, 147, 77, 23, 3, 2, 19, 80, 196, 308, 322, 224, 100, 26, 3, 2, 21, 99, 276, 504, 630, 546, 324, 126, 29, 3, 2, 23, 120, 375, 780, 1134, 1176, 870, 450, 155, 32, 3

Offset: 0

Mohammad K. Azarian

Reverse of A029618. - Philippe Deléham, Nov 21 2006
Triangle T(n,k), read by rows, given by (2,-1,0,0,0,0,0,0,0,...) DELTA (3,-2,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 10 2011
Row n: expansion of (2+3x)*(1+x)^(n-1), n>0. - Philippe Deléham, Oct 10 2011.
For n > 0: T(n,k) = A029635(n,k) + A007318(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
For n>0, row sums = 5*2^(n-1). Generally, for all (a,b)-Pascal triangles, row sums are (a+b)*2^(n-1), n>0. - Bob Selcoe, Mar 28 2015

			First few rows are:
  1;
  2, 3;
  2, 5,  3;
  2, 7,  8,  3;
  2, 9, 15, 11, 3;
...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened

Cf. A007318 (Pascal's triangle), A029618, A084938, A228196, A228576.

T:= function(n,k)
    if n=0 and k=0 then return 1;
    elif k=0 then return 2;
    elif k=n then return 3;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019

a029600 n k = a029600_tabl !! n !! k
a029600_row n = a029600_tabl !! n
a029600_tabl = [1] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [2,3]
-- Reinhard Zumkeller, Apr 08 2012

T:= proc(n, k) option remember;
      if k=0 and n=0 then 1
    elif k=0 then 2
    elif k=n then 3
    else T(n-1, k-1) + T(n-1, k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Nov 12 2019

T[n_, k_]:= T[n, k]= If[n==0 && k==0, 1, If[k==0, 2, If[k==n, 3, T[n-1, k-1] + T[n-1, k] ]]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)

T(n,k) = if(n==0 && k==0, 1, if(k==0, 2, if(k==n, 3, T(n-1, k-1) + T(n-1, k) ))); \\ G. C. Greubel, Nov 12 2019

@CachedFunction
def T(n, k):
    if (n==0 and k==0): return 1
    elif (k==0): return 2
    elif (k==n): return 3
    else: return T(n-1,k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

T(n,k) = T(n-1,k-1) + T(n-1,k) with T(0,0)=1, T(n,0)=2, T(n,n)=3; n, k > 0. - Boris Putievskiy, Sep 04 2013

G.f.: (-1-2*x*y-x)/(-1+x*y+x). - R. J. Mathar, Aug 11 2015

More terms from James Sellers

A161198 Triangle of polynomial coefficients related to the series expansions of (1-x)^((-1-2*n)/2).

Original entry on oeis.org

1, 1, 2, 3, 8, 4, 15, 46, 36, 8, 105, 352, 344, 128, 16, 945, 3378, 3800, 1840, 400, 32, 10395, 39048, 48556, 27840, 8080, 1152, 64, 135135, 528414, 709324, 459032, 160720, 31136, 3136, 128

Offset: 0

Johannes W. Meijer, Jun 08 2009, Jul 22 2011

The series expansion of (1-x)^((-1-2*n)/2) = sum(b(p)*x^p, p=0..infinity) for n = 0, 1, 2, .. can be described with b(p) = (F(p,n)/ (2*n-1)!!)*(binomial(2*p,p)/4^(p)) with F(x,n) = 2^n * product( x+(2*k-1)/2, k=1..n). The roots of the F(x,n) polynomials can be found at p = (1-2*k)/2 with k from 1 to n for n = 0, 1, 2, .. . The coefficients of the F(x,n) polynomials lead to the triangle given above. The triangle row sums lead to A001147.
Quite surprisingly we discovered that sum(b(p)*x^p, p=0..infinity) = (1-x)^(-1-2*n)/2, for n = -1, -2, .. . We assume that if m = n+1 then the value returned for product(f(k), k = m..n) is 1 and if m> n+1 then 1/product(f(k), k=n+1..m-1) is the value returned. Furthermore (1-2*n)!! = (-1)^(n+1)/(2*n-3)!! for n = 1, 2, 3 .. . This leads to b(p) = ((-1-2*n)!!/ G(p,n))*(binomial(2*p,p) /4^(p)) for n = -1, -2, .. . For the G(p,n) polynomials we found that G(p,n) = F(-p,-n). The roots of the G(p,n) polynomials can be found at p=(2*k-1)/2 with k from 1 to (-n) for n = -1, -2, .. . The coefficients of the G(p,n) polynomials lead to a second triangle that stands with its head on top of the first one. It is remarkable that the row sums lead once again to A001147.
These two triangles together look like an hourglass so we propose to call the F(p,n) and the G(p,n) polynomials the hourglass polynomials.
Triangle T(n,k), read by rows, given by (1, 2, 3, 4, 5, 6, 7, 8, 9, ...) DELTA (2, 0, 2, 0, 2, 0, 2, 0, 2, ...) where DELTA is the operator defined in A084938. Philippe Deléham, May 14 2015.

