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.

Showing 1-3 of 3 results.

A211955 Triangle of coefficients of a polynomial sequence related to the Morgan-Voyce polynomials A085478.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 6, 10, 4, 1, 10, 30, 28, 8, 1, 15, 70, 112, 72, 16, 1, 21, 140, 336, 360, 176, 32, 1, 28, 252, 840, 1320, 1056, 416, 64, 1, 36, 420, 1848, 3960, 4576, 2912, 960, 128, 1, 45, 660, 3696, 10296, 16016, 14560, 7680, 2176, 256
Offset: 0

Views

Author

Peter Bala, Apr 30 2012

Keywords

Comments

Let b(n,x) = Sum_{k = 0..n} binomial(n+k,2*k)*x^k denote the Morgan-Voyce polynomials of A085478. This triangle lists the coefficients (in ascending powers of x) of the related polynomial sequence R(n,x) := (1/2)*b(n,2*x) + 1/2. Several sequences already in the database are of the form (R(n,x))n>=0 for a fixed value of x. These include A101265 (x = 1), A011900 (x = 2), A182432 (x = 3), A054318 (x = 4) as well as signed versions of A133872 (x = -1), A109613(x = -2), A146983 (x = -3) and A084159 (x = -4).
The polynomials R(n,x) factorize in the ring Z[x] as R(n,x) = P(n,x)*P(n+1,x) for n >= 1: explicitly, P(2*n,x) = 1/2*(b(2*n,2*x) + 1)/b(n,2*x) and P(2*n+1,x) = b(n,2*x). The coefficients of P(n,x) occur in several tables in the database, although without the connection to the Morgan-Voyce polynomials being noted - see A211956 for more details. In terms of T(n,x), the Chebyshev polynomials of the first kind, we have P(2*n,x) = T(2*n,u) and P(2*n+1,x) = 1/u * T(2*n+1,u), where u = sqrt((x+2)/2). Hence R(n,x) = 1/u * T(n,u) * T(n+1,u).

Examples

			Triangle begins
.n\k.|..0....1....2....3....4....5....6
= = = = = = = = = = = = = = = = = = = =
..0..|..1
..1..|..1....1
..2..|..1....3....2
..3..|..1....6...10....4
..4..|..1...10...30...28....8
..5..|..1...15...70..112...72...16
..6..|..1...21..140..336..360..176...32
		

Crossrefs

Formula

T(n,0) = 1; T(n,k) = 2^(k-1)*binomial(n+k,2*k) for k > 0.
O.g.f. for column k (except column 0): 2^(k-1)*x^k/(1-x)^(2*k+1).
O.g.f.: (1-t*(x+2)+t^2)/((1-t)*(1-2*t(x+1)+t^2)) = 1 + (1+x)*t + (1+3*x+2*x^2)*t^2 + ....
Removing the first column from the triangle produces the Riordan array (x/(1-x)^3, 2*x/(1-x)^2).
The row polynomials R(n,x) := 1/2*b(n,2*x) + 1/2 = 1 + x*Sum_{k = 1..n} binomial(n+k,2*k)*(2*x)^(k-1).
Recurrence equation: R(n,x) = 2*(1+x)*R(n-1,x) - R(n-2,x) - x with initial conditions R(0,x) = 1, R(1,x) = 1+x.
Another recurrence is R(n,x)*R(n-2,x) = R(n-1,x)*(R(n-1,x) + x).
With P(n,x) as defined in the Comments section we have (x+2)/x - {Sum_{k = 0..2n} 1/R(k,x)}^2 = 2/(x*P(2*n+1,x)^2); (x+2)/x - {Sum_{k = 0..2n+1} 1/R(k,x)}^2 = (x+2)/(x*P(2*n+2,x)^2); consequently Sum_{k >= 0} 1/R(k,x) = sqrt((x+2)/x) for either x > 0 or x <= -2.
Row sums R(n,1) = A101265(n+1); Alt. row sums R(n,-1) = A133872(n+1);
R(n,2) = A011900(n); R(n,-2) = (-1)^n * A109613(n); R(n,3) = A182432;
R(n,-3) = (-1)^n * A146983(n); R(n,4) = A054318(n+1); R(n,-4) = (-1)^n * A084159(n).

A211956 Coefficients of a sequence of polynomials related to the Morgan-Voyce polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 4, 2, 1, 6, 4, 1, 9, 12, 4, 1, 12, 20, 8, 1, 16, 40, 32, 8, 1, 20, 60, 56, 16, 1, 25, 100, 140, 80, 16, 1, 30, 140, 224, 144, 32, 1, 36, 210, 448, 432, 192, 32, 1, 42, 280, 672, 720, 352, 64, 1, 49, 392, 1176, 1680, 1232, 448, 64
Offset: 0

Views

Author

Peter Bala, Apr 30 2012

Keywords

Comments

The row generating polynomials R(n,x) of A211955 factorize in the ring Z[x] as R(n,x) = P(n,x)*P(n+1,x) for n >= 1: explicitly, P(2*n,x) = 1/2*(b(2*n,2*x) + 1)/b(n,2*x) and P(2*n+1,x) = b(n,2*x), where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478. This triangle lists the coefficients in ascending powers of x of the polynomials P(n,x).
The odd numbered rows of the present triangle produce triangle A123519; the even numbered row entries are recorded separately in A211957 and appear to equal the unsigned and row reversed form of A204021. The even numbered rows with a factor of 2^(k-1) removed from the k-th column entries produce triangle A208513.

