A157897 -id:A157897 - cp's OEIS frontend

A000073 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = a(1) = 0 and a(2) = 1.

0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080, 98950096, 181997601, 334745777, 615693474, 1132436852

Offset: 0

N. J. A. Sloane

The name "tribonacci number" is less well-defined than "Fibonacci number". The sequence A000073 (which begins 0, 0, 1) is probably the most important version, but the name has also been applied to A000213, A001590, and A081172. - N. J. A. Sloane, Jul 25 2024
Also (for n > 1) number of ordered trees with n+1 edges and having all leaves at level three. Example: a(4)=2 because we have two ordered trees with 5 edges and having all leaves at level three: (i) one edge emanating from the root, at the end of which two paths of length two are hanging and (ii) one path of length two emanating from the root, at the end of which three edges are hanging. - Emeric Deutsch, Jan 03 2004
a(n) is the number of compositions of n-2 with no part greater than 3. Example: a(5)=4 because we have 1+1+1 = 1+2 = 2+1 = 3. - Emeric Deutsch, Mar 10 2004
Let A denote the 3 X 3 matrix [0,0,1;1,1,1;0,1,0]. a(n) corresponds to both the (1,2) and (3,1) entries in A^n. - Paul Barry, Oct 15 2004
Number of permutations satisfying -k <= p(i)-i <= r, i=1..n-2, with k=1, r=2. - Vladimir Baltic, Jan 17 2005
Number of binary sequences of length n-3 that have no three consecutive 0's. Example: a(7)=13 because among the 16 binary sequences of length 4 only 0000, 0001 and 1000 have 3 consecutive 0's. - Emeric Deutsch, Apr 27 2006
Therefore, the complementary sequence to A050231 (n coin tosses with a run of three heads). a(n) = 2^(n-3) - A050231(n-3) - Toby Gottfried, Nov 21 2010
Convolved with the Padovan sequence = row sums of triangle A153462. - Gary W. Adamson, Dec 27 2008
For n > 1: row sums of the triangle in A157897. - Reinhard Zumkeller, Jun 25 2009
a(n+2) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 0, 0, 1; 1, 0, 0] or [1, 1, 0; 1, 0, 1; 1, 0, 0] or [1, 1, 1; 1, 0, 0; 0, 1, 0] or [1, 0, 1; 1, 0, 0; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 0, 1; 1, 1, 1; 0, 1, 0], [0, 1, 0; 0, 1, 1; 1, 1, 0], [0, 0, 1; 1, 0, 1; 0, 1, 1] or [0, 1, 0; 0, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
Also row sums of A082601 and of A082870. - Reinhard Zumkeller, Apr 13 2014
Least significant bits are given in A021913 (a(n) mod 2 = A021913(n)). - Andres Cicuttin, Apr 04 2016
The nonnegative powers of the tribonacci constant t = A058265 are t^n = a(n)*t^2 + (a(n-1) + a(n-2))*t + a(n-1)*1, for  n >= 0, with  a(-1) = 1 and a(-2) = -1. This follows from the recurrences derived from t^3 = t^2 + t + 1. See the example in A058265 for the first nonnegative powers. For the negative powers see A319200. - Wolfdieter Lang, Oct 23 2018
The term "tribonacci number" was coined by Mark Feinberg (1963), a 14-year-old student in the 9th grade of the Susquehanna Township Junior High School in Pennsylvania. He died in 1967 in a motorcycle accident. - Amiram Eldar, Apr 16 2021
Andrews, Just, and Simay (2021, 2022) remark that it has been suggested that this sequence is mentioned in Charles Darwin's Origin of Species as bearing the same relation to elephant populations as the Fibonacci numbers do to rabbit populations. - N. J. A. Sloane, Jul 12 2022

			G.f. = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + 24*x^8 + 44*x^9 + 81*x^10 + ...

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

Cf. A000045, A000078, A000213, A000931, A001590 (first differences, also a(n)+a(n+1)), A001644, A008288 (tribonacci triangle), A008937 (partial sums), A021913, A027024, A027083, A027084, A046738 (Pisano periods), A050231, A054668, A062544, A063401, A077902, A081172, A089068, A118390, A145027, A153462, A230216.
A057597 is this sequence run backwards: A057597(n) = a(1-n).
Row 3 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
Partitions: A240844 and A117546.
Cf. also A092836 (subsequence of primes), A299399 = A092835 + 1 (indices of primes).

a:=[0,0,1];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Oct 24 2018

a000073 n = a000073_list !! n
a000073_list = 0 : 0 : 1 : zipWith (+) a000073_list (tail
                          (zipWith (+) a000073_list $ tail a000073_list))
-- Reinhard Zumkeller, Dec 12 2011

[n le 3 select Floor(n/3) else Self(n-1)+Self(n-2)+Self(n-3): n in [1..70]]; // Vincenzo Librandi, Jan 29 2016

a:= n-> (<<0|1|0>, <0|0|1>, <1|1|1>>^n)[1,3]:
seq(a(n), n=0..40);  # Alois P. Heinz, Dec 19 2016
# second Maple program:
A000073:=proc(n) option remember; if n <= 1 then 0 elif n=2 then 1 else procname(n-1)+procname(n-2)+procname(n-3); fi; end; # N. J. A. Sloane, Aug 06 2018

CoefficientList[Series[x^2/(1 - x - x^2 - x^3), {x, 0, 50}], x]
a[0] = a[1] = 0; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; Array[a, 36, 0] (* Robert G. Wilson v, Nov 07 2010 *)
LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, May 24 2011 *)
a[n_] := SeriesCoefficient[If[ n < 0, x/(1 + x + x^2 - x^3), x^2/(1 - x - x^2 - x^3)], {x, 0, Abs @ n}] (* Michael Somos, Jun 01 2013 *)
Table[-RootSum[-1 - # - #^2 + #^3 &, -#^n - 9 #^(n + 1) + 4 #^(n + 2) &]/22, {n, 0, 20}] (* Eric W. Weisstein, Nov 09 2017 *)

A000073[0]:0$
A000073[1]:0$
A000073[2]:1$
A000073[n]:=A000073[n-1]+A000073[n-2]+A000073[n-3]$
  makelist(A000073[n], n, 0, 40);  /* Emanuele Munarini, Mar 01 2011 */

{a(n) = polcoeff( if( n<0, x / ( 1 + x + x^2 - x^3), x^2 / ( 1 - x - x^2 - x^3) ) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, Sep 03 2007 */

my(x='x+O('x^99)); concat([0, 0], Vec(x^2/(1-x-x^2-x^3))) \\ Altug Alkan, Apr 04 2016

a(n)=([0,1,0;0,0,1;1,1,1]^n)[1,3] \\ Charles R Greathouse IV, Apr 18 2016, simplified by M. F. Hasler, Apr 18 2018

def a(n, adict={0:0, 1:0, 2:1}):
    if n in adict:
        return adict[n]
    adict[n]=a(n-1)+a(n-2)+a(n-3)
    return adict[n] # David Nacin, Mar 07 2012
from functools import cache
@cache
def A000073(n: int) -> int:
    if n <= 1: return 0
    if n == 2: return 1
    return A000073(n-1) + A000073(n-2) + A000073(n-3) # Peter Luschny, Nov 21 2022

