A193722 -id:A193722 - cp's OEIS frontend

A193842 Triangular array: the fission of the polynomial sequence ((x+1)^n: n >= 0) by the polynomial sequence ((x+2)^n: n >= 0). (Fission is defined at Comments.)

1, 1, 4, 1, 7, 13, 1, 10, 34, 40, 1, 13, 64, 142, 121, 1, 16, 103, 334, 547, 364, 1, 19, 151, 643, 1549, 2005, 1093, 1, 22, 208, 1096, 3478, 6652, 7108, 3280, 1, 25, 274, 1720, 6766, 17086, 27064, 24604, 9841, 1, 28, 349, 2542, 11926, 37384, 78322, 105796

Offset: 0

Clark Kimberling, Aug 07 2011

Suppose that p = p(n)*x^n + p(n-1)*x^(n-1) + ... + p(1)*x + p(0) is a polynomial and that Q is a sequence of polynomials:
...
q(k,x) = t(k,0)*x^k + t(k,1)*x^(k-1) + ... + t(k,k-1)*x + t(k,k),
...
for k = 0, 1, 2, ...  The Q-downstep of p is the polynomial given by
...
D(p) = p(n)*q(n-1,x) + p(n-1)*q(n-2,x) + ... + p(1)*q(0,x). (Note that p(0) does not appear. "Q-downstep" as just defined differs slightly from "Q-downstep" as defined for a different purpose at A193649.)
...
Now suppose that P = (p(n,x): n >= 0) and Q = (q(n,x): n >= 0) are sequences of polynomials, where n indicates degree.  The fission of P by Q, denoted by P^^Q, is introduced here as the sequence W = (w(n,x): n >= 0) of polynomials defined by w(0,x) = 1 and w(n,x) = D(p(n+1,x)).
...
Strictly speaking, ^^ is an operation on sequences of polynomials.  However, if P and Q are regarded as numerical triangles (of coefficients of polynomials), then ^^ can be regarded as an operation on numerical triangles.  In this case, row n of P^^Q, for n > 0, is given by the matrix product P(n+1)*QQ(n), where P(n+1) =(p(n+1,n+1), p(n+1,n), ..., p(n+1,2), p(n+1,1)) and QQ(n) is the (n+1)-by-(n+1) matrix given by
...
q(n,0) .. q(n,1)............. q(n,n-1) .... q(n,n)
0 ....... q(n-1,0)........... q(n-1,n-2)... q(n-1,n-1)
0 ....... 0.................. q(n-2,n-3) .. q(n-2,n-2)
...
0 ....... 0.................. q(1,0) ...... q(1,1)
0 ....... 0 ................. 0 ........... q(0,0).
Here, the polynomial q(k,x) is taken to be
q(k,0)*x^k + q(k,1)x^(k-1) + ... + q(k,k)*x + q(k,k);
i.e., "q" is used instead of "t".
...
Example:  Let p(n,x) = (x+1)^n and q(n,x) = (x+2)^n.  Then
...
w(0,x) = 1 by the definition of W,
w(1,x) = D(p(2,x)) = 1*(x+2) + 2*1 = x + 4,
w(2,x) = D(p(3,x)) = 1*(x^2+4*x+4) + 3*(x+2) + 3*1 = x^2 + 7*x + 13,
w(3,x) = D(p(4,x)) = 1*(x^3+6*x^2+12*x+8) + 4*(x^2+4x+4) + 6*(x+2) + 4*1 = x^3 + 10*x^2 + 34*x + 40.
...
From these first 4 polynomials in the sequence P^^Q, we can write the first 4 rows of P^^Q when P, Q, and P^^Q are regarded as triangles:
1
1...4
1...7....13
1...10...34...40
...
In the following examples, r(P^^Q) is the mirror of P^^Q, obtained by reversing the rows of P^^Q.  Let u denote the polynomial x^n + x^(n-1) + ... + x + 1.
...
..P........Q...........P^^Q........r(P^^Q)
(x+1)^n....(x+2)^n.....A193842.....A193843
(x+1)^n....(x+1)^n.....A193844.....A193845
(x+2)^n....(x+1)^n.....A193846.....A193847
(2x+1)^n...(x+1)^n.....A193856.....A193857
(x+1)^n....(2x+1)^n....A193858.....A193859
(x+1)^n.......u........A054143.....A104709
..u........(x+1)^n.....A074909.....A074909
..u...........u........A002260.....A004736
(x+2)^n.......u........A193850.....A193851
..u.........(x+2)^n....A193844.....A193845
(2x+1)^n......u........A193860.....A193861
..u.........(2x+1)^n...A115068.....A193862
...
Regarding A193842,
col 1 ...... A000012
col 2 ...... A016777
col 3 ...... A081271
w(n,n) ..... A003462
w(n,n-1) ... A014915

