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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A006498 a(n) = a(n-1) + a(n-3) + a(n-4), a(0) = a(1) = a(2) = 1, a(3) = 2.

Original entry on oeis.org

1, 1, 1, 2, 4, 6, 9, 15, 25, 40, 64, 104, 169, 273, 441, 714, 1156, 1870, 3025, 4895, 7921, 12816, 20736, 33552, 54289, 87841, 142129, 229970, 372100, 602070, 974169, 1576239, 2550409, 4126648, 6677056, 10803704, 17480761, 28284465, 45765225, 74049690, 119814916
Offset: 0

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Author

Keywords

Comments

Number of compositions of n into 1's, 3's and 4's. - Len Smiley, May 08 2001
The sum of any two alternating terms (terms separated by one term) produces a number from the Fibonacci sequence. (e.g. 4+9=13, 9+25=34, 6+15=21, etc.) Taking square roots starting from the first term and every other term after, we get the Fibonacci sequence. - Sreyas Srinivasan (sreyas_srinivasan(AT)hotmail.com), May 02 2002
(1 + x + 2*x^2 + x^3)/(1 - x - x^3 - x^4) = 1 + 2*x + 4*x^2 + 6*x^3 + 9*x^4 + 15*x^5 + 25*x^6 + 40*x^7 + ... is the g.f. for the number of binary strings of length where neither 101 nor 111 occur. [Lozansky and Rousseau] Or, equivalently, where neither 000 nor 010 occur.
Equivalently, a(n+2) is the number of length-n binary strings with no two set bits with distance 2; see fxtbook link. - Joerg Arndt, Jul 10 2011
a(n) is the number of words written with the letters "a" and "b", with the following restriction: any "a" must be followed by at least two letters, the second of which is a "b". - Bruno Petazzoni (bpetazzoni(AT)ac-creteil.fr), Oct 31 2005. [This is also equivalent to the previous two conditions.]
Let a(0) = 1, then a(n) = partial products of Product_{n>2} (F(n)/F(n-1))^2 = 1*1*2*2*(3/2)*(3/2)*(5/3)*(5/3)*(8/5)*(8/5)*.... E.g., a(7) = 15 = 1*1*1*2*2*(3/2)*(3/2)*(5/3). - Gary W. Adamson, Dec 13 2009
Number of permutations satisfying -k <= p(i) - i <= r and p(i)-i not in I, i=1..n, with k=1, r=3, I={1}. - Vladimir Baltic, Mar 07 2012
The 2-dimensional version, which counts sets of pairs no two of which are separated by graph-distance 2, is A273461. - Gus Wiseman, Nov 27 2019
a(n+1) is the number of multus bitstrings of length n with no runs of 4 ones. - Steven Finch, Mar 25 2020

Examples

			G.f. = 1 + x + x^2 + 2*x^3 + 4*x^4 + 6*x^5 + 9*x^6 + 15*x^7 + 25*x^8 + 40*x^9 + ...
From _Gus Wiseman_, Nov 27 2019: (Start)
The a(2) = 1 through a(7) = 15 subsets with no two elements differing by 2:
  {}  {}   {}     {}     {}     {}
      {1}  {1}    {1}    {1}    {1}
           {2}    {2}    {2}    {2}
           {1,2}  {3}    {3}    {3}
                  {1,2}  {4}    {4}
                  {2,3}  {1,2}  {5}
                         {1,4}  {1,2}
                         {2,3}  {1,4}
                         {3,4}  {1,5}
                                {2,3}
                                {2,5}
                                {3,4}
                                {4,5}
                                {1,2,5}
                                {1,4,5}
(End)
		

References

  • E. Lozansky and C. Rousseau, Winning Solutions, Springer, 1996; see pp. 157 and 172.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A060945 (for 1's, 2's and 4's). Essentially the same as A074677.
Diagonal sums of number triangle A059259.
Numbers whose binary expansion has no subsequence (1,0,1) are A048716.