G.f.: x^2/(1 - x - x^2 - x^3).
G.f.: x^2 / (1 - x / (1 - x / (1 + x^2 / (1 + x)))). - Michael Somos, May 12 2012
G.f.: Sum_{n >= 0} x^(n+2) *[ Product_{k = 1..n} (k + k*x + x^2)/(1 + k*x + k*x^2) ] = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + ... may be proved by the method of telescoping sums. - Peter Bala, Jan 04 2015
a(n+1)/a(n) -> A058265.  a(n-1)/a(n) -> A192918.
a(n) = central term in M^n * [1 0 0] where M = the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 1 1]. (M^n * [1 0 0] = [a(n-1) a(n) a(n+1)].) a(n)/a(n-1) tends to the tribonacci constant, 1.839286755... = A058265, an eigenvalue of M and a root of x^3 - x^2 - x - 1 = 0. - Gary W. Adamson, Dec 17 2004
a(n+2) = Sum_{k=0..n} T(n-k, k), where T(n, k) = trinomial coefficients (A027907). - Paul Barry, Feb 15 2005
A001590(n) = a(n+1) - a(n); A001590(n) = a(n-1) + a(n-2) for n > 1; a(n) = (A000213(n+1) - A000213(n))/2; A000213(n-1) = a(n+2) - a(n) for n > 0. - Reinhard Zumkeller, May 22 2006
Let C = the tribonacci constant, 1.83928675...; then C^n = a(n)*(1/C) + a(n+1)*(1/C + 1/C^2) + a(n+2)*(1/C + 1/C^2 + 1/C^3). Example: C^4 = 11.444...= 2*(1/C) + 4*(1/C + 1/C^2) + 7*(1/C + 1/C^2 + 1/C^3). - Gary W. Adamson, Nov 05 2006
a(n) = j*C^n + k*r1^n + L*r2^n where C is the tribonacci constant (C = 1.8392867552...), real root of x^3-x^2-x-1=0, and r1 and r2 are the two other roots (which are complex), r1 = m+p*i and r2 = m-p*i, where i = sqrt(-1), m = (1-C)/2 (m = -0.4196433776...) and p = ((3*C-5)*(C+1)/4)^(1/2) = 0.6062907292..., and where j = 1/((C-m)^2 + p^2) = 0.1828035330..., k = a+b*i, and L = a-b*i, where a = -j/2 = -0.0914017665... and b = (C-m)/(2*p*((C-m)^2 + p^2)) = 0.3405465308... . - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n+1) = 3*c*((1/3)*(a+b+1))^n/(c^2-2*c+4) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3), c=(586+102*sqrt(33))^(1/3). Round to the nearest integer. - Al Hakanson (hawkuu(AT)gmail.com), Feb 02 2009
a(n) = round(3*((a+b+1)/3)^n/(a^2+b^2+4)) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3).. - Anton Nikonov
Another form of the g.f.: f(z) = (z^2-z^3)/(1-2*z+z^4). Then we obtain a(n) as a sum: a(n) = Sum_{i=0..floor((n-2)/4)} ((-1)^i*binomial(n-2-3*i,i)*2^(n-2-4*i)) - Sum_{i=0..floor((n-3)/4)} ((-1)^i*binomial(n-3-3*i,i)*2^(n-3-4*i)) with natural convention: Sum_{i=m..n} alpha(i) = 0 for m > n. - Richard Choulet, Feb 22 2010
a(n+2) = Sum_{k=0..n} Sum_{i=k..n, mod(4*k-i,3)=0} binomial(k,(4*k-i)/3)*(-1)^((i-k)/3)*binomial(n-i+k-1,k-1). - Vladimir Kruchinin, Aug 18 2010
a(n) = 2*a(n-2) + 2*a(n-3) + a(n-4). - Gary Detlefs, Sep 13 2010
Sum_{k=0..2*n} a(k+b)*A027907(n,k) = a(3*n+b), b >= 0 (see A099464, A074581).
a(n) = 2*a(n-1) - a(n-4), with a(0)=a(1)=0, a(2)=a(3)=1. - Vincenzo Librandi, Dec 20 2010
Starting (1, 2, 4, 7, ...) is the INVERT transform of (1, 1, 1, 0, 0, 0, ...). - Gary W. Adamson, May 13 2013
G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1 - x*(4*k+1 + x + x^2)/( x*(4*k+3 + x + x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 09 2013
a(n+2) = Sum_{j=0..floor(n/2)} Sum_{k=0..j} binomial(n-2*j,k)*binomial(j,k)*2^k. - Tony Foster III, Sep 08 2017
Sum_{k=0..n} (n-k)*a(k) = (a(n+2) + a(n+1) - n - 1)/2. See A062544. - Yichen Wang, Aug 20 2020
a(n) = A008937(n-1) - A008937(n-2) for n >= 2. - Peter Luschny, Aug 20 2020
From Yichen Wang, Aug 27 2020: (Start)
Sum_{k=0..n} a(k) = (a(n+2) + a(n) - 1)/2. See A008937.
Sum_{k=0..n} k*a(k) = ((n-1)*a(n+2) - a(n+1) + n*a(n) + 1)/2. See A337282. (End)
For n > 1, a(n) = b(n) where b(1) = 1 and then b(n) = Sum_{k=1..n-1} b(n-k)*A000931(k+2). - J. Conrad, Nov 24 2022
Conjecture: the congruence a(n*p^(k+1)) + a(n*p^k) + a(n*p^(k-1)) == 0 (mod p^k) holds for positive integers k and n and for all the primes p listed in A106282. - Peter Bala, Dec 28 2022
Sum_{k=0..n} k^2*a(k) = ((n^2-4*n+6)*a(n+1) - (2*n^2-2*n+5)*a(n) + (n^2-2*n+3)*a(n-1) - 3)/2. - Prabha Sivaramannair, Feb 10 2024
a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(3*r^2-2*r-1). - Fabian Pereyra, Nov 23 2024

Minor edits by M. F. Hasler, Apr 18 2018

Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A059259 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-x-x*y-y^2) = 1/((1+y)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 2, 2, 0, 1, 3, 4, 2, 1, 1, 4, 7, 6, 3, 0, 1, 5, 11, 13, 9, 3, 1, 1, 6, 16, 24, 22, 12, 4, 0, 1, 7, 22, 40, 46, 34, 16, 4, 1, 1, 8, 29, 62, 86, 80, 50, 20, 5, 0, 1, 9, 37, 91, 148, 166, 130, 70, 25, 5, 1, 1, 10, 46, 128, 239, 314, 296, 200