			From _Gary W. Adamson_, Jul 19 2011: (Start)
The first few rows of matrix M are:
  1, 2,  0,  0, 0, ...
  1, 3,  2,  0, 0, ...
  1, 4,  5,  2, 0, ...
  1, 5,  9,  7, 2, ...
  1, 6, 14, 16, 9, ... (End)
The first few G(p,n) polynomials are:
  G(p,-3) = 15 - 46*p + 36*p^2 - 8*p^3
  G(p,-2) = 3 - 8*p + 4*p^2
  G(p,-1) = 1 - 2*p
The first few F(p,n) polynomials are:
  F(p,0) = 1
  F(p,1) = 1 + 2*p
  F(p,2) = 3 + 8*p + 4*p^2
  F(p,3) = 15 + 46*p + 36*p^2 + 8*p^3
The first few rows of the upper and lower hourglass triangles are:
  [15, -46, 36, -8]
  [3, -8, 4]
  [1, -2]
  [1]
  [1, 2]
  [3, 8, 4]
  [15, 46, 36, 8]

Cf. A001790 [(1-x)^(-1/2)], A001803 [(1-x)^(-3/2)], A161199 [(1-x)^(-5/2)] and A161201 [(1-x)^(-7/2)].
Cf. A002596 [(1-x)^(1/2)], A161200 [(1-x)^(3/2)] and A161202 [(1-x)^(5/2)].
A046161 gives the denominators of the series expansions of all (1-x)^((-1-2*n)/2).
A028338 is a scaled triangle version, A039757 is a scaled signed triangle version and A109692 is a transposed scaled triangle version.
A001147 is the first left hand column and equals the row sums.
A004041 is the second left hand column divided by 2, A028339 is the third left hand column divided by 4, A028340 is the fourth left hand column divided by 8, A028341 is the fifth left hand column divided by 16.
A000012, A000290, A024196, A024197 and A024198 are the first (n-m=0), second (n-m=1), third (n-m=2), fourth (n-m=3) and fifth (n-m=4) right hand columns divided by 2^m.
A074599 * A025549 is not always equals the second left hand column.
Cf. A029635. [Gary W. Adamson, Jul 19 2011]

nmax:=7; for n from 0 to nmax do a(n,n):=2^n: a(n,0):=doublefactorial(2*n-1) od: for n from 2 to nmax do for m from 1 to n-1 do a(n,m) := 2*a(n-1,m-1)+(2*n-1)*a(n-1,m) od: od: seq(seq(a(n,k), k=0..n), n=0..nmax);
nmax:=7: M := Matrix(1..nmax+1,1..nmax+1): A029635 := proc(n,k): binomial(n,k) + binomial(n-1,k-1) end: for i from 1 to nmax do for j from 1 to i+1 do M[i,j] := A029635(i,j-1) od: od: for n from 0 to nmax do B := M^n: for m from 0 to n do a(n,m):= B[1,m+1] od: od: seq(seq(a(n,m), m=0..n), n=0..nmax);
A161198 := proc(n,k) option remember; if k > n or k < 0 then 0 elif n = 0 and k = 0 then 1 else 2*A161198(n-1, k-1) + (2*n-1)*A161198(n-1, k) fi end:
seq(print(seq(A161198(n,k), k = 0..n)), n = 0..6);  # Peter Luschny, May 09 2013

nmax = 7; a[n_, 0] := (2*n-1)!!; a[n_, n_] := 2^n; a[n_, m_] := a[n, m] = 2*a[n-1, m-1]+(2*n-1)*a[n-1, m]; Table[a[n, m], {n, 0, nmax}, {m, 0, n}] // Flatten (* Jean-François Alcover, Feb 25 2014, after Maple *)

for(n=0,9, print(Vec(Ser( 2^n*prod( k=1,n, x+(2*k-1)/2 ),,n+1))))  \\ M. F. Hasler, Jul 23 2011