Programs

  • Haskell
    a006498 n = a006498_list !! n
    a006498_list = 1 : 1 : 1 : 2 : zipWith (+) (drop 3 a006498_list)
       (zipWith (+) (tail a006498_list) a006498_list)
    -- Reinhard Zumkeller, Apr 07 2012
  • Magma
    [ n eq 1 select 1 else n eq 2 select 1 else n eq 3 select 1 else n eq 4 select 2 else Self(n-1)+Self(n-3)+ Self(n-4): n in [1..40] ]; // Vincenzo Librandi, Aug 20 2011
    
  • Mathematica
    LinearRecurrence[{1,0,1,1},{1,1,1,2},50] (* Harvey P. Dale, Jul 13 2011 *)
    Table[Fibonacci[Floor[n/2] + 2]^Mod[n, 2]*Fibonacci[Floor[n/2] + 1]^(2 - Mod[n, 2]), {n, 0, 40}] (* David Nacin, Feb 29 2012 *)
    a[ n_] := Fibonacci[ Quotient[ n+2, 2]] Fibonacci[ Quotient[ n+3, 2]] (* Michael Somos, Jan 19 2014 *)
    Table[Length[Select[Subsets[Range[n]],!MatchQ[#,{_,x_,_,y_,_}/;x+2==y]&]],{n,10}] (* Gus Wiseman, Nov 27 2019 *)
  • PARI
    {a(n) = fibonacci( (n+2)\2 ) * fibonacci( (n+3)\2 )} /* Michael Somos, Mar 10 2004 */
    
  • PARI
    Vec(1/(1-x-x^3-x^4)+O(x^66))
    
  • Python
    def a(n, adict={0:1, 1:1, 2:1, 3:2}):
        if n in adict:
            return adict[n]
        adict[n]=a(n-1)+a(n-3)+a(n-4)
        return adict[n] # David Nacin, Mar 07 2012
    

Formula

G.f.: 1 / ((1 + x^2) * (1 - x - x^2)); a(2*n) = F(n+1)^2, a(2*n - 1) = F(n+1)*F(n). a(n) = a(-4-n) * (-1)^n. - Michael Somos, Mar 10 2004
The g.f. -(1+z+2*z^2+z^3)/((z^2+z-1)*(1+z^2)) for the truncated version 1, 2, 4, 6, 9, 15, 25, 40, ... was given in the Simon Plouffe thesis of 1992. [edited by R. J. Mathar, May 13 2008]
From Vladeta Jovovic, May 03 2002: (Start)
a(n) = round((-(1/5)*sqrt(5) - 1/5)*(-2*1/(-sqrt(5)+1))^n/(-sqrt(5)+1) + ((1/5)*sqrt(5) - 1/5)*(-2*1/( sqrt(5)+1))^n/(sqrt(5)+1)).
G.f.: 1/(1-x-x^2)/(1+x^2). (End)
a(n) = (-i)^n*Sum{k=0..n} U(n-2k, i/2) where i^2=-1. - Paul Barry, Nov 15 2003
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*F(n-2k+1). - Paul Barry, Oct 12 2007
F(floor(n/2) + 2)^(n mod 2)*F(floor(n/2) + 1)^(2 - (n mod 2)) where F(n) is the n-th Fibonacci number. - David Nacin, Feb 29 2012
a(2*n - 1) = A001654(n), a(2*n) = A007598(n+1). - Michael Somos, Mar 10 2004
a(n+1)*a(n+3) = a(n)*a(n+2) + a(n+1)*a(n+2) for all n in Z. - Michael Somos, Jan 19 2014
a(n) = round(1/(1/F(n+2) + 2/F(n+3))), where F(n) = A000045, and 0.5 is rounded to 1. - Richard R. Forberg, Aug 04 2014
5*a(n) = (-1)^floor(n/2)*A000034(n+1) + A000032(n+2). - R. J. Mathar, Sep 16 2017
a(n) = Sum_{j=0..floor(n/3)} Sum_{k=0..j} binomial(n-3j,k)*binomial(j,k)*2^k. - Tony Foster III, Sep 18 2017
E.g.f.: (2*cos(x) + sin(x) + exp(x/2)*(3*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2)))/5. - Stefano Spezia, Mar 12 2024