Offset: 0

N. J. A. Sloane, Jan 23 2001

This sequence provides the general solution to the recurrence a(n) = a(n-1) + k*(k+1)*a(n-2), a(0)=a(1)=1. The solution is (1, 1, k^2 + k + 1, 2*k^2 + 2*k + 1, ...) whose coefficients can be read from the rows of the triangle. The row sums of the triangle are given by the case k=1. These are the Jacobsthal numbers, A001045. Viewed as a square array, its first row is (1,0,1,0,1,...) with e.g.f. cosh(x), g.f. 1/(1-x^2) and subsequent rows are successive partial sums given by 1/((1-x)^n * (1-x^2)). - Paul Barry, Mar 17 2003
Conjecture: every second column of this triangle is identical to a column in the square array A071921. For example, column 4 of A059259 (1, 3, 9, 22, 46, ...) appears to be the same as column 3 of A071921; column 6 of A059259 (1, 4, 16, 50, 130, 296, ...) appears to be the same as column 4 of A071921; and in general column 2k of A059259 appears to be the same as column k+1 of A071921. Furthermore, since A225010 is a transposition of A071921 (ignoring the latter's top row and two leftmost columns), there appears to be a correspondence between column 2k of A059259 and row k of A225010. - Mathew Englander, May 17 2014
T(n,k) is the number of n-tilings of a (one-dimensional) board that use k (1,1)-fence tiles and n-k squares. A (1,1)-fence is a tile composed of two pieces of width 1 separated by a gap of width 1. - Michael A. Allen, Jun 25 2020
See the Edwards-Allen 2020 paper, page 14, for proof of Englander's conjecture. - Michael De Vlieger, Dec 10 2020

			Triangle begins:
  1;
  1,  0;
  1,  1,  1;
  1,  2,  2,   0;
  1,  3,  4,   2,   1;
  1,  4,  7,   6,   3,   0;
  1,  5, 11,  13,   9,   3,   1;
  1,  6, 16,  24,  22,  12,   4,   0;
  1,  7, 22,  40,  46,  34,  16,   4,  1;
  1,  8, 29,  62,  86,  80,  50,  20,  5,  0;
  1,  9, 37,  91, 148, 166, 130,  70, 25,  5, 1;
  1, 10, 46, 128, 239, 314, 296, 200, 95, 30, 6, 0;
...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Joseph Briggs, Alex Parker, Coy Schwieder, and Chris Wells, Frogs, hats and common subsequences, arXiv:2404.07285 [math.CO], 2024. See p. 28.
Kenneth Edwards and Michael A. Allen, New Combinatorial Interpretations of the Fibonacci Numbers Squared, Golden Rectangle Numbers, and Jacobsthal Numbers Using Two Types of Tile, arXiv:2009.04649 [math.CO], 2020.

See A059260 for an explicit formula.
Diagonals of this triangle are given by A006498.
Similar to the triangles A035317, A080242, A108561, A112555.
Cf. A123521, A157897.

read transforms; 1/(1-x-x*y-y^2); SERIES2(%,x,y,12); SERIES2TOLIST(%,x,y,12);

T[n_, 0]:= 1; T[n_, n_]:= (1+(-1)^n)/2; T[n_, k_]:= T[n, k] = T[n-1, k] + T[n-1, k-1]; Table[T[n, k], {n, 0, 10} , {k, 0, n}]//Flatten (* G. C. Greubel, Jan 03 2017 *)

{T(n,k) = if(k==0, 1, if(k==n, (1+(-1)^n)/2, T(n-1,k) +T(n-1,k-1)) )};
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Apr 29 2019

def A059259_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return (-1)^n
        if k==0: return 0
        return prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
    return [(-1)^(n-k+1)*prec(n+1, k) for k in (1..n)]
for n in (1..12): print(A059259_row(n)) # Peter Luschny, Mar 16 2016

G.f.: 1/(1 - x - x*y - y^2).
As a square array read by antidiagonals, this is T(n, k) = Sum_{i=0..n} (-1)^(n-i)*C(i+k, k). - Paul Barry, Jul 01 2003
T(2*n,n) = A026641(n). - Philippe Deléham, Mar 08 2007
T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = T(1,0) = T(2,0) = T(2,1) = T(2,2)=1, T(1,1)=0, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 24 2013
T(n,0) = 1, T(n,n) = (1+(-1)^n)/2, and T(n,k) = T(n-1,k) + T(n-1,k-1) for 0 < k < n. - Mathew Englander, May 24 2014
From Michael A. Allen, Jun 25 2020: (Start)
T(n,k) + T(n-1,k-1) = binomial(n,k) if n >= k > 0.
T(2*n-1,2*n-2) = T(2*n,2*n-1) = n, T(2*n,2*n-2) = n^2, T(2*n+1,2*n-1) = n*(n+1) for n > 0.
T(n,2) = binomial(n-2,2) + n - 1 for n > 1 and T(n,3) = binomial(n-3,3) + 2*binomial(n-2,2) for n > 2.
T(2*n-k,k) = A123521(n,k). (End)

A123521 Triangle read by rows: T(n,k)=number of tilings of a 2 X n grid with k pieces of 1 X 2 tiles (in horizontal position) and 2n-2k pieces of 1 X 1 tiles (0<=k<=n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 4, 4, 1, 6, 11, 6, 1, 1, 8, 22, 24, 9, 1, 10, 37, 62, 46, 12, 1, 1, 12, 56, 128, 148, 80, 16, 1, 14, 79, 230, 367, 314, 130, 20, 1, 1, 16, 106, 376, 771, 920, 610, 200, 25, 1, 18, 137, 574, 1444, 2232, 2083, 1106, 295, 30, 1, 1, 20, 172, 832, 2486, 4744, 5776, 4352, 1897, 420, 36

Offset: 0

Emeric Deutsch, Oct 16 2006

Also the triangle of the coefficients of the squares of the Fibonacci polynomials. Row n has 1+2*floor(n/2) terms. Sum of terms in row n = (Fibonacci(n+1))^2 (A007598).
From Michael A. Allen, Jun 24 2020: (Start)
T(n,k) is the number of tilings of an n-board (a board with dimensions n X 1) using k (1/2, 1/2)-fence tiles and 2*(n-k) half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal). A (1/2, 1/2)-fence is a tile composed of two 1/2 X 1 pieces separated by a gap of width 1/2.
T(n,k) is the (n, (n-k))-th entry of the (1/(1-x^2), x/(1-x)^2) Riordan array.
(-1)^(n+k)*T(n,k) is the (n, (n-k))-th entry of the (1/(1-x^2), x/(1+x)^2) Riordan array (A158454). (End)

			T(3,1)=4 because the 1 X 2 tile can be placed in any of the four corners of the 2 X 3 grid.
The irregular triangle begins as:
  1;
  1;
  1,  2,   1;
  1,  4,   4;
  1,  6,  11,   6,    1;
  1,  8,  22,  24,    9;
  1, 10,  37,  62,   46,   12,    1;
  1, 12,  56, 128,  148,   80,   16;
  1, 14,  79, 230,  367,  314,  130,   20,    1;
  1, 16, 106, 376,  771,  920,  610,  200,   25;
  1, 18, 137, 574, 1444, 2232, 2083, 1106,  295,  30,  1;
  1, 20, 172, 832, 2486, 4744, 5776, 4352, 1897, 420, 36;

Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.