			First six rows, for 0 <= k <= n and 0 <= n <= 5:
  1
  1...4
  1...7....13
  1...10...34....40
  1...13...64....142...121
  1...16...103...334...547...364

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Digital Library of Mathematical Functions, Hypergeometric function, analytic properties.
Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52(3) (2014), 195-202.

Cf. A193722 (fusion of P by Q), A193649 (Q-residue), A193843 (mirror of A193842).

[ (&+[3^(k-j)*Binomial(n-j,k-j): j in [0..k]]): k in [0..n], n in [0..10]]; // G. C. Greubel, Feb 18 2020

fission := proc(p, q, n) local d, k;
p(n+1,0)*q(n,x)+add(coeff(p(n+1,x),x^k)*q(n-k,x), k=1..n);
seq(coeff(%,x,n-k), k=0..n) end:
A193842_row := n -> fission((n,x) -> (x+1)^n, (n,x) -> (x+2)^n, n);
for n from 0 to 5 do A193842_row(n) od; # Peter Luschny, Jul 23 2014
# Alternatively:
p := (n,x) -> add(x^k*(1+3*x)^(n-k),k=0..n): for n from 0 to 7 do [n], PolynomialTools:-CoefficientList(p(n,x), x) od; # Peter Luschny, Jun 18 2017

(* First program *)
z = 10;
p[n_, x_] := (x + 1)^n;
q[n_, x_] := (x + 2)^n
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A193842 *)
TableForm[Table[h[n], {n, 0, z}]]  (* A193843 *)
Flatten[Table[h[n], {n, -1, z}]]
(* Second program *)
Table[SeriesCoefficient[((x+3)^(n+1) -1)/(x+2), {x,0,n-k}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 18 2020 *)

T(n,k) = sum(j=0,k, 3^(k-j)*binomial(n-j,k-j)); \\ G. C. Greubel, Feb 18 2020

from mpmath import mp, hyp2f1
mp.dps = 100; mp.pretty = True
def T(n,k):
    return 3^k*binomial(n,k)*hyp2f1(1,-k,-n,1/3)-0^(n-k)//2
for n in range(7):
    print([int(T(n,k)) for k in (0..n)]) # Peter Luschny, Jul 23 2014

# Second program using the 'fission' operation.
def fission(p, q, n):
    F = p(n+1,0)*q(n,x)+add(expand(p(n+1,x)).coefficient(x,k)*q(n-k,x) for k in (1..n))
    return [expand(F).coefficient(x,n-k) for k in (0..n)]
A193842_row = lambda k: fission(lambda n,x: (x+1)^n, lambda n,x: (x+2)^n, k)
for n in range(7): A193842_row(n) # Peter Luschny, Jul 23 2014

From Peter Bala, Jul 16 2013: (Start)
T(n,k) = Sum_{i = 0..k} 3^(k-i)*binomial(n-i,k-i).
O.g.f.: 1/((1 - x*t)*(1 - (1 + 3*x)*t)) = 1 + (1 + 4*x)*t + (1 + 7*x + 13*x^2)*t^2 + ....
The n-th row polynomial is R(n,x) = (1/(2*x + 1))*((3*x + 1)^(n+1) - x^(n+1)). (End)
T(n,k) = T(n-1,k) + 4*T(n-1,k-1) - T(n-2,k-1) - 3*T(n-2,k-2), T(0,0) = 1, T(1,0) = 1, T(1,1) = 4, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 17 2014
T(n,k) = 3^k * C(n,k) * hyp2F1(1, -k, -n, 1/3) with or without the additional term -0^(n-k)/2 depending on the exact definition of the hypergeometric function used. Compare formulas 15.2.5 and 15.2.6 in the DLMF reference. - Peter Luschny, Jul 23 2014

