A029907 -id:A029907 - cp's OEIS frontend

A119473 Triangle read by rows: T(n,k) is number of binary words of length n and having k runs of 0's of odd length, 0 <= k <= ceiling(n/2). (A run of 0's is a subsequence of consecutive 0's of maximal length.)

1, 1, 1, 2, 2, 3, 4, 1, 5, 8, 3, 8, 15, 8, 1, 13, 28, 19, 4, 21, 51, 42, 13, 1, 34, 92, 89, 36, 5, 55, 164, 182, 91, 19, 1, 89, 290, 363, 216, 60, 6, 144, 509, 709, 489, 170, 26, 1, 233, 888, 1362, 1068, 446, 92, 7, 377, 1541, 2580, 2266, 1105, 288, 34, 1, 610, 2662, 4830

Offset: 0

Emeric Deutsch, May 22 2006

Row n has 1+ceiling(n/2) terms.
T(n,0) = Fibonacci(n+1) = A000045(n+1).
T(n,1) = A029907(n).
Sum_{k>=0} k*T(n,k) = A059570(n).
Triangle, with zeros included, given by (1,1,-1,0,0,0,0,0,0,0,...) DELTA (1,-1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 07 2011
T(n,k) is the number of compositions of n+1 that have exactly k even parts. - Geoffrey Critzer, Mar 03 2012

			T(5,2)=8 because we have 00010, 01000, 01011, 01101, 01110, 10101, 10110 and 11010.
T(5,2)=8 because there are 8 compositions of 6 that have 2 even parts: 4+2, 2+4, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2. - _Geoffrey Critzer_, Mar 03 2012
Triangle starts:
  1;
  1,  1;
  2,  2;
  3,  4,  1;
  5,  8,  3;
  8, 15,  8,  1;
From _Philippe Deléham_, Dec 07 2011: (Start)
Triangle (1,1,-1,0,0,0...) DELTA (1,-1,0,0,0,...) begins:
   1;
   1,  1;
   2,  2,  0;
   3,  4,  1,  0;
   5,  8,  3,  0,  0;
   8, 15,  8,  1,  0,  0;
  13, 28, 19,  4,  0,  0,  0;
  21, 51, 42, 13,  1,  0,  0,  0;
  34, 92, 89, 36,  5,  0,  0,  0,  0; ... (End)

I. Goulden and D. Jackson, Combinatorial Enumeration, John Wiley and Sons, 1983, page 54.

Alois P. Heinz, Rows n = 0..200, flattened
Rigoberto Flórez, Javier González, Mateo Matijasevick, Cristhian Pardo, José Luis Ramírez, Lina Simbaqueba, and Fabio Velandia, Lattice paths in corridors and cyclic corridors, Contrib. Disc. Math. (2024) Vol. 19. No. 2, 36-55. See p. 17.
Ralph Grimaldi and Silvia Heubach, Binary strings without odd runs of zeros, Ars Combinatoria 75 (2005), 241-255.

Cf. A000045, A029907, A059570.

G:=(1+t*z)/(1-z-z^2-t*z^2): Gser:=simplify(series(G,z=0,18)): P[0]:=1: for n from 1 to 14 do P[n]:=sort(coeff(Gser,z^n)) od: for n from 0 to 14 do seq(coeff(P[n],t,j),j=0..ceil(n/2)) od; # yields sequence in triangular form
# second Maple program:
b:= proc(n) option remember; local j; if n=0 then 1
      else []; for j to n do zip((x, y)->x+y, %,
      [`if`(irem(j, 2)=0, 0, NULL), b(n-j)], 0) od; %[] fi
    end:
T:= n-> b(n+1):
seq(T(n), n=0..14);  # Alois P. Heinz, May 23 2013

f[list_] := Select[list, # > 0 &]; nn = 15; a = (x + y x^2)/(1 - x^2); Map[f, Drop[CoefficientList[Series[1/(1 - a), {x, 0, nn}], {x, y}], 1]] // Flatten  (* Geoffrey Critzer, Mar 03 2012 *)

