A228577 -id:A228577 - cp's OEIS frontend

A104578 A Padovan convolution triangle.

1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 2, 3, 0, 1, 2, 3, 3, 4, 0, 1, 2, 6, 6, 4, 5, 0, 1, 3, 7, 12, 10, 5, 6, 0, 1, 4, 12, 16, 20, 15, 6, 7, 0, 1, 5, 17, 30, 30, 30, 21, 7, 8, 0, 1, 7, 24, 45, 60, 50, 42, 28, 8, 9, 0, 1, 9, 36, 70, 95, 105, 77, 56, 36, 9, 10, 0, 1, 12, 50, 111, 160, 175, 168, 112, 72

Offset: 0

Paul Barry, Mar 16 2005

A Padovan convolution triangle. See A000931 for the Padovan sequence.
Row sums are tribonacci numbers A000073(n+2). Antidiagonal sums are A008346. The first columns are A000931(n+3), A228577.
From Wolfdieter Lang, Oct 30 2018: (Start)
The alternating row sums give A001057(n+1), for n >= 0.
The inverse of this Riordan triangle is given in A319203.
The row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, with R(-1, x) = 0, appear in the Cayley-Hamilton formula for nonnegative powers of a 3 X 3 matrix with Det M = sigma(3;3) = x1*x2*x3 = +1, sigma(3; 2) := x1*x2 + x1*x*3 + x2*x^3  = -1 and Tr M = sigma(3; 1) = x1 + x2 = x, where x1, x2, and  x3, are the eigenvalues of M, and sigma the elementary symmetric functions, as M^n = R(n-2, x)*M^2  + (R(n-3, x) + R(n-4, x))*M + R(n-3, x)*1_3, for n >= 3, where M^0 = 1_3 is the 3 X 3 unit matrix.
For the Cayley-Hamilton formula for 3 X 3 matrices with Det M = +1, sigma(3,2) = +1 and Tr(M) = x see A321196.
(End)

			From _Wolfdieter Lang_, Oct 30 2018: (Start)
The triangle T begins:
    n\k   0  1  2  3  4  5  6  7  8  9 10 ...
    --------------------------------------
    0:    1
    1:    0  1
    2:    1  0  1
    3:    1  2  0  1
    4:    1  2  3  0  1
    5:    2  3  3  4  0  1
    6:    2  6  6  4  5  0  1
    7:    3  7 12 10  5  6  0  1
    8:    4 12 16 20 15  6  7  0  1
    9:    5 17 30 30 30 21  7  8  0  1
   10:    7 24 45 60 50 42 28  8  9  0  1
   ...
Cayley-Hamilton formula for the tribonacci Q-matrix TQ(x) =[[x,1,1], [1,0,0], [0,1,0]] with Det(TQ) = +1, sigma(3, 2) = -1, and Tr(TQ) = x. For n = 3: TQ(x)^3 = R(1, x)*TQ(x)^2  + (R(0 x) + R(-1, x))*TQ(x) + R(0, x)*1_3 = x*TQ(x)^2 + TQ(x) + 1_3. For x = 1 see also A058265 (powers of the tribonacci constant).
Recurrence: T(6, 2) = T(5, 1) + T(4, 2) + T(3, 2) = 3 + 3 + 0 = 6.
Z- and A- recurrence with A319202 = {1, 0, 1, 1, -1, -3, 0, ...}:
  T(5, 0) = 0*1 + 1*2 + 1*3 + (-1)*0 + (-3)*1 = 2; T(5,2) = 1*2 + 0*3 + 1*0 + 1*1 = 3.
Boas-Buck type recurrence with b = {0, 2, 3, ...}: T(5, 2) = ((1+2)/(5-2)) * (3*1 + 2*0 + 0*3) = 1*3 = 3.
(End)

Tomislav Došlic and Luka Podrug, Tilings of a Honeycomb Strip and Higher Order Fibonacci Numbers, arXiv:2203.11761 [math.CO], 2022.
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A000073, A000931, A001057, A001608, A008346, A058265, A228577, A319202, A319203 (inverse), A321196.