Name and Comments edited by Petros Hadjicostas, Jun 05 2020

A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

Original entry on oeis.org

1, 1, 3, 5, 15, 33, 91, 221, 583, 1465, 3795, 9653, 24831, 63441, 162763, 416525, 1067575, 2733673, 7003971, 17938661, 45954543, 117709185, 301527355, 772364093, 1978473511

Offset: 0

Clark Kimberling, Aug 02 2011

nonn

Suppose that p=p(0)*x^n+p(1)*x^(n-1)+...+p(n-1)*x+p(n) is a polynomial of positive degree and that Q is a sequence of polynomials: q(k,x)=t(k,0)*x^k+t(k,1)*x^(k-1)+...+t(k,k-1)*x+t(k,k), for k=0,1,2,... The Q-downstep of p is the polynomial given by D(p)=p(0)*q(n-1,x)+p(1)*q(n-2,x)+...+p(n-1)*q(0,x)+p(n).

Since degree(D(p))

Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.

D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14

D(D(p))=2(x+1)+7(1)+14=2x+23

D(D(D(p)))=2(1)+23=25;

the Q-residue of p is 25.

We may regard the sequence Q of polynomials as the triangular array formed by coefficients:

t(0,0)

t(1,0)....t(1,1)

t(2,0)....t(2,1)....t(2,2)

t(3,0)....t(3,1)....t(3,2)....t(3,3)

and regard p as the vector (p(0),p(1),...,p(n)). If P is a sequence of polynomials [or triangular array having (row n)=(p(0),p(1),...,p(n))], then the Q-residues of the polynomials form a numerical sequence.

Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:

Q.....P...................Q-residue of P

1.....1...................A000079, 2^n

1....(x+1)^n..............A007051, (1+3^n)/2

1....(x+2)^n..............A034478, (1+5^n)/2

1....(x+3)^n..............A034494, (1+7^n)/2

1....(2x+1)^n.............A007582

1....(3x+1)^n.............A081186

1....(2x+3)^n.............A081342

1....(3x+2)^n.............A081336

1.....A040310.............A193649

1....(x+1)^n+(x-1)^n)/2...A122983

1....(x+2)(x+1)^(n-1).....A057198

1....(1,2,3,4,...,n)......A002064

1....(1,1,2,3,4,...,n)....A048495

1....(n,n+1,...,2n).......A087323

1....(n+1,n+2,...,2n+1)...A099035

1....p(n,k)=(2^(n-k))*3^k.A085350

1....p(n,k)=(3^(n-k))*2^k.A090040

1....A008288 (Delannoy)...A193653

1....A054142..............A101265

1....cyclotomic...........A193650

1....(x+1)(x+2)...(x+n)...A193651

1....A114525..............A193662

More examples:

Q...........P.............Q-residue of P

(x+1)^n...(x+1)^n.........A000110, Bell numbers

(x+1)^n...(x+2)^n.........A126390

(x+2)^n...(x+1)^n.........A028361

(x+2)^n...(x+2)^n.........A126443

(x+1)^n.....1.............A005001

(x+2)^n.....1.............A193660

A094727.....1.............A193657

(k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)

(k+1).....(x+1)^n.........A112091

(x+1)^n...(k+1)...........A029761

(k+1)......A049310........A193663

(In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)

A051162...(x+1)^n.........A193658

A094727...(x+1)^n.........A193659

A049310...(x+1)^n.........A193664

A075362....A075362........A193665

Changing the notation slightly leads to the Mathematica program below and the following formulation for the Q-downstep of p: first, write t(n,k) as q(n,k). Define r(k)=Sum{q(k-1,i)*r(k-1-i) : i=0,1,...,k-1} Then row n of D(p) is given by v(n)=Sum{p(n,k)*r(n-k) : k=0,1,...,n}.