G.f.: (1+t*z)/(1-z-z^2-t*z^2).
G.f. of column k (k>=1): z^(2*k-1)*(1-z^2)/(1-z-z^2)^(k+1).
T(n,k) = T(n-1,k) + T(n-2,k) + T(n-2,k-1). - Philippe Deléham, Dec 07 2011
Sum_{k=0..n} T(n,k)*x^k = A000045(n+1), A000079(n), A105476(n+1), A159612(n+1), A189732(n+1) for x = 0, 1, 2, 3, 4 respectively. - Philippe Deléham, Dec 07 2011
G.f.: (1+x*y)*T(0)/2, where T(k) = 1 + 1/(1 - (2*k+1+ x*(1+y))*x/((2*k+2+ x*(1+y))*x + 1/T(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 06 2013

A054453 Triangle of partial row sums of triangle A054450(n,m), n >= m >= 0.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 8, 5, 2, 1, 15, 10, 6, 2, 1, 28, 20, 12, 7, 2, 1, 51, 38, 26, 14, 8, 2, 1, 92, 71, 50, 33, 16, 9, 2, 1, 164, 130, 97, 64, 41, 18, 10, 2, 1, 290, 235, 180, 130, 80, 50, 20, 11, 2, 1, 509, 420, 332, 244, 171, 98, 60, 22, 12, 2, 1

Offset: 0

Wolfdieter Lang, Apr 27 2000 and May 08 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is ((1-z^2)*(Fib(z))^2)/(1-x*z/(1-z^2)) Fib(x)=1/(1-x-x^2) = g.f. for A000045(n+1) (Fibonacci numbers without 0).
This is the second member of the family of Riordan-type matrices obtained from the unsigned convolution matrix A049310(n,m) by repeated application of the partial row sums procedure.
The column sequences are A029907, A001629, A054454 for m=0..2.

			{1}; {2,1}; {4,2,1}; {8,5,2,1};...
Fourth row polynomial (n=3): p(3,x)= 8+5*x+2*x^2+x^3

Gregg Musiker, Nick Ovenhouse, and Sylvester W. Zhang, Double Dimers and Super Ptolemy Relations, Séminaire Lotharingien de Combinatoire XX, Proc. 35th Conf. Formal Power, Series and Algebraic Combinatorics (Davis) 2023, Art. #YY. See p. 12.

Cf. A049310, A054450, A000045, A029907, A001629. Row sums: A054455(n).

a(n, m)=sum(A054450(n, k), k=m..n), n >= m >= 0, a(n, m) := 0 if n

Column m recursion: a(n, m)= sum(a(j-1, m)*|A049310(n-j, 0)|, j=m..n) + A054450(n, m), n >= m >= 0, a(n, m) := 0 if n

G.f. for column m: ((1-x^2)*(Fib(x))^2)*(x/(1-x^2))^m, m >= 0, with Fib(x) G.f. for A000045(n+1).

A099573 Reverse of number triangle A054450.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 4, 5, 1, 1, 5, 5, 8, 8, 1, 1, 6, 6, 12, 12, 13, 1, 1, 7, 7, 17, 17, 21, 21, 1, 1, 8, 8, 23, 23, 33, 33, 34, 1, 1, 9, 9, 30, 30, 50, 50, 55, 55, 1, 1, 10, 10, 38, 38, 73, 73, 88, 88, 89, 1, 1, 11, 11, 47, 47, 103, 103, 138, 138, 144, 144, 1, 1, 12, 12, 57, 57, 141, 141, 211, 211, 232, 232, 233

Offset: 0

Paul Barry, Oct 23 2004

			First few rows of the array:
  1, 1, 2, 3,  5,  8, ... (A000045)
  1, 1, 3, 4,  8, 12, ... (A052952)
  1, 1, 4, 5, 12, 17, ... (A054451)
  1, 1, 5, 6, 17, 23, ... (A099571)
  1, 1, 6, 7, 23, 30, ... (A099572)
  ...
Triangle begins as:
  1;
  1, 1;
  1, 1, 2;
  1, 1, 3, 3;
  1, 1, 4, 4,  5;
  1, 1, 5, 5,  8,  8;
  1, 1, 6, 6, 12, 12, 13;
  1, 1, 7, 7, 17, 17, 21, 21;
  1, 1, 8, 8, 23, 23, 33, 33, 34;
  1, 1, 9, 9, 30, 30, 50, 50, 55, 55;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000045, A029907 (row sums), A052951, A052952, A054450, A054451, A052952.

Cf. A099571, A099572, A099574 (diagonal sums), A099575.