@CachedFunction
def A161198(n,k):
    if k > n or k < 0 : return 0
    if n == 0 and k == 0: return 1
    return 2*A161198(n-1,k-1)+(2*n-1)*A161198(n-1,k)
for n in (0..6): [A161198(n,k) for k in (0..n)]  # Peter Luschny, May 09 2013

a(n,m) := coeff(2^(n)*product((x+(2*k-1)/2),k=1..n), x, m) for n = 0, 1, .. ; m = 0, 1, .. .
a(n, m) = 2*a(n-1,m-1)+(2*n-1)*a(n-1,m) with a(n, n) = 2^n and a(n, 0) = (2*n-1)!!.
a(n,m) = the (m+1)-th term in the top row of M^n, where M is an infinite square production matrix; M[i,j] = A029635(i,j-1) = binomial(i, j-1) + binomial(i-1, j-2) with A029635 the (1.2)-Pascal triangle, see the examples and second Maple program. [Gary W. Adamson, Jul 19 2011]
T(n,k) = 2^k * A028338(n,k). - Philippe Deléham, May 14 2015

A095660 Pascal (1,3) triangle.

Original entry on oeis.org

3, 1, 3, 1, 4, 3, 1, 5, 7, 3, 1, 6, 12, 10, 3, 1, 7, 18, 22, 13, 3, 1, 8, 25, 40, 35, 16, 3, 1, 9, 33, 65, 75, 51, 19, 3, 1, 10, 42, 98, 140, 126, 70, 22, 3, 1, 11, 52, 140, 238, 266, 196, 92, 25, 3, 1, 12, 63, 192, 378, 504, 462, 288, 117, 28, 3, 1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3

Offset: 0

Wolfdieter Lang, May 21 2004

This is the third member, q=3, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with T(0,0)=2, not 1).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} T(n,m)*x^m is G(z,x) = g(z)/(1-x*z*f(z)). Here: g(x) = (3-2*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (3-2*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} T(n-1-k,k) = A000285(n-2), n>=2, with n=1 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Central terms: T(2*n,n) = A028329(n) = A100320(n) for n > 0, A028329 are the central terms of triangle A028326. - Reinhard Zumkeller, Apr 08 2012
Let P be Pascal's triangle, A007318 and R the Riordan array, A097805. Then Pascal triangle (1,q) = ((q-1) * R) + P. Example: Pascal triangle (1,3) = (2 * R) + P. - Gary W. Adamson, Sep 12 2015

			Triangle starts:
  3;
  1,  3;
  1,  4,  3;
  1,  5,  7,   3;
  1,  6, 12,  10,   3;
  1,  7, 18,  22,  13,   3;
  1,  8, 25,  40,  35,  16,   3;
  1,  9, 33,  65,  75,  51,  19,   3;
  1, 10, 42,  98, 140, 126,  70,  22,   3;
  1, 11, 52, 140, 238, 266, 196,  92,  25,   3;
  1, 12, 63, 192, 378, 504, 462, 288, 117,  28,  3;
  1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3;

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Wolfdieter Lang, First 10 rows.

Row sums: A000079(n+1), n>=1, 3 if n=0. Alternating row sums are [3, -2, followed by 0's].
Column sequences (without leading zeros) give for m=1..9 with n>=0: A000027(n+3), A055998(n+1), A006503(n+1), A095661, A000574, A095662, A095663, A095664, A095665.
Cf. A097805.

a095660 n k = a095660_tabl !! n !! k
a095660_row n = a095660_tabl !! n
a095660_tabl = [3] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,3]
-- Reinhard Zumkeller, Apr 08 2012

A095660:= func< n,k | n eq 0 select 3 else (1+2*k/n)*Binomial(n,k) >;
[A095660(n,k): k in [0..n], n in [1..12]]; // G. C. Greubel, May 02 2021

T(n,k):=piecewise(n=0,3,0Mircea Merca, Apr 08 2012

{3}~Join~Table[(1 + 2 k/n) Binomial[n, k], {n, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 14 2015 *)

def A095660(n,k): return 3 if n==0 else (1+2*k/n)*binomial(n,k)
flatten([[A095660(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 02 2021

Recursion: T(n, m)=0 if m>n, T(0, 0)= 3; T(n, 0)=1 if n>=1; T(n, m) = T(n-1, m) + T(n-1, m-1).
G.f. column m (without leading zeros): (3-2*x)/(1-x)^(m+1), m>=0.
T(n,k) = (1+2*k/n) * binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012
Closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013

A095666 Pascal (1,4) triangle.