G. C. Greubel, Rows n = 0..100 of the irregula triangle, flattened
Feryal Alayont and Evan Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs, J. Int. Seq. (2023) Vol. 26, Art. 23.9.4.
Kenneth Edwards and Michael A. Allen, New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile, arXiv:2009.04649 [math.CO], 2020.
Kenneth Edwards and Michael A. Allen, New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile, JIS 24 (2021) Article 21.3.8.

Cf. A007598, A158454.

Other triangles related to tiling using fences: A059259, A157897, A335964.

function A123521(n,k)
  if k eq 0 then return 1;
  elif k eq 1 then return 2*(n-1);
  else return A123521(n-2,k-2) + Binomial(2*n-k-1, 2*n-2*k-1);
  end if; return A123521;
end function;
[A123521(n,k): k in [0..2*Floor(n/2)], n in [0..14]]; // G. C. Greubel, Sep 01 2022

G:=(1-t*z)/(1+t*z)/(1-z-2*t*z+t^2*z^2): Gser:=simplify(series(G,z=0,14)): for n from 0 to 11 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 11 do seq(coeff(P[n],t,k),k=0..2*floor(n/2)) od; # yields sequence in triangular form

Block[{T}, T[0, 0]= T[1, 0]= 1; T[n_, k_]:= Which[k==0, 1, k==1, 2(n-1), True, T[n -2, k-2] + Binomial[2n-k-1, 2n-2k-1]]; Table[T[n, k], {n, 0, 14}, {k, 0, 2 Floor[n/2]}]] // Flatten (* Michael De Vlieger, Jun 24 2020 *)

@CachedFunction
def T(n,k): # T = A123521
    if (k==0): return 1
    elif (k==1): return 2*(n-1)
    else: return T(n-2, k-2) + binomial(2*n-k-1, 2*n-2*k-1)
flatten([[T(n,k) for k in (0..2*(n//2))] for n in (0..12)]) # G. C. Greubel, Sep 01 2022

G.f.: G = (1-t*z)/((1+t*z)*(1-z-2*t*z+t^2*z^2)). G = 1/(1-g), where g = z+t^2*z^2+2*t*z^2/(1-t*z) is the g.f. of the indecomposable tilings, i.e., of those that cannot be split vertically into smaller tilings. The row generating polynomials are P(n) = (Fibonacci(n))^2. They satisfy the recurrence relation P(n) = (1+t)*(P(n-1) + t*P(n-2)) - t^3*P(n-3).

T(n,k) = T(n-2,k-2) + binomial(2*n-k-1, 2*n-2*k-1). - Michael A. Allen, Jun 24 2020

A335964 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-3,k-1) + T(n-4,k-2) + delta(n,0)*delta(k,0), T(n,k<0) = T(n

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 2, 1, 0, 0, 1, 3, 2, 0, 0, 0, 1, 4, 4, 0, 0, 0, 0, 1, 5, 7, 2, 0, 0, 0, 0, 1, 6, 11, 6, 1, 0, 0, 0, 0, 1, 7, 16, 13, 3, 0, 0, 0, 0, 0, 1, 8, 22, 24, 9, 0, 0, 0, 0, 0, 0, 1, 9, 29, 40, 22, 3, 0, 0, 0, 0, 0, 0

Offset: 0

Michael A. Allen, Jul 01 2020

T(n,k) is the number of tilings of an n-board (a board with dimensions n X 1) using k (1,1)-fence tiles and n-2k square tiles. A (w,g)-fence tile is composed of two tiles of width w separated by a gap of width g.
Sum of n-th row = A006498(n).
T(2*j+r,k) is the coefficient of x^k in (f(j,x))^(2-r)*(f(j+1,x))^r for r=0,1 where f(n,x) is one form of a Fibonacci polynomial defined by f(n+1,x) = f(n,x) + x*f(n-1,x) where f(0,x)=1 and f(n<0,x)=0. - Michael A. Allen, Oct 02 2021

			Triangle begins:
  1;
  1,  0;
  1,  0,  0;
  1,  1,  0,  0;
  1,  2,  1,  0,  0;
  1,  3,  2,  0,  0,  0;
  1,  4,  4,  0,  0,  0,  0;
  1,  5,  7,  2,  0,  0,  0,  0;
  1,  6, 11,  6,  1,  0,  0,  0,  0;
  1,  7, 16, 13,  3,  0,  0,  0,  0,  0;
  1,  8, 22, 24,  9,  0,  0,  0,  0,  0,  0;
  1,  9, 29, 40, 22,  3,  0,  0,  0,  0,  0,  0;
  ...

Kenneth Edwards and Michael A. Allen, New Combinatorial Interpretations of the Fibonacci Numbers Squared, Golden Rectangle Numbers, and Jacobsthal Numbers Using Two Types of Tile, arXiv:2009.04649 [math.CO], 2020.
Kenneth Edwards and Michael A. Allen, New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile, J. Int. Seq. 24 (2021) Article 21.3.8.
John Konvalina, On the number of combinations without unit separation., Journal of Combinatorial Theory, Series A 31.2 (1981): 101-107. See Table I.

Other triangles related to tiling using fences: A059259, A123521, A157897, A158909.

Cf. A006498 (row sums), A011973, A348445.

Mathematica
```
T[n_,k_]:=If[n
				
```

TT(n,k) = if (nA059259
T(n,k) = TT(n-k,k);
\\ matrix(7,7,n,k, T(n-1,k-1)) \\ Michel Marcus, Jul 18 2020

T(n,k) = A059259(n-k,k).
From Michael A. Allen, Oct 02 2021: (Start)
G.f.: 1/((1 + x^2*y)(1 - x - x^2*y)) in the sense that T(n,k) is the coefficient of x^n*y^k in the expansion of the g.f.
T(n,0) = 1.
T(n,1) = n-2 for n>1.
T(n,2) = binomial(n-4,2) + n - 3 for n>3.
T(n,3) = binomial(n-6,3) + 2*binomial(n-5,2) for n>5.
T(4*m-3,2*m-2) = T(4*m-1,2*m-1) = m for m>0.
T(2*n+1,n-k) = A158909(n,k). (End)
T(n,k) = A348445(n-2,k) for n>1.

