A193737 -id:A193737 - cp's OEIS frontend

A330793 a(n) = A193737(2*n, n).

1, 2, 8, 36, 170, 826, 4088, 20496, 103752, 529100, 2714140, 13989560, 72393412, 375877684, 1957199120, 10216355632, 53443289946, 280101010170, 1470508417340, 7731675774900, 40706787482130, 214580612067690, 1132389348358320, 5981916549623040, 31629125981208600

Offset: 0

Peter Luschny, Jan 10 2020

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A193737.

[1] cat [n le 2 select 2*(3*n-2) else ( 2*(11*n-3)*(n-1)*Self(n-1) + 3*(3*n-4)*(3*n-5)*Self(n-2) )/(5*n*(n-1)): n in [1..30]]; // G. C. Greubel, Oct 24 2023

a := proc(n) option remember;
if n < 3 then return [1, 2, 8][n+1] fi;
((60-81*n+27*n^2)*a(n-2) + (22*n^2-28*n+6)*a(n-1))/(5*n*(n-1)) end:
seq(a(n), n=0..24);
# Alternative:
gf := x -> (16 + 8*hypergeom([2/3, 1/3], [1/2], (1+x)*27/32) +
sqrt(18*(1+x))*hypergeom([7/6, 5/6], [3/2], (1+x)*27/32))/48:
ser := series(gf(x), x, 32): evalf(%, 32):
seq(round(coeff(%, x, n)), n=0..24);
# Or:
Gf := x -> (1/(3*sqrt(5 - 27*x)))*(sqrt(5 - 27*x) +
2*sqrt(2)*cos((1/6)*arccos(1 - (27*(1 + x))/16)) +
2*sqrt(6)*sin((1/3)*arcsin((3/4)*sqrt(3/2)*sqrt(1 + x)))):
ser := series(Gf(x), x, 32): evalf(%, 32):
seq(round(coeff(%,x,n)), n=0..24);

a[n_]:= a[n]= If[n<3, 2^n*n!, (2*(n-1)*(11*n-3)*a[n-1] +3*(3*n-4)*(3*n -5)*a[n-2])/(5*n*(n-1))]; (* a=A330793 *)
Table[a[n], {n,0,40}] (* G. C. Greubel, Oct 24 2023 *)

@CachedFunction
def a(n): # a = A330793
    if n<3: return (1,2,8)[n]
    else: return (2*(n-1)*(11*n-3)*a(n-1) + 3*(3*n-4)*(3*n-5)*a(n-2))/(5*n*(n-1))
[a(n) for n in range(41)] # G. C. Greubel, Oct 24 2023

D-finite with recurrence a(n) = ( 2*(11*n-3)*(n-1)*a(n-1) + 3*(3*n - 4)*(3*n-5)*a(n-2) )/(5*n*(n-1)).
a(n) = [x^n] (16 + 8*hypergeometric2F1([2/3, 1/3], [1/2], (1+x)*27/32) + sqrt(18*(1+x))* hypergeometric2F1([7/6, 5/6], [3/2], (1+x)*27/32))/48.
a(n) = [x^n] (1/(3*sqrt(5 - 27*x)))*(sqrt(5 - 27*x) + 2*sqrt(2)*cos((1/6)*arccos(1 - (27*(1 + x))/16)) + 2*sqrt(6)*sin((1/3)*arcsin((3/4)*sqrt(3/2)*sqrt(1 + x)))).
a(n) ~ 2^(3/2) * 3^(3*n - 1/2) / (sqrt(Pi*n) * 5^(n + 1/2)). - Vaclav Kotesovec, Oct 24 2023

A330795 Evaluation of the polynomials given by the Riordan square of the Fibonacci sequence with a(0) = 1 (A193737) at 1/2 and normalized with 2^n.

Original entry on oeis.org

1, 3, 9, 39, 153, 615, 2457, 9831, 39321, 157287, 629145, 2516583, 10066329, 40265319, 161061273, 644245095, 2576980377, 10307921511, 41231686041, 164926744167, 659706976665, 2638827906663, 10555311626649, 42221246506599, 168884986026393, 675539944105575

Offset: 0

Peter Luschny, Jan 10 2020

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,4).

Cf. A006131, A015521, A193737, A321620, A324969 (Fibonacci with a(0)=1).