Original entry on oeis.org

4, 1, 4, 1, 5, 4, 1, 6, 9, 4, 1, 7, 15, 13, 4, 1, 8, 22, 28, 17, 4, 1, 9, 30, 50, 45, 21, 4, 1, 10, 39, 80, 95, 66, 25, 4, 1, 11, 49, 119, 175, 161, 91, 29, 4, 1, 12, 60, 168, 294, 336, 252, 120, 33, 4, 1, 13, 72, 228, 462, 630, 588, 372, 153, 37, 4, 1, 14, 85, 300, 690, 1092

Offset: 0

Wolfdieter Lang, Jun 11 2004

This is the fourth member, q=4, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A096940 (q=5), A096956 (q=6).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) := Sum_{m=0..n} a(n,m)*x^m is G(z,x) = g(z)/(1 - x*z*f(z)). Here: g(x) = (4-3*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (4-3*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022095(n-2), n >= 2, with n=1 value 4. [Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.]
T(2*n,n) = A029609(n) for n > 0, A029609 are the central terms of the Pascal (2,3) triangle A029600. - Reinhard Zumkeller, Apr 08 2012

			Triangle begins:
  [4];
  [1,4];
  [1,5,4];
  [1,6,9,4];
  [1,7,15,13,4];
  ...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Wolfdieter Lang, First 10 rows.

Row sums: A020714(n-1), n >= 1, 4 if n=0.
Alternating row sums are [4, -3, followed by 0's].
Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+4), A055999(n+1), A060488(n+3), A095667-71, A095819.

a095666 n k = a095666_tabl !! n !! k
a095666_row n = a095666_tabl !! n
a095666_tabl = [4] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,4]
-- Reinhard Zumkeller, Apr 08 2012

a(n,k):=(1+3*k/n)*binomial(n,k) # Mircea Merca, Apr 08 2012

A095666[n_, k_] := If[n == k,  4, (3*k/n + 1)*Binomial[n, k]];
Table[A095666[n, k], {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Apr 14 2025 *)

Recursion: a(n, m) = 0 if m > n, a(0, 0) = 4; a(n, 0) = 1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (4-3*x)/(1-x)^(m+1), m >= 0.
a(n,k) = (1 + 3*k/n)*binomial(n,k). - Mircea Merca, Apr 08 2012

A096940 Pascal (1,5) triangle.

Original entry on oeis.org

5, 1, 5, 1, 6, 5, 1, 7, 11, 5, 1, 8, 18, 16, 5, 1, 9, 26, 34, 21, 5, 1, 10, 35, 60, 55, 26, 5, 1, 11, 45, 95, 115, 81, 31, 5, 1, 12, 56, 140, 210, 196, 112, 36, 5, 1, 13, 68, 196, 350, 406, 308, 148, 41, 5, 1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5, 1, 15, 95, 345, 810, 1302

Offset: 0

Wolfdieter Lang, Jul 16 2004

This is the fifth member, q=5, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A095666 (q=4), A096956 (q=6).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(5-4*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(5-4*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022096(n-2), n>=2, with n=1 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins:
  5;
  1,  5;
  1,  6,  5;
  1,  7, 11,   5;
  1,  8, 18,  16,   5;
  1,  9, 26,  34,  21,   5;
  1, 10, 35,  60,  55,  26,   5;
  1, 11, 45,  95, 115,  81,  31,   5;
  1, 12, 56, 140, 210, 196, 112,  36,   5;
  1, 13, 68, 196, 350, 406, 308, 148,  41,  5;
  1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5; etc.

David A. Corneth, Table of n, a(n) for n = 0..9999
Wolfdieter Lang, First 10 rows.

Row sums: A007283(n-1), n>=1, 5 if n=0; g.f.: (5-4*x)/(1-2*x). Alternating row sums are [5, -4, followed by 0's].

