cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A210869 Triangle of coefficients of polynomials v(n,x) jointly generated with A210868; see the Formula section.

Original entry on oeis.org

1, 0, 2, 1, 0, 3, 0, 3, 0, 5, 1, 0, 7, 0, 8, 0, 4, 0, 15, 0, 13, 1, 0, 12, 0, 30, 0, 21, 0, 5, 0, 31, 0, 58, 0, 34, 1, 0, 18, 0, 73, 0, 109, 0, 55, 0, 6, 0, 54, 0, 162, 0, 201, 0, 89, 1, 0, 25, 0, 145, 0, 344, 0, 365, 0, 144, 0, 7, 0, 85, 0, 361, 0, 707, 0, 655, 0, 233, 1, 0
Offset: 1

Views

Author

Clark Kimberling, Mar 29 2012

Keywords

Comments

Row n starts with 1 or 0 and ends with F(n+1), where F=A000045 (Fibonacci numbers).
Row sums: 1,2,4,8,16,32,... (A000079)
Alternating row sums: 1, -2, 4, -8, 16,... (A122803)
For a discussion and guide to related arrays, see A208510.

Examples

			First six rows:
1
0...2
1...0...3
0...3...0...5
1...0...7...0....8
0...4...0...15...0...13
First three polynomials v(n,x): 1, 2x, 1 + 3x^2

Crossrefs

Cf. A210868, A208510.

Programs

  • Mathematica
    u[1,x_]:=1;v[1,x_]:=1;z=14;
u[n_,x_]:=u[n-1,x]+x*v[n-1,x];
v[n_,x_]:=(x+1)*u[n-1,x]+(x-1)*v[n-1,x];
Table[Expand[u[n,x]],{n,1,z/2}]
Table[Expand[v[n,x]],{n,1,z/2}]
cu=Table[CoefficientList[u[n,x],x],{n,1,z}];
TableForm[cu]
Flatten[%]   (* A210868 *)
cv=Table[CoefficientList[v[n,x],x],{n,1,z}];
TableForm[cv]
Flatten[%]   (* A210869 *)
Table[u[n,x]/.x->1,{n,1,z}]   (* A000079 *)
Table[v[n,x]/.x->1,{n,1,z}]   (* A000079 *)
Table[u[n,x]/.x->-1,{n,1,z}]  (* A151575 *)
Table[v[n,x]/.x->-1,{n,1,z}]  (* A122803 *)

Formula

u(n,x)=u(n-1,x)+x*v(n-1,x),
v(n,x)=(x+1)*u(n-1,x)+(x-1)*v(n-1,x),
where u(1,x)=1, v(1,x)=1.
T(n,k) = T(n-1,k-1) + T(n-2,k) + T(n-2,k-2), T(1,0) = 1, T(2,0) = 0, T(2,1) = 2 and T(n,k) = 0 if k<0 or if k>=n. - Philippe Deléham, Apr 02 2012

A136531 Coefficients of polynomials B(x,n) = ((1+a+b)*x - c)*B(x,n-1) - a*b*B(x,n-2) where B(x,0) = 1, B(x,1) = x, a=-b, b=1, c=1.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, -1, 3, -2, 1, 2, -5, 6, -3, 1, -3, 10, -13, 10, -4, 1, 5, -18, 29, -26, 15, -5, 1, -8, 33, -60, 65, -45, 21, -6, 1, 13, -59, 122, -151, 125, -71, 28, -7, 1, -21, 105, -241, 338, -321, 217, -105, 36, -8, 1, 34, -185, 468, -730, 784, -609, 350, -148, 45, -9, 1
Offset: 0

Views

Author

Roger L. Bagula, Mar 23 2008

Keywords

Examples

			Triangle begins
        k=0  k=1  k=2  k=3  k=4  k=5  k=6
  n=0:   1;
  n=1:   0,   1;
  n=2:   1,  -1,   1;
  n=3:  -1,   3,  -2,   1;
  n=4:   2,  -5,   6,  -3,   1;
  n=5:  -3,  10, -13,  10,  -4,   1;
  n=6:   5, -18,  29, -26,  15,  -5,   1;

Links

Crossrefs

Cf. A000027, A000217, A008728, A010049, A039834.
Cf. A055243, A078008, A136526, A151575, A212804.