[1] cat [3*(4^n -(-1)^n)/5: n in [1..30]]; // G. C. Greubel, Sep 14 2023

gf := (4*x^2 - 1)/(x*(4*x + 3) - 1): ser := series(gf, x, 32):
seq(coeff(ser, x, n), n=0.. 25);
# Alternative:
gf:= (3/5)*exp(-x)*(exp(5*x) - 1) + 1: ser := series(gf, x, 32):
seq(n!*coeff(ser, x, n), n=0.. 25);
# Or:
a := proc(n) option remember; if n < 3 then return [1, 3, 9][n + 1] fi;
4*a(n-2) + 3*a(n-1) end: seq(a(n), n=0..25);

LinearRecurrence[{3,4}, {1,3,9}, 31] (* G. C. Greubel, Sep 14 2023 *)

[3*(4^n -(-1)^n)//5 + int(n==0) for n in range(31)] # G. C. Greubel, Sep 14 2023

a(n) = 2^n*Sum_{k=0..n} A193737(n,k)/2^k.
a(n) = [x^n] (1 - 4*x^2)/(1 - x*(3 + 4*x)).
a(n) = n! [x^n] (3/5)*exp(-x)*(exp(5*x) - 1) + 1.
a(n) = 4*a(n-2) + 3*a(n-1).
a(n) = 3*A015521(n), n>0. - R. J. Mathar, Aug 19 2022

A193722 Triangular array: the fusion of (x+1)^n and (x+2)^n; see Comments for the definition of fusion.

Original entry on oeis.org

1, 1, 2, 1, 5, 6, 1, 8, 21, 18, 1, 11, 45, 81, 54, 1, 14, 78, 216, 297, 162, 1, 17, 120, 450, 945, 1053, 486, 1, 20, 171, 810, 2295, 3888, 3645, 1458, 1, 23, 231, 1323, 4725, 10773, 15309, 12393, 4374, 1, 26, 300, 2016, 8694, 24948, 47628, 58320, 41553, 13122

Offset: 0

Clark Kimberling, Aug 04 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-upstep of p is the polynomial given by
...
U(p) = p(n)*q(n+1,x) + p(n-1)*q(n,x) + ... + p(0)*q(1,x); note that q(0,x) does not appear.
...
Now suppose that P=(p(n,x)) and Q=(q(n,x)) are sequences of polynomials, where n indicates degree.  The fusion of P by Q, denoted by P**Q, is introduced here as the sequence W=(w(n,x)) of polynomials defined by w(0,x)=1 and w(n+1,x)=U(p(n,x)).
...
Strictly speaking, ** is an operation on sequences of polynomials.  However, if P and Q are regarded as numerical triangles (e.g., coefficients of polynomials), then ** can be regarded as an operation on numerical triangles.  In this case, row (n+1) of P**Q, for n >= 0, is given by the matrix product P(n)*QQ(n), where P(n)=(p(n,n)...p(n,n-1)......p(n,1), p(n,0)) and QQ(n) is the (n+1)-by-(n+2) matrix given by
...
q(n+1,0) .. q(n+1,1)........... q(n+1,n) .... q(n+1,n+1)
0 ......... q(n,0)............. q(n,n-1) .... q(n,n)
0 ......... 0.................. q(n-1,n-2) .. q(n-1,n-1)
...
0 ......... 0.................. q(2,1) ...... q(2,2)
0 ......... 0 ................. q(1,0) ...... q(1,1);
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-1); i.e., "q" is used instead of "t".
...
If s=(s(1),s(2),s(3),...) is a sequence, then the infinite square matrix indicated by
s(1)...s(2)...s(3)...s(4)...s(5)...
..0....s(1)...s(2)...s(3)...s(4)...
..0......0....s(1)...s(2)...s(3)...
..0......0.......0...s(1)...s(2)...
is the self-fusion matrix of s; e.g., A202453, A202670.
...
Example:  let p(n,x)=(x+1)^n and q(n,x)=(x+2)^n.  Then
  ...
w(0,x) = 1 by definition of W
w(1,x) = U(p(0,x)) = U(1) = p(0,0)*q(1,x) = 1*(x+2) = x+2;
w(2,x) = U(p(1,x)) = U(x+1) = q(2,x) + q(1,x) = x^2+5x+6;
w(3,x) = U(p(2,x)) = U(x^2+2x+1) = q(3,x) + 2q(2,x) + q(1,x) = x^3+8x^2+21x+18;
  ...
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,  2;
  1,  5,  6;
  1,  8, 21, 18;
  ...
Generally, if P and Q are the sequences given by p(n,x)=(ax+b)^n and q(n,x)=(cx+d)^n, then P**Q is given by (cx+d)(bcx+a+bd)^n.
...
In the following examples, r(P**Q) is the mirror of P**Q, obtained by reversing the rows of P**Q.
...
..P...........Q.........P**Q.......r(P**Q)
(x+1)^n.....(x+1)^n.....A081277....A118800 (unsigned)
(x+1)^n.....(x+2)^n.....A193722....A193723
(x+2)^n.....(x+1)^n.....A193724....A193725
(x+2)^n.....(x+2)^n.....A193726....A193727
(x+2)^n.....(2x+1)^n....A193728....A193729
(2x+1)^n....(x+1)^n.....A038763....A136158
(2x+1)^n....(2x+1)^n....A193730....A193731
(2x+1)^n,...(x+1)^n.....A193734....A193735
...
Continuing, let u denote the polynomial x^n+x^(n-1)+...+x+1, and let Fibo[n,x] denote the n-th Fibonacci polynomial.
...
P.............Q.........P**Q.......r(P**Q)
Fib[n+1,x]...(x+1)^n....A193736....A193737
u.............u.........A193738....A193739
u**u..........u**u......A193740....A193741
...
Regarding A193722:
col 1 ..... A000012
col 2 ..... A016789
col 3 ..... A081266
w(n,n) .... A025192
w(n,n-1) .. A081038
...
Associated with "upstep" as defined above is "downstep" defined at A193842 in connection with fission.