Column sequences (without leading zeros) give for m=1..9, with n>=0: A000027(n+5), A056000(n-1), A096941-7.

a(n,k):=piecewise(n=0,5,0Mircea Merca, Apr 08 2012

a(n) = {if(n <= 1, return(5 - 4*(n==1))); my(m = (sqrtint(8*n + 1) - 1)\2, t = n - binomial(m + 1, 2)); (1+4*t/m)*binomial(m,t)} \\ David A. Corneth, Aug 28 2019

Recursion: a(n, m)=0 if m>n, a(0, 0)= 5; a(n, 0)=1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (5-4*x)/(1-x)^(m+1), m>=0.
a(n,k) = (1+4*k/n)*binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012

A143491 Unsigned 2-Stirling numbers of the first kind.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 24, 26, 9, 1, 120, 154, 71, 14, 1, 720, 1044, 580, 155, 20, 1, 5040, 8028, 5104, 1665, 295, 27, 1, 40320, 69264, 48860, 18424, 4025, 511, 35, 1, 362880, 663696, 509004, 214676, 54649, 8624, 826, 44, 1, 3628800, 6999840, 5753736, 2655764

Offset: 2

Peter Bala, Aug 20 2008

Essentially the same as A136124 but with column numbers differing by one. See A049444 for a signed version of this array. The unsigned 2-Stirling numbers of the first kind count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the elements 1 and 2 belong to distinct cycles. This is the particular case r = 2 of the unsigned r-Stirling numbers of the first kind, which count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the numbers 1, 2, ..., r belong to distinct cycles. The case r = 1 gives the usual unsigned Stirling numbers of the first kind, abs(A008275); for other cases see A143492 (r = 3) and A143493 (r = 4). The corresponding 2-Stirling numbers of the second kind can be found in A143494.
In general, the lower unitriangular array of unsigned r-Stirling numbers of the first kind (with suitable offsets in the row and column indexing) equals the matrix product St1 * P^(r-1), where St1 is the array of unsigned Stirling numbers of the first kind, abs(A008275) and P is Pascal's triangle, A007318. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For details of the related r-Lah numbers see A143497.
This sequence also represents the number of permutations in the alternating group An of length k, where the length is taken with respect to the generators set {(12)(ij)}. For a bijective proof of the relation between these numbers and the 2-Stirling numbers of the first kind see the Rotbart link. - Aviv Rotbart, May 05 2011
With offset n=0,k=0 : triangle T(n,k), read by rows, given by [2,1,3,2,4,3,5,4,6,5,...] DELTA [1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 29 2011
With offset n=0 and k=0, this is the Sheffer triangle (1/(1-x)^2,-log(1-x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(-2*t),1-exp(-t))). See the e.g.f. given below. Compare also with the e.g.f. for the signed version A049444. - Wolfdieter Lang, Oct 10 2011
Reversed rows correspond to the Betti numbers of the moduli space M(0,n) of smooth Riemann surfaces (see Murri link). - Tom Copeland, Sep 19 2012

			Triangle begins
  n\k|.....2.....3.....4.....5.....6.....7
  ========================================
  2..|.....1
  3..|.....2.....1
  4..|.....6.....5.....1
  5..|....24....26.....9.....1
  6..|...120...154....71....14.....1
  7..|...720..1044...580...155....20.....1
  ...
T(4,3) = 5. The permutations of {1,2,3,4} with 3 cycles such that 1 and 2 belong to different cycles are: (1)(2)(3 4), (1)(3)(24), (1)(4)(23), (2)(3)(14) and (2)(4)(13). The remaining possibility (3)(4)(12) is not allowed.
From _Aviv Rotbart_, May 05 2011: (Start)
Example of the alternating group permutations numbers:
Triangle begins
  n\k|.....0.....1.....2.....3.....4.....5.....6.....7
  ====================================================
  2..|.....1
  3..|.....1.....2
  4..|.....1.....5.....6
  5..|.....1.....9....26....24
  6..|.....1....14....71...154...120
  7..|.....1....20...155...580..1044..720
A(n,k) = number of permutations in An of length k, with respect to the generators set {(12)(ij)}. For example, A(2,0)=1 (only the identity is there), for A4, the generators are {(12)(13),(12)(14),(12,23),(12)(24),(12)(34)}, thus we have A(4,1)=5 (exactly 5 generators), the permutations of length 2 are:
   (12)(13)(12)(13) = (312)
   (12)(13)(12)(14) = (41)(23)
   (12)(13)(12)(24) = (432)(1)
   (12)(13)(12)(34) = (342)(1)
   (12)(23)(12)(24) = (13)(24)
   (12)(14)(12)(14) = (412)(3)
Namely, A(4,2)=6. Together with the identity [=(12)(12), of length 0. therefore A(4,0)=1] we have 12 permutations, comprising all A4 (4!/2=12). (End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Olivier Bodini, Antoine Genitrini, and Mehdi Naima, Ranked Schröder Trees, arXiv:1808.08376 [cs.DS], 2018.
Andrei Z. Broder, The r-Stirling numbers, Discrete Math. 49, 241-259 (1984)
A. Dzhumadildaev and D. Yeliussizov, Path decompositions of digraphs and their applications to Weyl algebra, arXiv preprint arXiv:1408.6764 [math.CO], 2014. [Version 1 contained many references to the OEIS, which were removed in Version 2. - _N. J. A. Sloane_, Mar 28 2015]
Askar Dzhumadil’daev and Damir Yeliussizov, Walks, partitions, and normal ordering, Electronic Journal of Combinatorics, 22(4) (2015), #P4.10.
Neuwirth Erich, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 1-3, 33-51 (2001)
R. Murri, Fatgraph Algorithms and the Homology of the Kontsevich Complex, arXiv:1202.1820 [math.AG], 2012. (see Table 1, p. 3)
Aviv Rotbart, Generator Sets for the Alternating Group, Séminaire Lotharingien de Combinatoire 65 (2011), Article B65b, 16pp.
Michael J. Schlosser and Meesue Yoo, Elliptic Rook and File Numbers, Electronic Journal of Combinatorics, 24(1) (2017), #P1.31.
M. Shattuck, Generalized r-Lah numbers, arXiv:1412.8721 [math.CO], 2014.