			First five rows of Q, coefficients of Fibonacci polynomials (A049310):
1
1...0
1...0...1
1...0...2...0
1...0...3...0...1
To obtain a(4)=15, downstep four times:
D(x^4+3*x^2+1)=(x^3+x^2+x+1)+3(x+1)+1: (1,1,4,5) [coefficients]
DD(x^4+3*x^2+1)=D(1,1,4,5)=(1,2,11)
DDD(x^4+3*x^2+1)=D(1,2,11)=(1,14)
DDDD(x^4+3*x^2+1)=D(1,14)=15.

Cf. A192872 (polynomial reduction), A193091 (polynomial augmentation), A193722 (the upstep operation and fusion of polynomial sequences or triangular arrays).

q[n_, k_] := 1;
r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}];
f[n_, x_] := Fibonacci[n + 1, x];
p[n_, k_] := Coefficient[f[n, x], x, k]; (* A049310 *)
v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
Table[v[n], {n, 0, 24}]    (* A193649 *)
TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
Table[r[k], {k, 0, 8}]  (* 2^k *)
TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]

Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - R. J. Mathar, Feb 19 2015

A118800 Triangle read by rows: T satisfies the matrix products: CTC = T^-1 and TCT = C^-1, where C is Pascal's triangle.

Original entry on oeis.org

1, 1, -1, 2, -3, 1, 4, -8, 5, -1, 8, -20, 18, -7, 1, 16, -48, 56, -32, 9, -1, 32, -112, 160, -120, 50, -11, 1, 64, -256, 432, -400, 220, -72, 13, -1, 128, -576, 1120, -1232, 840, -364, 98, -15, 1, 256, -1280, 2816, -3584, 2912, -1568, 560, -128, 17, -1, 512, -2816, 6912, -9984, 9408, -6048, 2688, -816, 162, -19, 1

Offset: 0

Paul D. Hanna, May 02 2006

The matrix square, T^2, consists of columns that are all the same.
Matrix inverse is triangle A118801. Row sums form {0^n, n>=0}.
Unsigned row sums equal A025192(n) = 2*3^(n-1), n>=1.
Row squared sums equal A051708.
Antidiagonal sums equals all 1's.
Unsigned antidiagonal sums form A078057 (with offset).
Antidiagonal squared sums form A002002(n) = Sum_{k=0..n-1} C(n,k+1)*C(n+k,k), n>=1.
From Paul Barry, Nov 10 2008: (Start)
T is [1,1,0,0,0,...] DELTA [ -1,0,0,0,0,...] or C(1,n) DELTA -C(0,n). (DELTA defined in A084938).
The positive matrix T_p is [1,1,0,0,0,...] DELTA [1,0,0,0,0,...]. T_p*C^-1 is
[0,1,0,0,0,....] DELTA [1,0,0,0,0,...] which is C(n-1,k-1) for n,k>=1. (End)
The triangle formed by deleting the minus signs is the mirror of the self-fusion of Pascal's triangle; see Comments at A081277 and A193722. - Clark Kimberling, Aug 04 2011
Riordan array ( (1 - x)/(1 - 2*x), -x/(1 - 2*x) ). Cf. A209149. The matrix square is the Riordan array ( (1 - x)^2/(1 - 2*x), x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013
From Peter Bala, Feb 23 2019: (Start)
There is a 1-parameter family of solutions to the simultaneous equations C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle. Let T(k) denote the Riordan array ( (1 - k*x)/(1 - (k + 1)*x), -x/(1 - (k + 1)*x) ) so that T(1) = T. Then C*T(k)*C = T(k)^-1 and T(k)*C*T(k) = C^-1, for arbitrary k. For arbitrary m, the Riordan arrays (T(k)*C^m)^2 and (C^m*T(k))^2 both belong to the Appell subgroup of the Riordan group.
More generally, given a fixed m, we can ask for a lower triangular array X solving the simultaneous equations (C^m)*X*(C^m) = X^-1 and X*(C^m)*X = C^(-m). A 1-parameter family of solutions is given by the Riordan arrays X = ( (1 - m*k*x)/(1 - m*(k + 1)*x), -x/(1 - m*(k + 1)*x) ). The Riordan arrays X^2 , (X*C^n)^2 and (C^n*X)^2, for arbitrary n, all belong to the Appell subgroup of the Riordan group. (End)