Programs

  • Magma
    C := ComplexField(); // T = A136531
T:= func< n,k | k eq n select 1 else Round(i^(k-n-1)*(i*Evaluate(GegenbauerPolynomial(n-k, k+1), 1/(2*i)) - Evaluate(GegenbauerPolynomial(n-k-1, k+1), 1/(2*i)))) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 26 2022
  • Mathematica
    (* First program *)
a = -b; c = 1; b = 1;
B[x_, n_]:= B[x, n]= If[n<2, x^n, ((1+a+b)*x -c)*B[x, n-1] -a*b*B[x, n-2]];
Table[CoefficientList[B[x,n], x], {n,0,10}]//Flatten
(* Second program *)
B[x_, n_]:= (-1)^n*(Fibonacci[n+1, 1-x] - Fibonacci[n, 1-x]);
Table[CoefficientList[B[x, n], x], {n,0,16}]//Flatten (* G. C. Greubel, Sep 22 2022 *)
  • SageMath
    def T(n,k): # T = A136531
    if k==n: return 1
    else: return i^(k-n-1)*(i*gegenbauer(n-k, k+1, 1/(2*i)) - gegenbauer(n-k-1, k+1, 1/(2*i)))
flatten([[T(n,k) for k in range(n+1)] for n in range(12)]) # G. C. Greubel, Sep 26 2022

Formula

G.f.: (1+y) / (1 + (1-x)*y - y^2). - Kevin Ryde, Sep 21 2022
From G. C. Greubel, Sep 22 2022: (Start)
T(n, k) = coefficients of i^n*(ChebyshevU(n, (x-1)/(2*i)) - i*ChebyshevU(n-1, (x-1)/(2*i))).
T(n, k) = coefficients of (-1)^n*( Fibonacci(n+1, 1-x) - Fibonacci(n, 1-x) ).
T(n, k) = i^(k-n-1)*(i*GegenbauerC(n-k, k+1, 1/(2*i)) - GegenbauerC(n-k-1, k+1, 1/(2*i))).
T(n, 0) = Fibonacci(1-n) = (-1)^n*A212804(n) = A039834(n-1).
T(n, 1) = (-1)^(n-1)*A010049(n), n >= 1.
T(n, 2) = (-1)^n*A055243(n-2), n >= 2.
T(n, n) = 1.
T(n, n-1) = -(n-1).
T(n, n-2) = A000217(n-1), n >= 2.
T(n, n-3) = -A008728(n-3), n >= 3.
Sum_{k=0..n-2} T(n, k) = A000027(n-1), n >= 2.
Sum_{k=0..n} T(n, k) = 1.
Sum_{k=0..floor(n/2)} T(n-k, k) = A151575(n) = (-1)^n*A078008(n). (End)

Extensions

Offset corrected by Kevin Ryde, Sep 21 2022

A210868 Triangle of coefficients of polynomials u(n,x) jointly generated with A210869; see the Formula section.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 2, 3, 1, 2, 5, 3, 5, 1, 3, 5, 10, 5, 8, 1, 3, 9, 10, 20, 8, 13, 1, 4, 9, 22, 20, 38, 13, 21, 1, 4, 14, 22, 51, 38, 71, 21, 34, 1, 5, 14, 40, 51, 111, 71, 130, 34, 55, 1, 5, 20, 40, 105, 111, 233, 130, 235, 55, 89, 1, 6, 20, 65, 105, 256, 233, 474
Offset: 1

Views

Author

Clark Kimberling, Mar 29 2012

Keywords

Comments

In row n the first two terms are 1 and floor(n/2), and the last two, for n>1, are F(n-1) and F(n), where F = A000045 (Fibonacci numbers).
Row sums: 1,2,4,8,16,32,...; A000079.
Alternating row sums: A151575
For a discussion and guide to related arrays, see A208510.
Subtriangle of the triangle given by (1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Apr 02 2012

Examples

			First six rows:
  1
  1...1
  1...1...2
  1...2...2...3
  1...2...5...3....5
  1...3...5...10...5...8
First three polynomials u(n,x): 1, 1 + x, 1 + x + 2x^2.
(1, 0, -1, 0, 0, 0, 0, ...) DELTA (0, 1, 1, -1, 0, 0, 0, ...) begins :
1
1, 0
1, 1, 0
1, 1, 2, 0
1, 2, 2, 3, 0
1, 2, 5, 3, 5, 0
1, 3, 5, 10, 5, 8, 0. - _Philippe Deléham_, Apr 02 2012

Crossrefs

Cf. A210867, A208510.

Programs

  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 14;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
v[n_, x_] := (x + n)*u[n - 1, x] + x*v[n - 1, x] - x;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A210866 *)
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A210867 *)

Formula

u(n,x)=u(n-1,x)+x*v(n-1,x),
v(n,x)=(x+1)*u(n-1,x)+(x-1)*v(n-1,x),
where u(1,x)=1, v(1,x)=1.
From Philippe Deléham, Apr 02 2012: (Start)
As DELTA-triangle T(n,k) with 0<=k<=n :
G.f.: (1+x-y*x-y^2*x^2)/(1-y*x-y^2*x^2-x^2).
T(n,k) = T(n-1,k-1) + T(n-2,k) + T(n-2,k-2), T(0,0) = T(1,0) = T(2,0) = T(2,1) = 1, T(1,1) = T(2,2) = 0 and T(n,k) = 0 if k<0 or if k>n. (End)

A140950 a(n) = A140944(n+1) - 3*A140944(n).