			First six rows:
  1;
  1,   2;
  1,   5,   6;
  1,   8,  21,  18;
  1,  11,  45,  81,  54;
  1,  14,  78, 216, 297, 162;

Jinyuan Wang, Rows n=0..101 of triangle, flattened
Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52 (2014), 195-202.

Cf. A081277, A084938, A118800, A193649, A193723-A193741, A202453.

Flat(List([0..10], n-> List([0..n], k-> 3^(k-1)*( Binomial(n-1,k) + 2*Binomial(n,k) ) ))); # G. C. Greubel, Feb 18 2020

[3^(k-1)*( Binomial(n-1,k) + 2*Binomial(n,k) ): k in [0..n], n in [0..10]]; // G. C. Greubel, Feb 18 2020

fusion := 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-1);
[1,seq(coeff(%,x,n-1-k), k=0..n-1)] end:
p := (n, x) -> (x + 1)^n; q := (n, x) -> (x + 2)^n;
A193722_row := n -> fusion(p, q, n);
for n from 0 to 5 do A193722_row(n) od; # Peter Luschny, Jul 24 2014

(* First program *)
z = 9; a = 1; b = 1; c = 1; d = 2;
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, 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}]] (* A193722 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]] (* A193723 *)
(* Second program *)
Table[3^(k-1)*(Binomial[n-1,k] +2*Binomial[n,k]), {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 18 2020 *)

T(n,k) = 3^(k-1)*(binomial(n-1,k) +2*binomial(n,k)); \\ G. C. Greubel, Feb 18 2020

def fusion(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-1))
    return [1]+[expand(F).coefficient(x,n-1-k) for k in (0..n-1)]
A193842_row = lambda k: fusion(lambda n,x: (x+1)^n, lambda n,x: (x+2)^n, k)
for n in range(7): A193842_row(n) # Peter Luschny, Jul 24 2014

Triangle T(n,k), read by rows, given by [1,0,0,0,0,0,0,0,...] DELTA [2,1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 04 2011
T(n,k) = 3*T(n-1,k-1) + T(n-1,k) with T(0,0)=T(1,0)=1 and T(1,1)=2. - Philippe Deléham, Oct 05 2011
T(n, k) = 3^(k-1)*( binomial(n-1,k) + 2*binomial(n,k) ). - G. C. Greubel, Feb 18 2020

A119915 Number of ternary words of length n and having exactly one run of 0's of odd length.