[(&+[Binomial(n-j,j): j in [0..Floor(k/2)]]): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

T[n_, k_]:= Sum[Binomial[n-j,j], {j,0,Floor[k/2]}];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 25 2022 *)

def A099573(n,k): return sum(binomial(n-j, j) for j in (0..(k//2)))
flatten([[A099573(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 25 2022

Number triangle T(n, k) = Sum_{j=0..floor(k/2)} binomial(n-j, j) if k <= n, 0 otherwise.
T(n, n) = A000045(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A099574(n).
Sum_{k=0..n} T(n, k) = A029907(n+1).
Antidiagonals of the following array: the first row equals the Fibonacci numbers, (1, 1, 2, 3, 5, ...), and the (n+1)-st row is obtained by the matrix-vector product A128174 * n-th row. - Gary W. Adamson, Jan 19 2011
From G. C. Greubel, Jul 25 2022: (Start)
T(n, n-1) = A052952(n-1), n >= 1.
T(n, n-2) = A054451(n-2), n >= 2.
T(n, n-3) = A099571(n-3), n >= 3.
T(n, n-4) = A099572(n-4), n >= 4. (End)

A099576 Row sums of triangle A099575.

Original entry on oeis.org

1, 2, 6, 12, 35, 72, 210, 440, 1287, 2730, 8008, 17136, 50388, 108528, 319770, 692208, 2042975, 4440150, 13123110, 28614300, 84672315, 185122080, 548354040, 1201610592, 3562467300, 7821594872, 23206929840, 51037462560, 151532656696

Offset: 0

Paul Barry, Oct 23 2004

Andrew Howroyd, Table of n, a(n) for n = 0..200

Cf. A029907, A099575, A099577.

[(&+[Binomial(n+Floor(k/2)+1, Floor(k/2)+1)*(1+Floor(k/2))/(n+1): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Jul 24 2022

seq(op([(1+n/(n+1))*binomial(3*n+1,n),2*binomial(3*n+3,n)]),n=0..20);

a[n_] := Sum[Binomial[n + j, j], {k, 0, n}, {j, 0, k/2}];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jul 06 2018 *)
a[n_] := Binomial[2*n+2, n]*Hypergeometric2F1[-n, n+1, -2*n-2, -1]; Flatten[Table[a[n], {n, 0, 28}]] (* Detlef Meya, Dec 25 2023 *)

a(n) = sum(k=0, n, sum(j=0, floor(k/2), binomial(n+j, j))); \\ Andrew Howroyd, Feb 13 2018

[sum( binomial(n+(k//2)+1, (k//2)+1)*(1+(k//2))/(n+1) for k in (0..n) ) for n in (0..40)] # G. C. Greubel, Jul 24 2022

a(n) = Sum_{k=0..n} Sum_{j=0..floor(k/2)} binomial(n+j, j).
Conjecture: 4*n*(n-1)*(3*n+2)*(n+2)*a(n) - 36*(n-1)*(n+1)*a(n-1) - 3*n*(3*n+5)*(3*n-1)*(3*n-2)*a(n-2) = 0. - R. J. Mathar, Nov 28 2014
From Robert Israel, May 08 2018: (Start)
a(2*n) = (1+n/(n+1))*binomial(3*n+1,n).
a(2*n+1) = 2*binomial(3*n+3,n).
The conjecture follows from this. (End)
a(n) = (1/(n+1))*Sum_{k=0..n} binomial(n + floor(k/2) + 1, floor(k/2) + 1)*(1 + floor(k/2)). - G. C. Greubel, Jul 24 2022
a(n) = binomial(2*n+2, n)*hypergeom([-n, n+1], [-2*n-2], -1). - Detlef Meya, Dec 25 2023

A193736 Triangular array: the fusion of polynomial sequences P and Q given by p(n,x) = (n+1)-st Fibonacci polynomial and q(n,x) = (x+1)^n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 4, 2, 1, 4, 8, 8, 3, 1, 5, 13, 19, 15, 5, 1, 6, 19, 36, 42, 28, 8, 1, 7, 26, 60, 91, 89, 51, 13, 1, 8, 34, 92, 170, 216, 182, 92, 21, 1, 9, 43, 133, 288, 446, 489, 363, 164, 34, 1, 10, 53, 184, 455, 826, 1105, 1068, 709, 290, 55, 1, 11, 64, 246, 682, 1414, 2219, 2619, 2266, 1362, 509, 89

Offset: 0

Clark Kimberling, Aug 04 2011

See A193722 for the definition of fusion of two sequences of polynomials or triangular arrays.