			Triangle begins:
     1;
     1,    -1;
     2,    -3,     1;
     4,    -8,     5,     -1;
     8,   -20,    18,     -7,     1;
    16,   -48,    56,    -32,     9,     -1;
    32,  -112,   160,   -120,    50,    -11,     1;
    64,  -256,   432,   -400,   220,    -72,    13,    -1;
   128,  -576,  1120,  -1232,   840,   -364,    98,   -15,    1;
   256, -1280,  2816,  -3584,  2912,  -1568,   560,  -128,   17,   -1;
   512, -2816,  6912,  -9984,  9408,  -6048,  2688,  -816,  162,  -19,  1;
  1024, -6144, 16640, -26880, 28800, -21504, 11424, -4320, 1140, -200, 21, -1;
  ...
The matrix square, T^2, equals:
   1;
   0,  1;
   1,  0,  1;
   2,  1,  0,  1;
   4,  2,  1,  0,  1;
   8,  4,  2,  1,  0,  1;
  16,  8,  4,  2,  1,  0,  1;
  32, 16,  8,  4,  2,  1,  0,  1;
  64, 32, 16,  8,  4,  2,  1,  0,  1; ...
where all columns are the same.

Paul D. Hanna, Rows 0..45 of triangle, flattened.
Florian Luca, Attila Pethő, and László Szalay, Duplications in the k-generalized Fibonacci sequences, New York J. Math. (2021) Vol. 27, 1115-1133.

Cf. A118801 (inverse), A025192 (unsigned row sums), A051708 (row squared sums), A078057 (unsigned antidiagonal sums), A002002 (antidiagonal squared sums).

Cf. A135387, A209149, A000012, A097805, A130595, A167374, A200139, A239473, A321620.

(* This program generates A118800 as the mirror of the self-fusion of Pascal's triangle. *)
z = 8; a = 1; b = 1; c = 1; d = 1;
p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n;
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, 0];
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, _] = 1;
g[n_] := CoefficientList[w[n, -x], x];
TableForm[Table[Reverse[Abs@g[n]], {n, -1, z}]]
Flatten[Table[Reverse[Abs@g[n]], {n, -1, z}]] (* A081277 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]] (* A118800 *)
(* Clark Kimberling, Aug 04 2011 *)
T[ n_, k_] := If[ n<0 || k<0, 0, (-1)^k 2^(n-k) (Binomial[ n, k] + Binomial[ n-1, n-k]) / 2]; (* Michael Somos, Nov 25 2016 *)

{T(n,k)=if(n==0&k==0,1,(-1)^k*2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2)}
for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

/* Chebyshev Polynomials as Antidiagonals: */
{T(n,k)=local(Ox=x*O(x^(2*k))); polcoeff(((1+sqrt(1-x^2+Ox))^(n+k)+(1-sqrt(1-x^2+Ox))^(n+k))/2,2*k,x)}
for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

# uses[riordan_square from A321620]
# Computes the unsigned triangle.
riordan_square((1-x)/(1-2*x), 8) # Peter Luschny, Jan 03 2019

T(n,k) = (-1)^k * 2^(n-k) * ( C(n,k) + C(n-1,k-1) )/2 for n>=k>=0 with T(0,0) = 1. Antidiagonals form the coefficients of Chebyshev polynomials: T(n,k) = [x^(2*n)] [(1+sqrt(1-x^2))^(n+k) + (1-sqrt(1-x^2))^(n+k)]/2.
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 09 2007
O.g.f.: (1 - t)/(1 + t*(x - 2)) = 1 + (1 - x)*t + (2 - 3*x + x^2)^t^2 + (4 - 8*x + 5*x^2 - x^3)*t^3 + ....  Row polynomial R(n,x) = (1 - x)*(2 - x)^(n-1) for n >= 1. - Peter Bala, Jul 17 2013
T(n,k)=2*T(n-1,k)-T(n-1,k-1) with T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 25 2013
G.f. for row n (n>=1): Sum_{k=0..n} T(n,k)*x^k = (1-x)*(2-x)^(n-1). - Philippe Deléham, Nov 25 2013
From Tom Copeland, Nov 15 2016: (Start)
E.g.f. is [1 + (1-x)e^((2-x)t)]/(2-x), so the row polynomials are p_n(x) = (1-q,(x))^n, umbrally, where (q.(x))^k = q_k(x) are the row polynomials of A239473, or, equivalently, T = M*A239473, where M is the inverse Pascal matrix C^(-1) = A130595 with the odd rows negated, i.e., M(n,k) = (-1)^n C^(-1)(n,k) with e.g.f. exp[(1-x)t]. Cf. A200139: A200139(n,k) = (-1)^k* A118800(n,k).
TCT = C^(-1) = A130595 and A239473 = A000012*C^(-1) = S*C^(-1) imply (M*S)^2 = Identity matrix, i.e., M*S = (M*S)^(-1) = S^(-1)*M^(-1) = A167374*M^(-1). Note that M = M^(-1). Cf. A097805. (End)