A350110 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) + T(n-3,k-1) + T(n-3,k-2) + T(n-3,k-3) - T(n-4,k-3) - T(n-4,k-4) + delta(n,0)delta(k,0) - delta(n,1)delta(k,1), T(n

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 2, 3, 2, 0, 1, 3, 5, 4, 0, 0, 1, 4, 8, 8, 4, 2, 1, 1, 5, 12, 16, 13, 9, 3, 0, 1, 6, 17, 28, 30, 22, 9, 0, 0, 1, 7, 23, 45, 58, 51, 27, 9, 3, 1, 1, 8, 30, 68, 103, 108, 78, 40, 18, 4, 0, 1, 9, 38, 98, 171, 211, 187, 123, 58, 16, 0, 0

Offset: 0

Michael A. Allen, Dec 21 2021

This is the m=3 member in the sequence of triangles A007318, A059259, A350110, A350111, A350112 which give the number of tilings of an (n+k) X 1 board using k (1,m-1)-fences and n-k unit square tiles. A (1,g)-fence is composed of two unit square tiles separated by a gap of width g.
It is also the m=3, t=2 member of a two-parameter family of triangles such that T(n,k) is the number of tilings of an (n+(t-1)*k) X 1 board using k (1,m-1;t)-combs and n-k unit square tiles. A (1,g;t)-comb is composed of a line of t unit square tiles separated from each other by gaps of width g. - Michael A. Allen, Dec 27 2021
T(3*j+r-k,k) is the coefficient of x^k in (f(j,x))^(3-r)*(f(j+1,x))^r for r=0,1,2 where f(n,x) is one form of a Fibonacci polynomial defined by f(n+1,x)=f(n,x)+x*f(n-1,x) where f(0,x)=1 and f(n<0,x)=0.
T(n+3-k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 3.
Sum of (n+3)-th antidiagonal (counting initial 1 as the 0th) is A006500(n).

			Triangle begins:
   1;
   1,   0;
   1,   0,   0;
   1,   1,   1,   1;
   1,   2,   3,   2,   0;
   1,   3,   5,   4,   0,   0;
   1,   4,   8,   8,   4,   2,   1;
   1,   5,  12,  16,  13,   9,   3,   0;
   1,   6,  17,  28,  30,  22,   9,   0,   0;
   1,   7,  23,  45,  58,  51,  27,   9,   3,   1;
   1,   8,  30,  68, 103, 108,  78,  40,  18,   4,   0;
   1,   9,  38,  98, 171, 211, 187, 123,  58,  16,   0,   0;
   1,  10,  47, 136, 269, 382, 399, 310, 176,  64,  16,   4,   1;

Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
Michael A. Allen, On A Two-Parameter Family of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.9.8.
Michael A. Allen and Kenneth Edwards, On Two Families of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.7.1.

Other members of the two-parameter family of triangles: A007318 (m=1,t=2), A059259 (m=2,t=2), A350111 (m=4,t=2), A350112 (m=5,t=2), A354665 (m=2,t=3), A354666 (m=2,t=4), A354667 (m=2,t=5), A354668 (m=3,t=3).

Other triangles related to tiling using fences: A123521, A157897, A335964.

Mathematica
```
T[n_, k_]:=If[k<0 || n
				
```

T(n,0) = 1.
T(n,n) = delta(n mod 3,0).
T(n,1) = n-2 for n>1.
T(3*j-r,3*j-p) = 0 for j>0, p=1,2, and r=1,...,p.
T(3*(j-1)+p,3*(j-1)) = T(3*j,3*j-p) = j^p for j>0 and p=0,1,2,3.
T(3*j+1,3*j-1) = 3*j(j+1)/2 for j>0.
T(3*j+2,3*j-2) = 3*(C(j+2,4) + C(j+1,2)^2) for j>1.
G.f. of row sums: (1-x)/((1-2*x)*(1+x^2-x^3)).
G.f. of antidiagonal sums: (1-x^2)/((1-x-x^2)*(1+x^3-x^6)).
T(n,k) = T(n-1,k) + T(n-1,k-1) for n>=2*k+1 if k>=0.

A350111 Triangle read by rows: T(n,k) is the number of tilings of an (n+k)-board using k (1,3)-fences and n-k squares.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 3, 4, 2, 0, 1, 3, 6, 7, 4, 0, 0, 1, 4, 9, 12, 8, 0, 0, 0, 1, 5, 13, 20, 16, 8, 4, 2, 1, 1, 6, 18, 32, 36, 28, 19, 12, 3, 0, 1, 7, 24, 50, 69, 69, 58, 31, 9, 0, 0, 1, 8, 31, 74, 120, 144, 127, 78, 27, 0, 0, 0

Offset: 0

Michael A. Allen, Dec 22 2021

This is the m=4 member in the sequence of triangles A007318, A059259, A350110, A350111, A350112 which give the number of tilings of an (n+k) X 1 board using k (1,m-1)-fences and n-k unit square tiles. A (1,g)-fence is composed of two unit square tiles separated by a gap of width g.
It is also the m=4, t=2 member of a two-parameter family of triangles such that T(n,k) is the number of tilings of an (n+(t-1)*k) X 1 board using k (1,m-1;t)-combs and n-k unit square tiles. A (1,g;t)-comb is composed of a line of t unit square tiles separated from each other by gaps of width g.
T(4*j+r-k,k) is the coefficient of x^k in (f(j,x))^(4-r)*(f(j+1,x))^r for r=0,1,2,3 where f(n,x) is one form of a Fibonacci polynomial defined by f(n+1,x)=f(n,x)+x*f(n-1,x) where f(0,x)=1 and f(n<0,x)=0.
T(n+4-k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 4.
Sum of (n+3)-th antidiagonal (counting initial 1 as the 0th) is A031923(n).

			Triangle begins:
  1;
  1,   0;
  1,   0,   0;
  1,   0,   0,   0;
  1,   1,   1,   1,   1;
  1,   2,   3,   4,   2,   0;
  1,   3,   6,   7,   4,   0,   0;
  1,   4,   9,  12,   8,   0,   0,   0;
  1,   5,  13,  20,  16,   8,   4,   2,   1;
  1,   6,  18,  32,  36,  28,  19,  12,   3,   0;
  1,   7,  24,  50,  69,  69,  58,  31,   9,   0,   0;
  1,   8,  31,  74, 120, 144, 127,  78,  27,   0,   0,   0;
  1,   9,  39, 105, 195, 264, 265, 189,  81,  27,   9,   3,   1;
  1,  10,  48, 144, 300, 458, 522, 432, 270, 132,  58,  24,   4,   0;

Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
Michael A. Allen, On A Two-Parameter Family of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.9.8.
Michael A. Allen and Kenneth Edwards, On Two Families of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.7.1.