			First six rows:
  1;
  1,  1;
  1,  2,  1;
  1,  3,  4,  2;
  1,  4,  8,  8,  3;
  1,  5, 13, 19, 15,  5;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000007, A005314 (diagonal sums), A052542 (row sums), A077962.

Cf. A029907, A193722, A193737, A324969.

function T(n,k) // T = A193736
  if k lt 0 or n lt 0 then return 0;
  elif n lt 3 then return Binomial(n,k);
  else return T(n-1, k) + T(n-1, k-1) + T(n-2, k-2);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 24 2023

(* First program *)
p[0, x_] := 1
p[n_, x_] := Fibonacci[n + 1, x] /; n > 0
q[n_, x_] := (x + 1)^n
t[n_, k_] := Coefficient[p[n, x], x^(n - k)];
t[n_, n_] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n - k + 1, 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}]] (* A193736 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]]          (* A193737 *)
(* Second program *)
T[n_, k_]:= T[n, k]= If[n<3, Binomial[n, k], T[n-1,k] +T[n-1,k-1] +T[n -2,k-2]];
Table[T[n, k], {n,0,12}, {k,0,n}]//TableForm (* G. C. Greubel, Oct 24 2023 *)

def T(n,k): # T = A193736
    if (n<3): return binomial(n,k)
    else: return T(n-1,k) +T(n-1,k-1) +T(n-2,k-2)
flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 24 2023

T(0,0) = T(1,0) = T(1,1) = T(2,0) = T(2,2) = 1; T(n,k) = 0 if k<0 or k>n; T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k-2). - Philippe Deléham, Feb 13 2020
From G. C. Greubel, Oct 24 2023: (Start)
T(n, k) = A193737(n, n-k).
T(n, n) = Fibonacci(n) + [n=0] = A324969(n+1).
T(n, n-1) = A029907(n).
Sum_{k=0..n} T(n, k) = A052542(n).
Sum_{k=0..n} (-1)^k * T(n, k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A005314(n) + [n=0].
Sum_{k=0..floor(n/2)} (-1)^k * T(n-k, k) = [n=0] + A077962(n-1). (End)

A207611 Triangle of coefficients of polynomials v(n,x) jointly generated with A207610; see Formula section.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 5, 4, 2, 1, 8, 8, 5, 2, 1, 13, 15, 11, 6, 2, 1, 21, 28, 23, 14, 7, 2, 1, 34, 51, 47, 32, 17, 8, 2, 1, 55, 92, 93, 70, 42, 20, 9, 2, 1, 89, 164, 181, 148, 97, 53, 23, 10, 2, 1, 144, 290, 346, 306, 217, 128, 65, 26, 11, 2, 1, 233, 509, 653, 619, 472

Offset: 1

Clark Kimberling, Feb 19 2012

Column 1:  Fibonacci numbers, A000045
Column 2:  A029907
Row sums:  A003945.
For a discussion and guide to related arrays, see A208510.
Subtriangle of the triangle given by (0, 2, -1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 25 2012

			First five rows:
  1;
  2, 1;
  3, 2, 1;
  5, 4, 2, 1;
  8, 8, 5, 2, 1;
From _Philippe Deléham_, Mar 25 2012: (Start)
(0, 2, -1/2, -1/2, 0, 0, ...) DELTA (1, 0, -1, 1, 0, 0, ...) begins:
  1;
  0,  1;
  0,  2,  1;
  0,  3,  2,  1;
  0,  5,  4,  2,  1;
  0,  8,  8,  5,  2,  1;
  0, 13, 15, 11,  6,  2,  1;
  0, 21, 28, 23, 14,  7,  2,  1; (End)

Cf. A207610, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := u[n - 1, x] + x*v[n - 1, x] + 1
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A207610 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A207611 *)
T[ n_, k_] := Which[k<0 || n<0, 0, n<2, Boole[k<=n] + Boole[k==0&&n==1], True, T[n, k] = T[n-1, k] + T[n-1, k-1] + T[n-2, k] - T[n-2, k-1] ]; (* Michael Somos, Sep 19 2024 *)