A038763 Triangular matrix arising in enumeration of catafusenes, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 3, 1, 7, 15, 9, 1, 10, 36, 54, 27, 1, 13, 66, 162, 189, 81, 1, 16, 105, 360, 675, 648, 243, 1, 19, 153, 675, 1755, 2673, 2187, 729, 1, 22, 210, 1134, 3780, 7938, 10206, 7290, 2187, 1, 25, 276, 1764, 7182, 19278, 34020, 37908, 24057, 6561, 1, 28, 351, 2592, 12474, 40824, 91854, 139968, 137781, 78732, 19683

Offset: 0

N. J. A. Sloane, May 03 2000

Triangle T(n,k), 0<=k<=n, read by rows, given by [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 10 2005
Triangle read by rows, n-th row = X^(n-1) * [1, 1, 0, 0, 0, ...] where X = an infinite bidiagonal matrix with (1,1,1,...) in the main diagonal and (3,3,3,...) in the subdiagonal; given row 0 = 1. - Gary W. Adamson, Jul 19 2008
Fusion of polynomial sequences P and Q given by p(n,x)=(x+2)^n and q(n,x)=(2x+1)^n; see A193722 for the definition of fusion of two sequences of polynomials or triangular arrays. - Clark Kimberling, Aug 04 2011

			Triangle begins:
  1;
  1,  1;
  1,  4,   3;
  1,  7,  15,   9;
  1, 10,  36,  54,   27;
  1, 13,  66, 162,  189,   81;
  1, 16, 105, 360,  675,  648,  243;
  1, 19, 153, 675, 1755, 2673, 2187, 729;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
S. J. Cyvin, B. N. Cyvin, and J. Brunvoll, Unbranched catacondensed polygonal systems containing hexagons and tetragons, Croatica Chem. Acta, 69 (1996), 757-774.

Cf. A000007, A000012, A001296, A006138, A006234, A016777, A024462.
Cf. A051836, A062741, A080420, A080421, A080422, A080423, A081294.
Cf. A084938, A098399, A110523, A133494, A136158, A193722.

A038763:= func< n,k | n eq 0 select 1 else 3^(k-1)*(3*n-2*k)*Binomial(n,k)/n >;
[A038763(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 27 2023

A038763[n_,k_]:= If[n==0, 1, 3^(k-1)*(3*n-2*k)*Binomial[n,k]/n];
Table[A038763[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 27 2023 *)

T(n,k) = if ((n<0) || (k<0), return(0)); if ((n==0) && (k==0), return(1)); if (n==1, if (k<=1, return(1))); T(n-1,k) + 3*T(n-1,k-1);
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", "))); \\ Michel Marcus, Jul 25 2023

def A038763(n,k): return 1 if (n==0) else 3^(k-1)*(3*n-2*k)*binomial(n,k)/n
flatten([[A038763(n, k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Dec 27 2023