Sums of antidiagonals: A031923
Other members of the two-parameter family of triangles: A007318 (m=1,t=2), A059259 (m=2,t=2), A350110 (m=3,t=2), A350112 (m=5,t=2), A354665 (m=2,t=3), A354666 (m=2,t=4), A354667 (m=2,t=5), A354668 (m=3,t=3).
Other triangles related to tiling using fences: A123521, A157897, A335964.

f[n_]:=If[n<0,0,f[n-1]+x*f[n-2]+KroneckerDelta[n,0]];
T[n_, k_]:=Module[{j=Floor[(n+k)/4],r=Mod[n+k,4]},
  Coefficient[f[j]^(4-r)*f[j+1]^r,x,k]];
Flatten@Table[T[n,k], {n, 0, 13}, {k, 0, n}]
(* or *)
T[n_,k_]:=If[k<0 || n

T(n,k) = T(n-1,k) + T(n-2,k-1) - T(n-3,k-1) + T(n-3,k-2) + T(n-4,k-1) + T(n-4,k-3) + 2*T(n-4,k-4) + T(n-5,k-2) + 2*T(n-5,k-3) - T(n-5,k-4) - T(n-6,k-3)-T(n-6,k-5) - T(n-7,k-4)-T(n-7,k-5) - T(n-7,k-6) - T(n-8,k-7)-T(n-8,k-8) + delta(n,0)*delta(k,0) - delta(n,2)*delta(k,1) - delta(n,3)*delta(k,2) - delta(n,4)*delta(k,4) with T(n

T(n,0) = 1.

T(n,n) = delta(n mod 4,0).

T(n,1) = n-3 for n>2.

T(4*j-r,4*j-p) = 0 for j>0, p=1,2,3, and r=1,...,p.

T(4*(j-1)+p,4*(j-1)) = T(4*j,4*j-p) = j^p for j>0 and p=0,1,2,3,4.

T(4*j+1,4*j-1) = 4*j(j+1)/2 for j>0.

T(4*j+2,4*j-2) = 4*C(j+2,4) + 6*C(j+1,2)^2 for j>1.

G.f. of row sums: (1-x-x^3)/((1-2*x)*(1-x^2)*(1+2*x^2+x^3+x^4)).

G.f. of antidiagonal sums: (1-x^2-x^3+x^4-x^6)/((1-x-x^2)*(1-x^4)*(1+3*x^4+x^8)).

T(n,k) = T(n-1,k) + T(n-1,k-1) for n>=3*k+1 if k>=0.

A350112 Triangle read by rows: T(n,k) is the number of tilings of an (n+k)-board using k (1,4)-fences and n-k squares.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 2, 0, 1, 3, 6, 10, 9, 4, 0, 0, 1, 4, 10, 16, 16, 8, 0, 0, 0, 1, 5, 14, 25, 28, 16, 0, 0, 0, 0, 1, 6, 19, 38, 48, 32, 16, 8, 4, 2, 1, 1, 7, 25, 56, 80, 80, 60, 40, 25, 15, 3, 0, 1, 8, 32, 80, 136, 166, 157, 128, 95, 40, 9, 0, 0

Offset: 0

Michael A. Allen, Dec 22 2021

This is the m=5 member in the sequence of triangles A007318, A059259, A350110, A350111, A350112 which give the number of tilings of an (n+k) X 1 board using k (1,m-1)-fences and n-k unit square tiles. A (1,g)-fence is composed of two unit square tiles separated by a gap of width g.
It is also the m=5, t=2 member of a two-parameter family of triangles such that T(n,k) is the number of tilings of an (n+(t-1)*k) X 1 board using k (1,m-1;t)-combs and n-k unit square tiles. A (1,g;t)-comb is composed of a line of t unit square tiles separated from each other by gaps of width g.
T(5*j+r-k,k) is the coefficient of x^k in (f(j,x))^(5-r)*(f(j+1,x))^r for r=0,1,2,3,4 where f(n,x) is one form of a Fibonacci polynomial defined by f(n+1,x)=f(n,x)+x*f(n-1,x) where f(0,x)=1 and f(n<0,x)=0.
T(n+5-k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 5.
Sum of (5j+r)-th antidiagonal (counting initial 1 as the 0th) is f(j)^(5-r)*f(j+1)^r where j=0,1,..., r=0,1,2,3,4, and f(n) is the Fibonacci number A000045(n+1).

			Triangle begins:
  1;
  1,   0;
  1,   0,   0;
  1,   0,   0,   0;
  1,   0,   0,   0,   0;
  1,   1,   1,   1,   1,   1;
  1,   2,   3,   4,   5,   2,   0;
  1,   3,   6,  10,   9,   4,   0,   0;
  1,   4,  10,  16,  16,   8,   0,   0,   0;
  1,   5,  14,  25,  28,  16,   0,   0,   0,   0;
  1,   6,  19,  38,  48,  32,  16,   8,   4,   2,   1;
  1,   7,  25,  56,  80,  80,  60,  40,  25,  15,   3,   0;
  1,   8,  32,  80, 136, 166, 157, 128,  95,  40,   9,   0,   0;
  1,   9,  40, 112, 217, 309, 346, 330, 223, 105,  27,   0,   0,   0;

Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
Michael A. Allen, On A Two-Parameter Family of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.9.8.
Michael A. Allen and Kenneth Edwards, On Two Families of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.7.1.

Other members of the two-parameter family of triangles: A007318 (m=1,t=2), A059259 (m=2,t=2), A350110 (m=3,t=2), A350111 (m=4,t=2), A354665 (m=2,t=3), A354666 (m=2,t=4), A354667 (m=2,t=5), A354668 (m=3,t=3).

Other triangles related to tiling using fences: A123521, A157897, A335964.

f[n_]:=If[n<0,0,f[n-1]+x*f[n-2]+KroneckerDelta[n,0]];
T[n_, k_]:=Module[{j=Floor[(n+k)/5], r=Mod[n+k,5]},
  Coefficient[f[j]^(5-r)*f[j+1]^r,x,k]];
Flatten@Table[T[n,k], {n, 0, 13}, {k, 0, n}]

T(n,0) = 1.
T(n,n) = delta(n mod 5,0).
T(n,1) = n-4 for n>3.
T(5*j-r,5*j-p) = 0 for j>0, p=1,2,3,4, and r=1,...,p.
T(5*(j-1)+p,5*(j-1)) = T(5*j,5*j-p) = j^p for j>0 and p=0,1,...,5.
T(5*j+1,5*j-1) = 5*j(j+1)/2 for j>0.
T(5*j+2,5*j-2) = 5*C(j+2,4) + 10*C(j+1,2)^2 for j>1.
T(n,k) = T(n-1,k) + T(n-1,k-1) for n >= 4*k+1 if k >= 0.