Cf. A001705 - A001709 (column 3..7), A001710 (row sums), A008275, A049444 (signed version), A136124, A143492, A143493, A143494, A143497.

Cf. A094638.

with combinat: T := (n, k) -> (n-2)! * add((n-j-1)*abs(stirling1(j,k-2))/j!,j = k-2..n-2): for n from 2 to 10 do seq(T(n, k), k = 2..n) end do;

t[n_, k_] := (n-2)!*Sum[(n-j-1)*Abs[StirlingS1[j, k-2]]/j!, {j, k-2, n-2}]; Table[t[n, k], {n, 2, 11}, {k, 2, n}] // Flatten (* Jean-François Alcover, Apr 16 2013, after Maple *)

T(n,k) = (n-2)! * Sum_{j = k-2 .. n-2} (n-j-1)*|stirling1(j,k-2)|/j!.
Recurrence relation: T(n,k) = T(n-1,k-1) + (n-1)*T(n-1,k) for n > 2, with boundary conditions: T(n,1) = T(1,n) = 0, for all n; T(2,2) = 1; T(2,k) = 0 for k > 2.
Special cases: T(n,2) = (n-1)!; T(n,3) = (n-1)!*(1/2 + 1/3 + ... + 1/(n-1)).
T(n,k) = Sum_{2 <= i_1 < ... < i_(n-k) < n} (i_1*i_2*...*i_(n-k)). For example, T(6,4) = Sum_{2 <= i < j < 6} (i*j) = 2*3 + 2*4 + 2*5 + 3*4 + 3*5 + 4*5 = 71.
Row g.f.: Sum_{k = 2..n} T(n,k)*x^k = x^2*(x+2)*(x+3)*...*(x+n-1).
E.g.f. for column (k+2): Sum_{n>=k} T(n+2,k+2)*x^n/n! = (1/k!)*(1/(1-x)^2)*(log(1/(1-x)))^k.
E.g.f.: (1/(1-t))^(x+2) = Sum_{n>=0} Sum_{k = 0..n} T(n+2,k+2)*x^k*t^n/n! = 1 + (2+x)*t/1! + (6+5*x+x^2)*t^2/2! + ... .
This array is the matrix product St1 * P, where St1 denotes the lower triangular array of unsigned Stirling numbers of the first kind, abs(A008275) and P denotes Pascal's triangle, A007318. The row sums are n!/2 ( A001710 ). The alternating row sums are (n-2)!.
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a - j), then T(n-1,i) = |f(n,i,2)|, for n=1,2,...; i=0..n. - Milan Janjic, Dec 21 2008
From Gary W. Adamson, Jul 19 2011: (Start)
n-th row of the triangle = top row of M^(n-2), M = a reversed variant of the (1,2) Pascal triangle (Cf. A029635); as follows:
  2, 1, 0, 0, 0, 0, ...
  2, 3, 1, 0, 0, 0, ...
  2, 5, 4, 1, 0, 0, ...
  2, 7, 9, 5, 1, 0, ...
  ... (End)
The reversed, row polynomials of this entry multiplied by (1+x) are the row polynomials of A094638. E.g., (1+x)(1+5x+6x^2) = (1+6x+11x^2+6x^3). - Tom Copeland, Dec 11 2016

A096956 Pascal (1,6) triangle.