T(n, 0)=1; T(1, 1)=1; T(n, k)=0 for k>n; T(n, k) = T(n-1, k-1)*3 + T(n-1, k) for n >= 2.
Sum_{k=0..n} T(n,k) = A081294(n). - Philippe Deléham, Sep 22 2006
T(n, k) = A136158(n, n-k). - Philippe Deléham, Dec 17 2007
G.f.: (1-2*x*y)/(1-(3*y+1)*x). - R. J. Mathar, Aug 11 2015
From G. C. Greubel, Dec 27 2023: (Start)
T(n, 0) = A000012(n).
T(n, 1) = A016777(n-1).
T(n, 2) = A062741(n-1).
T(n, 3) = 9*A002411(n-2).
T(n, 4) = 27*A001296(n-3).
T(n, 5) = 81*A051836(n-4).
T(n, n) = A133494(n).
T(n, n-1) = A006234(n+2).
T(n, n-2) = A080420(n-2).
T(n, n-3) = A080421(n-3).
T(n, n-4) = A080422(n-4).
T(n, n-5) = A080423(n-5).
T(2*n, n) = 4*A098399(n-1) + (2/3)*[n=0].
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A006138(n-1) + (2/3)*[n=0].
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = A110523(n-1) + (4/3)*[n=0]. (End)

More terms from Michel Marcus, Jul 25 2023

A193815 Triangular array: the fusion of polynomial sequences P and Q given by p(n,x) = x^n + x^(n-1) + ... + x+1 and q(n,x)=(x+1)^n.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 4, 6, 3, 1, 5, 10, 10, 4, 1, 6, 15, 20, 15, 5, 1, 7, 21, 35, 35, 21, 6, 1, 8, 28, 56, 70, 56, 28, 7, 1, 9, 36, 84, 126, 126, 84, 36, 8, 1, 10, 45, 120, 210, 252, 210, 120, 45, 9, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 10, 1, 12, 66, 220, 495

Offset: 0

Clark Kimberling, Aug 06 2011

See A193722 for the definition of fusion of two sequences of polynomials or triangular arrays.
Triangle T(n,k), read by rows, given by (1,0,-1,1,0,0,0,0,0,0,0,...) DELTA (1,1,-1,1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 08 2011
Row sums are A095121. - Philippe Deléham, Nov 24 2011

			First six rows:
  1;
  1,  1;
  1,  3,  2;
  1,  4,  6,  3;
  1,  5, 10, 10,  4;
  1,  6, 15, 20, 15,  5;

Branko Malesevic, Yue Hu, Cristinel Mortici, Accurate Estimates of (1+x)^{1/x} Involved in Carleman Inequality and Keller Limit, arXiv:1801.04963 [math.CA], 2018.

Cf. A193722, A153861, A193818.

z = 10; c = 1; d = 1;
p[0, x_] := 1
p[n_, x_] := x*p[n - 1, x] + 1; p[n_, 0] := p[n, x] /. x -> 0;
q[n_, x_] := (c*x + d)^n
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n, -1, z}]]  (* A193815 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]]   (* A153861 *)
t[0, 0] = t[1, 0] = t[1, 1] = t[2, 0] = 1; t[2, 1] = 3; t[2, 2] = 2; t[n_, k_] /; k<0 || k>n = 0; t[n_, k_] := t[n, k] = t[n-1, k]+2*t[n-1, k-1]-t[n-2, k-1]-t[n-2, k-2]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 16 2013, after Philippe Deléham *)

T(n,k) = A153861(n,n-k). - Philippe Deléham, Oct 08 2011
G.f.: (1-y*x+y*(y+1)*x^2)/((1-y*x)*(1-(y+1)*x)). - Philippe Deléham, Nov 24 2011
Sum_{k=0..n} T(n,k)*x^k = (x+1)*((x+1)^n - x^n) + 0^n. - Philippe Deléham, Nov 24 2011
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(0,0)=T(1,0)=T(1,1)=T(2,0)=1, T(2,1)=3, T(2,2)=2, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Dec 15 2013

A115262 Correlation triangle for n+1.

Original entry on oeis.org

1, 2, 2, 3, 5, 3, 4, 8, 8, 4, 5, 11, 14, 11, 5, 6, 14, 20, 20, 14, 6, 7, 17, 26, 30, 26, 17, 7, 8, 20, 32, 40, 40, 32, 20, 8, 9, 23, 38, 50, 55, 50, 38, 23, 9, 10, 26, 44, 60, 70, 70, 60, 44, 26, 10, 11, 29, 50, 70, 85, 91, 85, 70, 50, 29, 11