A354665 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) + T(n-2,k-2) + T(n-3,k-1) - T(n-3,k-3) + delta(n,0)delta(k,0) - delta(n,1)delta(k,1), T(n

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 1, 2, 0, 1, 2, 4, 0, 1, 1, 3, 6, 3, 3, 0, 1, 4, 9, 8, 9, 0, 1, 1, 5, 13, 17, 18, 6, 4, 0, 1, 6, 18, 30, 36, 20, 16, 0, 1, 1, 7, 24, 48, 66, 55, 40, 10, 5, 0, 1, 8, 31, 72, 114, 120, 100, 40, 25, 0, 1, 1, 9, 39, 103, 186

Offset: 0

Michael A. Allen, Jun 04 2022

This is the m=2, t=3 member of a two-parameter family of triangles such that T(n,k) is the number of tilings of an (n+(t-1)*k) X 1 board using k (1,m-1;t)-combs and n-k unit square tiles. A (1,g;t)-comb is composed of a line of t unit square tiles separated from each other by gaps of width g.
T(2*j+r-2*k,k) is the coefficient of x^k in (f(j,x))^(2-r)*(f(j+1,x))^r for  r=0,1, where f(n,x) is a Narayana's cows polynomial defined by f(n,x)=f(n-1,x)+x*f(n-3,x)+delta(n,0) where f(n<0,x)=0.
T(n+4-2*k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 2 or 4.

			Triangle begins:
  1;
  1,   0;
  1,   0,   1;
  1,   1,   2,   0;
  1,   2,   4,   0,   1;
  1,   3,   6,   3,   3,   0;
  1,   4,   9,   8,   9,   0,   1;
  1,   5,  13,  17,  18,   6,   4,   0;
  1,   6,  18,  30,  36,  20,  16,   0,   1;
  1,   7,  24,  48,  66,  55,  40,  10,   5,   0;
  1,   8,  31,  72, 114, 120, 100,  40,  25,   0,   1;
  1,   9,  39, 103, 186, 234, 221, 135,  75,  15,   6,   0;
...

Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
Michael A. Allen, On A Two-Parameter Family of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.9.8.

Row sums are A011782.
Sums over k of T(n-2*k,k) are A224809.
Other members of the family of triangles: A007318 (m=1,t=2), A059259 (m=2,t=2), A350110 (m=3,t=2), A350111 (m=4,t=2), A350112 (m=5,t=2), A354666 (m=2,t=4), A354667 (m=2,t=5), A354668 (m=3,t=3).
Other triangles related to tiling using combs: A059259, A123521, A157897, A335964.

Mathematica
```
T[n_, k_]:=If[k<0 || n
				
```

T(n,0) = 1.
T(n,n) = delta(n mod 2,0).
T(n,1) = n-2 for n>1.
T(2*j-r,2*j-1) = 0 for j>0, r=0,1.
T(2*(j-1)+p,2*(j-1)) = j^p for j>0 and p=0,1,2.
T(2*(j-1)+3,2*(j-1)) = j^2*(j+1)/2 for j>0.
T(2*j+p,2*j-p) = C(j+1,2)^p for j>0 and p=0,1,2.
G.f. of row sums: (1-x)/(1-2*x).
G.f. of sums of T(n-2*k,k) over k: (1-x^3)/((1-x-x^3)*(1+x^4-x^6)).
T(n,k) = T(n-1,k) + T(n-1,k-1) for n>=2*k+1 if k>=0.

A354666 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-2,k-1) + 2T(n-2,k-2) - T(n-3,k-1) - T(n-3,k-2) + T(n-4,k-1) + T(n-4,k-2) - T(n-4,k-3) - T(n-4,k-4) + delta(n,0)delta(k,0) - delta(n,2)*(delta(k,1) + delta(k,2)), T(n

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 0, 2, 0, 1, 1, 4, 0, 1, 1, 2, 6, 0, 3, 0, 1, 3, 9, 4, 9, 0, 1, 1, 4, 12, 10, 18, 0, 4, 0, 1, 5, 16, 21, 36, 10, 16, 0, 1, 1, 6, 21, 36, 60, 30, 40, 0, 5, 0, 1, 7, 27, 57, 100, 81, 100, 20, 25, 0, 1, 1, 8, 34, 84, 158, 168

Offset: 0

Michael A. Allen, Jun 04 2022

This is the m=2, t=4 member of a two-parameter family of triangles such that T(n,k) is the number of tilings of an (n+(t-1)*k) X 1 board using k (1,m-1;t)-combs and n-k unit square tiles. A (1,g;t)-comb is composed of a line of t unit square tiles separated from each other by gaps of width g.
T(2*j+r-3*k,k) is the coefficient of x^k in (f(j,x))^(2-r)*(f(j+1,x))^r for r=0,1, where f(n,x) is a (1,4)-bonacci polynomial defined by f(n,x)=f(n-1,x)+x*f(n-4,x)+delta(n,0) where f(n<0,x)=0.
T(n+6-3*k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 2, 4, or 6.

			Triangle begins:
  1;
  1,   0;
  1,   0,   1;
  1,   0,   2,   0;
  1,   1,   4,   0,   1;
  1,   2,   6,   0,   3,   0;
  1,   3,   9,   4,   9,   0,   1;
  1,   4,  12,  10,  18,   0,   4,   0;
  1,   5,  16,  21,  36,  10,  16,   0,   1;
  1,   6,  21,  36,  60,  30,  40,   0,   5,   0;
  1,   7,  27,  57, 100,  81, 100,  20,  25,   0,   1;
  1,   8,  34,  84, 158, 168, 200,  70,  75,   0,   6,   0;
  1,   9,  42, 118, 243, 322, 400, 231, 225,  35,  36,   0,   1;
...

Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
Michael A. Allen, On A Two-Parameter Family of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.9.8.

Row sums are A099163.
Sums over k of T(n-3*k,k) are A224808.
Other members of the family of triangles: A007318 (m=1,t=2), A059259 (m=2,t=2), A350110 (m=3,t=2), A350111 (m=4,t=2), A350112 (m=5,t=2), A354665 (m=2,t=3), A354667 (m=2,t=5), A354668 (m=3,t=3).
Other triangles related to tiling using combs: A059259, A123521, A157897, A335964.