Original entry on oeis.org

1, -3, -1, 5, -6, 3, -11, 10, -12, -5, 21, -22, 20, -24, 11, -43, 42, -44, 40, -48, -21, 85, -86, 84, -88, 80, -96, 43, -171, 170, -172, 168, -176, 160, -192, -85, 341, -342, 340, -344, 336, -352, 320, -384, 171, -683, 682, -684, 680, -688
Offset: 0

Views

Author

Paul Curtz, Jul 25 2008

Keywords

Comments

Jacobsthal numbers appear twice: 1) A001045(n+2) signed, terms 0, 1, 3, 6, 10 (A000217); 2) A001045(n+1) signed, terms 0, 2, 5, 9 (n*(n+3)/2=A000096); between them are -3; 5, -6; -11, 10, -12; which appear (opposite sign) by rows in A140503 (1, -1, 2, 3, -2, 4) square.
Consider the permutation of the nonnegative numbers
0, 2, 5, 9, 14, 20, 27,
1, 3, 6, 10, 15, 21, 28,
4, 7, 11, 16, 22, 29,
8, 12, 17, 23, 30,
13, 18, 24, 31,
19, 25, 32,
26, 33,
34, etc.
The corresponding distribution of a(n) is
1, -1, 3, -5, 11, -21, 43,
-3, 5, -11, 21, -43, 85, -171,
-6, 10, -22, 42, -86, 170,
-12, 20, -44, 84, -172,
-24, 40, -88, 168,
-48, 80, -176,
-96, 160,
-192, etc.
Column sums: -2, -2, -10, -10, -42, -42, -170, ... duplicate of a bisection of -A078008(n+2).
b(n)= 1, -1, 3, -5, 11, 21, ... = (-1)^n*A001045(n+1) = A077925(n). Every row is b(n) or b(n+2) multiplied by 1, -1, -2, -4, -8, -16, ..., essentially -A011782(n).

Crossrefs

Cf. A000096, A000217, A005015, A001045, A007283, A020988, A077925, A078008, A140503, A146523, A151575, A175805, A084247.

Programs

  • Mathematica
    T[0, 0] = 0; T[1, 0] = T[0, 1] = 1; T[0, n_] := T[0, n] = T[0, n - 1] + 2*T[0, n - 2]; T[d_, d_] = 0; T[d_, n_] := T[d, n] = T[d - 1, n + 1] - T[d - 1, n]; A140944 = Table[T[d, n], {d, 0, 10}, {n, 0, d}] // Flatten; a[n_] := A140944[[n + 2]] - 3*A140944[[n + 1]]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Dec 18 2014 *)

Extensions

More terms and a(19)=-48 instead of 42 corrected by Jean-François Alcover, Dec 22 2014

A307688 a(n) = 2*a(n-1)-2*a(n-2)+a(n-3)+2*a(n-4) with a(0)=a(1)=0, a(2)=2, a(3)=3.

Original entry on oeis.org

0, 0, 2, 3, 2, 0, 3, 14, 26, 27, 22, 44, 123, 234, 310, 363, 586, 1224, 2259, 3382, 4642, 7227, 13070, 23092, 36555, 54450, 85022, 143883, 245282, 396720, 616803, 973214, 1600106, 2664027, 4334662, 6887804, 10970523, 17828154, 29272390, 47634603, 76493626
Offset: 0

Views

Author

Jean-François Alcover and Paul Curtz, Apr 22 2019

Keywords

Comments

This is an autosequence of the second kind, the companion to A192395.
The array D(n, k) of successive differences begins:
0, 0, 2, 3, 2, 0, 3, 14, 26, 27, ...
0, 2, 1, -1, -2, 3, 11, 12, 1, -5, ...
2, -1, -2, -1, 5, 8, 1, -11, -6, 27, ...
-3, -1, 1, 6, 3, -7, -12, 5, 33, 30, ...
2, 2, 5, -3, -10, -5, 17, 28, -3, -55, ...
0, 3, -8, -7, 5, 22, 11, -31, -52, 13, ...
...
The main diagonal (0,2,-2,6,-10,22,...) is essentially the same as A151575.
It can be seen that abs(D(n, 1)) = D(1, n).
The diagonal starting from the third 0 is -(-1)^n*11*A001045(n), inverse binomial transform of 11*A001045(n).

Links

Crossrefs

Cf. A151575, A192395.
Cf. A001045 (first and fifth upper diagonals), A014551 (second upper diagonal), A115102 (third), A155980 (fourth).

Programs

  • Mathematica
    a[0] = a[1] = 0; a[2] = 2; a[3] = 3; a[n_] := a[n] = 2*a[n-1] - 2*a[n-2] + a[n-3] + 2*a[n-4]; Table[a[n], {n, 0, 40}]
LinearRecurrence[{2,-2,1,2},{0,0,2,3},50] (* Harvey P. Dale, Oct 01 2021 *)
  • PARI
    concat([0,0], Vec(x^2*(2 - x) / ((1 - x - x^2)*(1 - x + 2*x^2)) + O(x^40))) \\ Colin Barker, Apr 22 2019

Formula

G.f.: x^2*(2 - x) / ((1 - x - x^2)*(1 - x + 2*x^2)). - Colin Barker, Apr 22 2019
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