Original entry on oeis.org

6, 1, 6, 1, 7, 6, 1, 8, 13, 6, 1, 9, 21, 19, 6, 1, 10, 30, 40, 25, 6, 1, 11, 40, 70, 65, 31, 6, 1, 12, 51, 110, 135, 96, 37, 6, 1, 13, 63, 161, 245, 231, 133, 43, 6, 1, 14, 76, 224, 406, 476, 364, 176, 49, 6, 1, 15, 90, 300, 630, 882, 840, 540, 225, 55, 6, 1, 16, 105, 390, 930

Offset: 0

Wolfdieter Lang, Aug 13 2004

Except for the first row this is the row reversed (6,1)-Pascal triangle A093563.
This is the sixth member, q=6, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A095666 (q=4), A096940 (q=5).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(6-5*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(6-5*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k) = A022097(n-2), n >= 2, with n=1 value 6. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins:
  [0]  6;
  [1]  1,  6;
  [2]  1,  7,  6;
  [3]  1,  8, 13,  6;
  [4]  1,  9, 21, 19,  6;
  [5]  1, 10, 30, 40, 25,  6;
  ...

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).
Wolfdieter Lang, First 10 rows.

Row sums: A005009(n-1), n>=1, 6 if n=0; g.f.: (6-5*x)/(1-2*x). Alternating row sums are [6, -5, followed by 0's].

Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+6), A056115, A096957-9, A097297-A097300.

a(n,k):=piecewise(n=0,6,0Mircea Merca, Apr 08 2012

A096956[n_, k_] := If[n == k, 6, (5*k/n + 1)*Binomial[n, k]];
Table[A096956[n, k], {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Apr 14 2025 *)

Recursion: a(n,m)=0 if m > n, a(0,0) = 6; a(n,0) = 1 if n >= 1; a(n,m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (6-5*x)/(1-x)^(m+1), m >= 0.
a(n,k) = (1+5*k/n)*binomial(n,k), for n > 0. - Mircea Merca, Apr 08 2012

A114496 a(n) = Sum of binomial(n,k)*binomial(2n+k,k) over all k.

Original entry on oeis.org

1, 4, 26, 190, 1462, 11584, 93536, 765314, 6323270, 52638760, 440815036, 3709445084, 31340292076, 265683004240, 2258793820988, 19251776923210, 164440378882630, 1407266585304760, 12063701803046300, 103571977632247076

Offset: 0

Eric Rowland, Dec 01 2005

Modification of A001850 inspired by the Apéry numbers A005259.
From Paul Barry, Feb 17 2009: (Start)
Central coefficient of (1 + 4x + 5x^2 + 2x^3)^n. The coefficients are the 4th row of A029635.
The third row of A029635 corresponds to the central Delannoy numbers A001850. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
P. Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), #13.5.4.

Cf. A114497, A114498, A003168, A156894.

Cf. A156886. - Paul Barry, Feb 17 2009

Table[Sum[Binomial[n, k]*Binomial[2n+k, k], {k, 0, n}], {n,0,25}]

a(n) = sum(k=0, n, 2^(n-k)*binomial(2*n,k)*binomial(n,k));
vector(50, n, a(n-1)) \\ Altug Alkan, Oct 05 2015

a(n) = Sum_{k=0..n} (binomial(n,k)*binomial(2n+k,k)).
Recurrence: 20*n*(2*n - 1)*a(n) = (371*n^2 - 411*n + 120)*a(n-1) -2*(81*n^2 - 299*n + 278)*a(n-2) + 4*(n-2)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ sqrt(1734 + 442*sqrt(17))*((71 + 17*sqrt(17))/16)^n/(68*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 19 2012
From Peter Bala, Oct 05 2015: (Start)
a(n) = Sum_{i = 0..n} 2^(n-i)*binomial(2*n,i)*binomial(n,i).
4*n*(2*n - 1)*(17*n - 23)*a(n) = (1207*n^3 - 2840*n^2 + 1897*n - 360)*a(n-1) - 2*(n - 1)*(17*n - 6)*(2*n - 3)*a(n-2) with a(0) = 1 and a(1) = 4.
1 + x*exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 4*x^2 + 21*x^3 + 126*x^4 + ... is the o.g.f. for A003168. (End)