Mathematica
```
T[n_,k_]:=If[k<0 || n
				
```

T(n,0) = 1.
T(n,n) = delta(n mod 2,0).
T(n,1) = n-3 for n>2.
T(2*j-r,2*j-1) = 0 for j>0, r=-1,0,1.
T(2*(j-1)+p,2*(j-1)) = j^p for j>0 and p=0,1,2.
T(2*j+p,2*(j-1)) = j^2*((j+1)/2)^p for j>0 and p=1,2.
T(2*j+3,2*(j-1)) = (j*(j+1))^2*(j+2)/12 for j>0.
T(2*(j+p),2*j-p) = C(j+2,3)^p for j>0 and p=0,1,2.
G.f. of row sums: (1-2*x^2)/(1-x-3*x^2+2*x^3).
G.f. of sums of T(n-3*k,k) over k: (1-x^5-x^8)/(1-x-x^5+x^6-x^7-2*x^8+x^9-x^10+x^13+x^16).
T(n,k) = T(n-1,k) + T(n-1,k-1) for n>=3*k+1 if k>=0.

A354667 Triangle read by rows: T(n,k) is the number of tilings of an (n+4*k) X 1 board using k (1,1;5)-combs and n-k squares.

Original entry on oeis.org

1, 1, 0, 1, 0, 1, 1, 0, 2, 0, 1, 0, 4, 0, 1, 1, 1, 6, 0, 3, 0, 1, 2, 9, 0, 9, 0, 1, 1, 3, 12, 5, 18, 0, 4, 0, 1, 4, 16, 12, 36, 0, 16, 0, 1, 1, 5, 20, 25, 60, 15, 40, 0, 5, 0, 1, 6, 25, 42, 100, 42, 100, 0, 25, 0, 1, 1, 7, 31, 66, 150, 112, 200

Offset: 0

Michael A. Allen, Jun 05 2022

This is the m=2, t=5 member of a two-parameter family of triangles such that T(n,k) is the number of tilings of an (n+(t-1)*k) X 1 board using k (1,m-1;t)-combs and n-k unit square tiles. A (1,g;t)-comb is composed of a line of t unit square tiles separated from each other by gaps of width g.
T(2*j+r-4*k,k) is the coefficient of x^k in (f(j,x))^(2-r)*(f(j+1,x))^r for r=0,1, where f(n,x) is a (1,5)-bonacci polynomial defined by f(n,x)=f(n-1,x)+x*f(n-5,x)+delta(n,0) where f(n<0,x)=0.
T(n+8-4*k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 2, 4, 6, or 8.

			Triangle begins:
  1;
  1,   0;
  1,   0,   1;
  1,   0,   2,   0;
  1,   0,   4,   0,   1;
  1,   1,   6,   0,   3,   0;
  1,   2,   9,   0,   9,   0,   1;
  1,   3,  12,   5,  18,   0,   4,   0;
  1,   4,  16,  12,  36,   0,  16,   0,   1;
  1,   5,  20,  25,  60,  15,  40,   0,   5,   0;
  1,   6,  25,  42, 100,  42, 100,   0,  25,   0,   1;
  1,   7,  31,  66, 150, 112, 200,  35,  75,   0,   6,   0;
...

Michael A. Allen, On a Two-Parameter Family of Generalizations of Pascal's Triangle, arXiv:2209.01377 [math.CO], 2022.
Michael A. Allen, On A Two-Parameter Family of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.9.8.

Row sums are A005578.
Sums over k of T(n-4*k,k) are A224811.
Other members of the family of triangles: A007318 (m=1,t=2), A059259 (m=2,t=2), A350110 (m=3,t=2), A350111 (m=4,t=2), A350112 (m=5,t=2), A354665 (m=2,t=3), A354666 (m=2,t=4), A354668 (m=3,t=3).
Other triangles related to tiling using combs: A059259, A123521, A157897, A335964.

Mathematica
```
T[n_,k_]:=If[k<0 || n
				
```

T(n,k) = T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) + 2*T(n-2,k-2) + T(n-3,k-1) - T(n-3,k-2) - 2*T(n-3,k-3) - T(n-4,k-1) + T(n-4,k-2) + T(n-4,k-3) - T(n-4,k-4) + T(n-5,k-1) - 2*T(n-5,k-3) + T(n-5,k-5) + delta(n,0)*delta(k,0) - delta(n,1)*delta(k,1) - delta(n,2)*delta(k,2) - delta(n,3)*(delta(k,1) - delta(k,3)) with T(n,k<0) = T(n

T(n,0) = 1.

T(n,n) = delta(n mod 2,0).

T(n,1) = n-4 for n>3.

T(2*j+r,2*j-1) = 0 for j>0, r=-1,0,1,2.

T(n,2*j) = C(n/2,j)^2 for j>0 and n even and 2*j <= n <= 2*j+8.

T(n,2*j) = C((n-1)/2,j)*C((n+1)/2,j) for j>0 and n odd and 2*j < n < 2*j+8.

T(2*j+3*p,2*j-p) = C(j+3,4)^p for j>0 and p=0,1,2.

G.f. of row sums: (1-x-x^2)/(1-2*x-x^2+2*x^3).

G.f. of sums of T(n-4*k,k) over k: (1-x^5-x^7-x^10+x^15)/(1-x-x^5+x^6-x^7+x^8-x^9-2*x^10+x^11-x^12+2*x^15-x^16+2*x^17+x^20-x^25).

T(n,k) = T(n-1,k) + T(n-1,k-1) for n>=4*k+1 if k>=0.

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cp's OEIS Frontend

A000073 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = a(1) = 0 and a(2) = 1.

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A059259 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-x-x*y-y^2) = 1/((1+y)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

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A123521 Triangle read by rows: T(n,k)=number of tilings of a 2 X n grid with k pieces of 1 X 2 tiles (in horizontal position) and 2n-2k pieces of 1 X 1 tiles (0<=k<=n).

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A335964 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-3,k-1) + T(n-4,k-2) + delta(n,0)*delta(k,0), T(n,k<0) = T(n

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A350110 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) + T(n-3,k-1) + T(n-3,k-2) + T(n-3,k-3) - T(n-4,k-3) - T(n-4,k-4) + delta(n,0)*delta(k,0) - delta(n,1)*delta(k,1), T(n

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Mathematica

Formula

A350111 Triangle read by rows: T(n,k) is the number of tilings of an (n+k)-board using k (1,3)-fences and n-k squares.

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A350110 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) + T(n-3,k-1) + T(n-3,k-2) + T(n-3,k-3) - T(n-4,k-3) - T(n-4,k-4) + delta(n,0)delta(k,0) - delta(n,1)delta(k,1), T(n

A354665 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-1,k-1) - T(n-2,k-1) + T(n-2,k-2) + T(n-3,k-1) - T(n-3,k-3) + delta(n,0)delta(k,0) - delta(n,1)delta(k,1), T(n

A354666 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-2,k-1) + 2T(n-2,k-2) - T(n-3,k-1) - T(n-3,k-2) + T(n-4,k-1) + T(n-4,k-2) - T(n-4,k-3) - T(n-4,k-4) + delta(n,0)delta(k,0) - delta(n,2)*(delta(k,1) + delta(k,2)), T(n