Original entry on oeis.org

0, 1, 4, 13, 40, 117, 332, 921, 2512, 6761, 18004, 47525, 124536, 324317, 840092, 2166065, 5562272, 14232273, 36300196, 92321085, 234192584, 592695109, 1496810732, 3772761289, 9492450672, 23844342073, 59804611060, 149787196117

Offset: 0

Emeric Deutsch, May 29 2006

Column 1 of A119914.

			a(3) = 13 because we have 000, 011, 012, 021, 022, 101, 102, 110, 120, 201, 202, 210 and 220 (for example, 001, 020 do not qualify).

Index entries for linear recurrences with constant coefficients, signature (4,-2,-4,-1).

Cf. A119914, A193737, A006645.

g := z*(1-z^2)/(1-2*z-z^2)^2:
gser := series(g,z=0,34):
seq(coeff(gser,z,n), n=0..30);

LinearRecurrence[ {4, -2, -4, -1}, {0, 1, 4, 13}, 28] (* Peter Luschny, Jan 14 2020 *)

a(n) = [z^n] z*(1 - z^2)/(1 - 2*z - z^2)^2.
a(n) = A006645(n+1) - A006645(n-1). - R. J. Mathar, Aug 07 2015
From Peter Luschny, Jan 14 2020: (Start)
a(n) = Sum_{k=0..n} A193737(n, k)*k.
Let h(k) = (1 + k)*exp((1 + k)*x)*(1 + x - 1/k)/4 then
a(n) = n!*[x^n](h(sqrt(2)) + h(-sqrt(2))).  (End)

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)

A331321 a(n) = [x^n] ((x^2 - 1)(x^2 + x - 1))/(x^2 + 2x - 1)^2.

Original entry on oeis.org

1, 3, 8, 23, 64, 175, 472, 1259, 3328, 8731, 22760, 59007, 152256, 391239, 1001656, 2556115, 6503936, 16505651, 41788616, 105571303, 266181440, 669923039, 1683255448, 4222878651, 10579130112, 26467818315, 66138242984, 165077936207, 411584855488, 1025162759287

Offset: 0

Peter Luschny, Jan 14 2020

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-2,-4,-1).

Cf. A193737 (Fibonacci (with a(0)=1) triangle), A331319, A331320.

R:=PowerSeriesRing(Integers(), 33); Coefficients(R!( (1 - x)*(1 + x)*(1-x-x^2) / (1-2*x-x^2)^2 )); // Marius A. Burtea, Jan 15 2020

gf := ((x^2 - 1)*(x^2 + x - 1))/(x^2 + 2*x - 1)^2:
ser := series(gf, x, 32): seq(coeff(ser, x, n), n=0..29);

LinearRecurrence[{4,-2,-4,-1},{1,3,8,23,64},40] (* Harvey P. Dale, Feb 01 2022 *)

Vec((1 - x)*(1 + x)*(1 - x - x^2) / (1 - 2*x - x^2)^2 + O(x^30)) \\ Colin Barker, Jan 14 2020

a(n) = Sum_{k=0..n} A193737(n, k)*(1 + k).
Let h(k) = (1 + k)*exp((1 + k)*x)*(2*x + 10 - 5*k)/8 then
a(n) = n!*[x^n](h(sqrt(2)) + h(-sqrt(2)) + 1).
From Colin Barker, Jan 14 2020: (Start)
a(n) = 4*a(n-1) - 2*a(n-2) - 4*a(n-3) - a(n-4) for n>4.
a(n) = (-5*sqrt(2)*((1-sqrt(2))^n - (1+sqrt(2))^n) + 2*((1-sqrt(2))^n + (1+sqrt(2))^n)*n) / 8 for n>0.
(End)

cp's OEIS Frontend

A330793 a(n) = A193737(2*n, n).

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

SageMath

Formula

A330795 Evaluation of the polynomials given by the Riordan square of the Fibonacci sequence with a(0) = 1 (A193737) at 1/2 and normalized with 2^n.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

SageMath

Formula

A193722 Triangular array: the fusion of (x+1)^n and (x+2)^n; see Comments for the definition of fusion.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

A119915 Number of ternary words of length n and having exactly one run of 0's of odd length.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

A331321 a(n) = [x^n] ((x^2 - 1)*(x^2 + x - 1))/(x^2 + 2*x - 1)^2.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A331321 a(n) = [x^n] ((x^2 - 1)(x^2 + x - 1))/(x^2 + 2x - 1)^2.