A112857 Triangle T(n,k) read by rows: number of Green's R-classes in the semigroup of order-preserving partial transformations (of an n-element chain) consisting of elements of height k (height(alpha) = |Im(alpha)|).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 15, 17, 7, 1, 1, 31, 49, 31, 9, 1, 1, 63, 129, 111, 49, 11, 1, 1, 127, 321, 351, 209, 71, 13, 1, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1

Offset: 0

Abdullahi Umar, Aug 25 2008

Sum of rows of T(n, k) is A007051; T(n,k) = |A118801(n,k)|.
Row-reversed variant of A119258. - R. J. Mathar, Jun 20 2011
Pairwise sums of row terms starting from the right yields triangle A038207. - Gary W. Adamson, Feb 06 2012
Riordan array (1/(1 - x), x/(1 - 2*x)). - Philippe Deléham, Jan 17 2014
Appears to coincide with the triangle T(n,m) (n >= 1, 1 <= m <= n) giving number of set partitions of [n], avoiding 1232, with m blocks [Crane, 2015]. See also A250118, A250119. - N. J. A. Sloane, Nov 25 2014
(A007318)^2 = A038207 = T*|A167374|. See A118801 for other relations to the Pascal matrix. - Tom Copeland, Nov 17 2016

			T(3,2) = 5 because in a regular semigroup of transformations the Green's R-classes coincide with convex partitions of subsets of {1,2,3} with convex classes (modulo the subsets): {1}, {2}/{1}, {3}/{2}, {3}/{1,2}, {3}/{1}, {2,3}
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,    1;
  1,    3,    1;
  1,    7,    5,    1;
  1,   15,   17,    7,    1;
  1,   31,   49,   31,    9,    1;
  1,   63,  129,  111,   49,   11,    1;
  1,  127,  321,  351,  209,   71,   13,   1;
  1,  255,  769, 1023,  769,  351,   97,  15,   1;
  1,  511, 1793, 2815, 2561, 1471,  545, 127,  17,  1;
  1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1;
  ...
As to matrix M, top row of M^3 = (1, 7, 5, 1, 0, 0, 0, ...)

Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
Harry Crane, Left-right arrangements, set partitions, and pattern avoidance, Australasian Journal of Combinatorics, 61(1) (2015), 57-72.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), 04.3.8.

Cf. A007051, A118801, A135233, columns: A000225, A000337, A055580, A027608.
Cf. also A038207, A250118, A250119.
Cf. A007318, A167374, A145661, A029635.

A112857 := proc(n,k) if k=0 or k=n then 1; elif k <0 or k>n then 0; else 2*procname(n-1,k)+procname(n-1,k-1) ; end if; end proc: # R. J. Mathar, Jun 20 2011

Table[Abs[1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]] - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

T(n,k) = Sum_{j = k..n} C(n,j)*C(j-1,k-1).
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) for n >= 2 and 1 <= k <= n-1 with T(n,0) = 1 = T(n,n) for n >= 0.
n-th row = top row of M^n, deleting the zeros, where M is an infinite square production matrix with (1,1,1,...) as the superdiagonal and (1,2,2,2,...) as the main diagonal. - Gary W. Adamson, Feb 06 2012
From Peter Bala, Mar 05 2018 (Start):
The following remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,2*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(2*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,2*x). For example, when n = 3 we have exp(x)*(1 + 7*(2*x) + 5*(2*x)^2/2! + (2*x)^3/3!) = 1 + 15*x + 49*x^2/2! + 111*x^3/3! + 209*x^4/4! + ....
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 P(n,x) is the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = x^4 + 15*x^3 + 17*x^2 + 7*x + 1 + O(x^5).
See A118801 for a signed version of this triangle and A145661 for a signed version of the row reversed triangle. (End)
Bivariate o.g.f.: Sum_{n,k>=0} T(n,k)*x^n*y^k = (1 - 2*x)/((1 - x)*(1 - 2*x - x*y)). - Petros Hadjicostas, Feb 14 2021
The matrix inverse of the Lucas triangle A029635 is -T(n, k)/(-2)^(n-k+1). - Peter Luschny, Dec 22 2024

cp's OEIS Frontend

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

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A029600 Numbers in the (2,3)-Pascal triangle (by row).

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A161198 Triangle of polynomial coefficients related to the series expansions of (1-x)^((-1-2*n)/2).

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A095660 Pascal (1,3) triangle.

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A095666 Pascal (1,4) triangle.

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A096940 Pascal (1,5) triangle.

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