T[n_, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + T[n-2, k] + T[n-3, k]; T[0, 0] = 1; T[, ] = 0; Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Jean-François Alcover, Jun 11 2019 *)

# uses[riordan_array from A256893]
riordan_array( 1/(1 - x^2 - x^3), x/(1 - x^2 - x^3), 8) # Peter Luschny, Nov 09 2018

Riordan array (1/(1 - x^2 - x^3), x/(1 - x^2 - x^3)).
T(n,k) = T(n-1,k-1) + T(n-2,k) + T(n-3,k), T(0,0)=1, T(n,k)=0 if k > n or if k < n. - Philippe Deléham, Jan 08 2014
From Wolfdieter Lang, Oct 30 2018: (Start)
The Riordan property  T = (G(x), x*G(x)) with G(x)= 1/(1-x^2-x^3) implies the following.
G.f. of row polynomials R(n, x) is G(x,z) = 1/(1- x*z - z^2 - z^3).
G.f. of column sequence k:  x^k/(1 - x^2 - x^3)^(k+1), k >= 0.
Boas-Buck recurrence (see the Aug 10 2017 remark in A046521, also for the reference):
T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} b(n-1-j)*T(j, k), for n >= 1, k = 0,1, ..., n-1, and input T(n,n) = 1, for n >= 0. Here b(n) = [x^n]*(d/dx)log(G(x)) = A001608(n+1), for n >= 0.
Recurrences from the A- and Z- sequences (see the W. Lang link under A006232 with references), which are A(n) = A319202(n) and Z(n) = A(n+1).
  T(0, 0) = 1, T(n, k) = 0 for n < k, and
  T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and
  T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), for n >= m >= 1.
(End)

A281791 Ways to tile a 5 X (2n+1) floor with tatami mats, including one monomer.

Original entry on oeis.org

3, 18, 10, 8, 18, 24, 32, 52, 68, 100, 142, 196, 280, 388, 542, 756, 1046, 1452, 2006, 2768, 3816, 5248, 7212, 9896, 13562, 18568, 25392, 34692, 47354, 64580, 88002, 119824, 163034, 221672, 301200, 409004, 555060, 752844, 1020550, 1382732, 1872520, 2534596, 3429206, 4637556, 6269070

Offset: 0

Yasutoshi Kohmoto, Jan 30 2017

Apart from a single 1 X 1 monomer, the area is tiled with 2 X 1 mats. No four mats are permitted to meet at a point.

			For n=0, the 5X1 floor allows the monomer to be placed at one of the two ends or in the middle: a(n=0)=3.

Index entries for linear recurrences with constant coefficients, signature (0,2,2,-1,-2,-1).

Cf. A271786 [3X(2n+1) floor]. 2nd column of A272474.

s1(n)=my(s); forstep(k=(n%4!=1),(n-1)\6,2, s+=((n+3)/4-k/2)*((n-1)/4-k/2)!/(k!*((n-1)/4-3*k/2)!)); 2*s
s3(n)=my(s); forstep(k=(n%4==1),(n-3)\6,2, s+=((n-3)/4-k/2)!/(k!*((n-3)/4-3*k/2)!)); 2*s
s5(n)=my(s); forstep(k=(n%4!=1),(n-5)\6,2, s+=((n+7)/4-k/2)*((n-5)/4-k/2)!/(k!*((n-5)/4-3*k/2)!)); 2*s
a(n)=s1(n) + s3(n) + s5(n) \\ Charles R Greathouse IV, Feb 20 2017

a(n) = S_1(2n+1) + S_5(2n+1) + S_3(2n+1) for n>1 where
S_1(n) = 2* Sum_{k= 0<=k<=[(n-1)/6]} ((n+3)/4-1/2*k) *((n-1)/4-1/2*k)!/(k!*((n-1)/4-3/2*k)!). The sum is over even k if n==1 (mod 4), else over odd k.
S_5(n) = 2* Sum_{0<=k<=[(n-5)/6]} ((n+7)/4-1/2*k) *((n-5)/4-1/2*k)!/(k!*((n-5)/4-3/2*k)!). The sum is over even k if n==1 (mod 4) else over odd k.
S_3(n) = 2* Sum_{0<=k<=[(n-3)/6]} 2*((n-3)/4-1/2*k)!/(k!*((n-3)/4-3/2*k)!). The sum is over odd k if n==1 (mod 4), else over even k.
Where [m] is floor(m).
G.f. x +14*x^3 +2*x*(1 +2*x^2 +3*x^4 -2*x^6 -4*x^8 -2*x^10)/ (1-x^4-x^6)^2. (Includes zeros for even floor widths).- R. J. Mathar, Apr 10 2017
a(n) = 2*(A228577(n-1)+A228577(n+1))+4*(A182097(n-2)+A182097(n-1)), n>1. - R. J. Mathar, Apr 10 2017

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A104578 A Padovan convolution triangle.

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