{T(n, k) = if(k<0 || n<0, 0, n<2, (k<=n) + (k==0 && n==1), T(n-1, k) + T(n-1, k-1) + T(n-2, k) - T(n-2, k-1) )}; /* Michael Somos, Sep 19 2024 */

from sympy import Poly
from sympy.abc import x
def u(n, x): return 1 if n==1 else u(n - 1, x) + v(n - 1, x)
def v(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x) + 1
def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017

u(n,x) = u(n-1,x) + v(n-1,x), v(n,x) = u(n-1,x) + x*v(n-1,x)+1, where u(1,x)=1, v(1,x)=1.

T(n,k) = T(n-1,k) + (n-1,k-1) + T(n-2,k) - T(n-2,k-1), T(1,0) = T(2,1) = 1, T(2,0) = 2 and T(n,k) = 0 if k < 0 or if k >= n.

A239342 Number of 1's in all compositions of n into odd parts.

Original entry on oeis.org

0, 1, 2, 3, 6, 11, 20, 36, 64, 113, 198, 345, 598, 1032, 1774, 3039, 5190, 8839, 15016, 25452, 43052, 72685, 122502, 206133, 346346, 581136, 973850, 1630011, 2725254, 4551683, 7594748, 12660660, 21087448, 35094377, 58360134, 96979089, 161042110, 267248664

Offset: 0

Geoffrey Critzer, Mar 16 2014

nonn

a(n+1) is the number of ways to tile a strip of length n+1 using white tiles of only odd lengths, with total length n, and one red square of length one. - Gregory L. Simay, Aug 14 2016
A029907, the number of compositions of n with exactly one even part, is equal to a(n+1-2) + a(n+1-4) + a(n+1-6) + ... - Gregory L. Simay, Aug 14 2016
Apart from the initial 0 and 1, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = (1 - S)^2. See A291219. - Clark Kimberling, Sep 02 2017

			a(5) = 11 because in the compositions of 5 into odd parts there are a total of 11 1's: 5, 3+1+1, 1+3+1, 1+1+3, 1+1+1+1+1.
Let r represent the red square and 1,3,5 represent the possible odd lengths of the white squares for n=5. Then a(5+1) = a(6) = 20 because r combined with a tile of length 5 generates 2 compositions; r combined with 3,1,1 generates 12 compositions; and r combined with 1,1,1,1,1 generates 6 compositions. 2+12+6 = 20. - _Gregory L. Simay_, Aug 14 2016

S. Heubach and T. Mansour, Combinatorics of Compositions and Words, Chapman and Hall, 2010, page 70.

Michael De Vlieger, Table of n, a(n) for n = 0..4771
Mengmeng Liu, Andrew Yezhou Wang, The Number of Designated Parts in Compositions with Restricted Parts, J. Int. Seq., Vol. 23 (2020), Article 20.1.8.

Cf. A000045, A029907.

nn=30;CoefficientList[Series[x (1-x^2)^2/(1-x-x^2)^2,{x,0,nn}],x]
(* or *)
Table[Count[Flatten[Level[Map[Permutations,IntegerPartitions[n,n,Table[2k+1,{k,0,n/2}]]],{2}]],1],{n,0,30}]

For n >= 4, a(n) = a(n-1) + a(n-2) + A000045(n-2).

G.f.: x*(1 - x^2)^2/(1 - x - x^2)^2.

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A131333 A131332 * A000012.

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A268103 T(n,k)=Number of nXk 0..k arrays with every repeated value in every row and column greater than or equal to the previous repeated value.

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A119473 Triangle read by rows: T(n,k) is number of binary words of length n and having k runs of 0's of odd length, 0 <= k <= ceiling(n/2). (A run of 0's is a subsequence of consecutive 0's of maximal length.)

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A205575 Triangle read by rows, related to Pascal's triangle, starting with rows 1; 1,0.

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A054453 Triangle of partial row sums of triangle A054450(n,m), n >= m >= 0.

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A099573 Reverse of number triangle A054450.

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A099576 Row sums of triangle A099575.

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Mathematica

PARI

SageMath

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A193736 Triangular array: the fusion of polynomial sequences P and Q given by p(n,x) = (n+1)-st Fibonacci polynomial and q(n,x) = (x+1)^n.

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