Examples

			Triangle begins
.n\k.|..0....1....2....3....4
= = = = = = = = = = = = = = =
..0..|..1
..1..|..1
..2..|..1....1
..3..|..1....2
..4..|..1....4....2
..5..|..1....6....4
..6..|..1....9...12....4
..7..|..1...12...20....8
..8..|..1...16...40...32....8
..9..|..1...20...60...56...16
...
		

Crossrefs

Formula

T(n,0) = 1; for k > 0, T(2*n,k) = 2^k * binomial(n+k,2*k) = A123519(n,k);
for k > 0, T(2*n-1,k) = n/(n+k)*(2^k)*binomial(n+k,2*k) = 2^(k-1)*A208513(n,k).
O.g.f.: ((1+t)*(1-t^2)-t^2*x)/((1-t^2)^2-2*t^2*x) = 1 + t + (1+x)*t^2 + (1+2*x)*t^3 + (1+4*x+2*x^2)*t^4 + ....
Row generating polynomials: P(2*n,x) := 1/2*(b(2*n,2*x)+1)/b(n,2*x) and P(2*n+1,x) := b(n,2*x), where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478.
The product P(n,x)*P(n+1,x) is the n-th row polynomial of A211955.
In terms of T(n,x), the Chebyshev polynomials of the first kind, we have P(2*n,x) = T(2*n,u) and P(2*n+1,x) = 1/u*T(2*n+1,u), where u = sqrt((x+2)/2).
Other representations for the row polynomials include
P(2*n,x) = 1/2*(1+x+sqrt(x^2+2*x))^n + 1/2*(1+x-sqrt(x^2+2*x))^n;
P(2*n,x) = n*Sum_{k = 0..n}(-1)^(n-k)/(n+k) * binomial(n+k,2*k) * (2*x+4)^k for n >= 1;
P(2*n+1,x) = (2*n+1)*Sum_{k=0..n} (-1)^(n-k)/(n+k+1) * binomial(n+k+1,2*k+1) * (2*x+4)^k.
Recurrence equation: P(n+1,x)*P(n-2,x) - P(n,x)*P(n-1,x) = x.
Row sums A005246(n+2).

A123520 Number of vertical dominoes in all possible tilings of a 2n X 3 grid by dominoes.

Original entry on oeis.org

4, 28, 152, 744, 3436, 15284, 66224, 281424, 1178196, 4874444, 19973192, 81189688, 327817404, 1316035940, 5257118560, 20909651104, 82849544868, 327163551612, 1288036695544, 5057236343176, 19807689093644, 77408388584724
Offset: 1

Views

Author

Emeric Deutsch, Oct 16 2006

Keywords

Examples

			a(1) = 4 because a 2 X 3 grid can be tiled in 3 ways with dominoes: 3 horizontal dominoes, 1 horizontal domino above two adjacent vertical dominoes and 1 horizontal domino below two adjacent vertical dominoes; these have altogether 4 vertical dominoes.
		

Crossrefs

Programs

  • Maple
    a:=n->sum(k*2^(k+1)*binomial(n+k,2*k),k=0..n): seq(a(n),n=1..24);
  • Mathematica
    FullSimplify[Table[(2+Sqrt[3])^n*((1+Sqrt[3])*n+1/Sqrt[3])/3 + (2-Sqrt[3])^n*((1-Sqrt[3])*n-1/Sqrt[3])/3,{n,1,20}]] (* Vaclav Kotesovec, Nov 29 2012 *)
    Table[Sum[2^(k + 1)*k*Binomial[n + k, 2 k], {k, 0, n}], {n, 0, 50}] (* G. C. Greubel, Oct 14 2017 *)
  • PARI
    z='z+O('z^50); Vec(4*z*(1-z)/(1-4*z+z^2)^2) \\ G. C. Greubel, Oct 14 2017

Formula

a(n) = Sum_{k=0..n} 2^(k+1) * k * C(n+k,2*k).
a(n) = Sum_{k=0..n} k * A123519(n,k).
G.f.: 4*z*(1-z)/(1-4*z+z^2)^2.
a(n) = (2+sqrt(3))^n*((1+sqrt(3))*n+1/sqrt(3))/3 + (2-sqrt(3))^n*((1-sqrt(3))*n-1/sqrt(3))/3. - Vaclav Kotesovec, Nov 29 2012
Showing 1-3 of 3 results.