Offset: 0

Paul Barry, Jan 18 2006

This sequence (formatted as a square array) gives the counts of all possible squares in an m X n rectangle. For example, 11 = 8 (1 X 1 squares) + 3 (2 X 2 square) in 4 X 2 rectangle. - Philippe Deléham, Nov 26 2009
From Clark Kimberling, Feb 07 2011: (Start)
Also the accumulation array of min{n,k}, when formatted as a rectangle.
This is the accumulation array of the array M=A003783 given by M(n,k)=min{n,k}; see A144112 for the definition of accumulation array.
The accumulation array of A115262 is A185957. (End)
From Clark Kimberling, Dec 22 2011: (Start)
As a square matrix, A115262 is the self-fusion matrix of A000027 (1,2,3,4,...).  See A193722 for the definition of fusion and A202673 for characteristic polynomials associated with A115622. (End)

			Triangle begins
  1;
  2,  2;
  3,  5,  3;
  4,  8,  8,  4;
  5, 11, 14, 11,  5;
  6, 14, 20, 20, 14,  6;
  ...
When formatted as a square matrix:
  1,  2,  3,  4,  5, ...
  2,  5,  8, 11, 14, ...
  3,  8, 14, 20, 26, ...
  4, 11, 20, 30, 40, ...
  5, 14, 26, 40, 55, ...
  ...

Cf. A000027, A202673, A271916.
For the triangular version: row sums are A001752. Diagonal sums are A097701. T(2n,n) is A000330(n+1).
Diagonals (1,5,...): A000330 (square pyramidal numbers),
diagonals (2,8,...): A007290,
diagonals (3,11,...): A051925,
diagonals (4,14,...): A159920,
antidiagonal sums: A001752.

U = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[k, {k, 1, 12}]];
L = Transpose[U]; M = L.U; TableForm[M]
m[i_, j_] := M[[i]][[j]];
Flatten[Table[m[i, n + 1 - i], {n, 1, 12}, {i, 1, n}]]
(* Clark Kimberling, Dec 22 2011 *)

Let f(m,n) = m*(m-1)*(3*n-m-1)/6. This array is (with a different offset) the infinite square array read by antidiagonals U(m,n) = f(n,m) if m < n, U(m,n) = f(m,n) if m <= n. See A271916. - N. J. A. Sloane, Apr 26 2016
G.f.: 1/((1-x)^2*(1-x*y)^2*(1-x^2*y)).
Number triangle T(n, k) = Sum_{j=0..n} [j<=k]*(k-j+1)[j<=n-k]*(n-k-j+1).
T(2n,n) - T(2n,n+1) = n+1.

A196661 Expansion of g.f. (1-2x)/(1-7x).

Original entry on oeis.org

1, 5, 35, 245, 1715, 12005, 84035, 588245, 4117715, 28824005, 201768035, 1412376245, 9886633715, 69206436005, 484445052035, 3391115364245, 23737807549715, 166164652848005, 1163152569936035, 8142067989552245, 56994475926865715, 398961331488060005

Offset: 0

Philippe Deléham, Oct 05 2011

Index entries for linear recurrences with constant coefficients, signature (7).

Cf. A002001, A193577 (which is the same except for the initial 1), A193722.

a(n)=if(n,5*7^(n-1),1) \\ Charles R Greathouse IV, Nov 21 2011

a(0) = 1, a(n) = 5*7^(n-1) for n>0.
a(n) = Sum_{k=0..n} A193722(n,k)*2^k.
From Elmo R. Oliveira, Mar 18 2025: (Start)
E.g.f.: (5*exp(7*x) + 2)/7.
a(n) = 7*a(n-1). (End)

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A193723 Mirror of the fusion triangle A193722.

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A202453 Fibonacci self-fusion matrix, by antidiagonals.

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A193842 Triangular array: the fission of the polynomial sequence ((x+1)^n: n >= 0) by the polynomial sequence ((x+2)^n: n >= 0). (Fission is defined at Comments.)

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A081277 Square array of unsigned coefficients of Chebyshev polynomials of the first kind.

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A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

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A118800 Triangle read by rows: T satisfies the matrix products: C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle.

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A038763 Triangular matrix arising in enumeration of catafusenes, read by rows.

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A118800 Triangle read by rows: T satisfies the matrix products: CTC = T^-1 and TCT = C^-1, where C is Pascal's triangle.

A196661 Expansion of g.f. (1-2x)/(1-7x).