A059260 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

Original entry on oeis.org

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

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Author

N. J. A. Sloane, Jan 23 2001

Keywords

Comments

Coefficients of the (left, normalized) shifted cyclotomic polynomial. Or, coefficients of the basic n-th q-series for q=-2. Indeed, let Y_n(x) = Sum_{k=0..n} x^k, having as roots all the n-th roots of unity except for 0; then coefficients in x of (-1)^n Y_n(-x-1) give exactly the n-th row of A059260 and a practical way to compute it. - Olivier Gérard, Jul 30 2002
The maximum in the (2n)-th row is T(n,n), which is A026641; also T(n,n) ~ (2/3)*binomial(2n,n). The maximum in the (2n-1)-th row is T(n-1,n), which is A014300 (but T does not have the same definition as in A026637); also T(n-1,n) ~ (1/3)*binomial(2n,n). Here is a generalization of the formula given in A026641: T(i,j) = Sum_{k=0..j} binomial(i+k-x,j-k)*binomial(j-k+x,k) for all x real (the proof is easy by induction on i+j using T(i,j) = T(i-1,j) + T(i,j-1)). - Claude Morin, May 21 2002
The second greatest term in the (2n)-th row is T(n-1,n+1), which is A014301; the second greatest term in the (2n+1)-th row is T(n+1,n) = 2*T(n-1,n+1), which is 2*A014301. - Claude Morin
Diagonal sums give A008346. - Paul Barry, Sep 23 2004
Riordan array (1/(1-x^2), x/(1-x)). As a product of Riordan arrays, factors into the product of (1/(1+x),x) and (1/(1-x),1/(1-x)) (binomial matrix). - Paul Barry, Oct 25 2004
Signed version is A239473 with relations to partial sums of sequences. - Tom Copeland, Mar 24 2014
From Robert Coquereaux, Oct 01 2014: (Start)
Columns of the triangle (cf. Example below) give alternate partial sums along nw-se diagonals of the Pascal triangle, i.e., sequences A000035, A004526, A002620 (or A087811), A002623 (or A173196), A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808, etc.
The dimension of the space of closed currents (distributional forms) of degree p on Gr(n), the Grassmann algebra with n generators, equivalently, the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence, is V(n,p) = 2^n T(p,n-1) - (-1)^p.
If p is odd V(n,p) is also the dimension of the cyclic cohomology group of order p of the Z2 graded algebra Gr(n).
If p is even the dimension of this cohomology group is V(n,p)+1.
Cf. A193844. (End)
From Peter Bala, Feb 07 2024: (Start)
The following remarks assume the row indexing starts at n = 1.
The sequence of row polynomials R(n,x), beginning R(1,x) = 1, R(2,x) = x, R(3,x) = 1 + x + x^2 , ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd( R(n,x), R(m,x)) = R(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the polynomial sequence {R(n,x): n >= 1} is a divisibility sequence; that is, if n divides m then R(n,x) divides R(m,x) in Z[x]. (End)
From Miquel A. Fiol, Oct 04 2024: (Start)
For j>=1, T(i,j) is the independence number of the (i-j)-supertoken graph FF_(i-j)(S_j) of the star graph S_j with j points.
(Given a graph G on n vertices and an integer k>=1, the k-supertoken (or reduced k-th power) FF_k(G) of G has vertices representing configurations of k indistinguishable tokens in the (not necessarily different) vertices of G, with two configurations being adjacent if one can be obtained from the other by moving one token along an edge. See an example below.)
Following the suggestion of Peter Munn, the k-supertoken graph FF_k(S_j) can also be defined as follows: Consider the Lattice graph L(k,j), whose vertices are the k^j j-vectors with elements in the set {0,..,k-1}, two being adjacent if they differ in just one coordinate by one unity. Then, FF_k(S_j) is the subgraph of L(k+1,j) induced by the vertices at distance at most k from (0,..,0). (End)

Examples

			Triangle begins
  1;
  0,  1;
  1,  1,  1;
  0,  2,  2,  1;
  1,  2,  4,  3,  1;
  0,  3,  6,  7,  4,  1;
  1,  3,  9, 13, 11,  5,  1;
  0,  4, 12, 22, 24, 16,  6,  1;
  1,  4, 16, 34, 46, 40, 22,  7,  1;
  0,  5, 20, 50, 80, 86, 62, 29,  8,  1;
Sequences obtained with _Miquel A. Fiol_'s Sep 30 2024 formula of A(n,c1,c2) for other values of (c1,c2). (In the table, rows are indexed by c1=0..6 and columns by c2=0..6):
A000007  A000012  A000027  A025747  A000292* A000332* A000389*
A059841  A008619  A087811* A002623  A001752  A001753  A001769
A193356  A008794* A005993  A005994  -------  -------  -------
-------  -------  -------  A005995  A018210  -------  A052267
-------  -------  -------  -------  A018211  A018212  -------
-------  -------  -------  -------  -------  A018213  A018214
-------  -------  -------  -------  -------  -------  A062136
*requires offset adjustment.
The 2-supertoken FF_2(S_3) of the star graph S_3 with central vertex 1 and peripheral vertices 2,3,4. (The vertex `ij' of FF_2(S_3) represents the configuration of one token in `ì' and the other token in `j'). The T(5,3)=7 independent vertices are 22, 24, 44, 23, 11, 34, and 33.
     22--12---24---14---44
          | \    / |
         23   11   34
            \  |  /
              13
               |
              33
		

Crossrefs

Cf. A059259. Row sums give A001045.
Seen as a square array read by antidiagonals this is the coefficient of x^k in expansion of 1/((1-x^2)*(1-x)^n) with rows A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808 etc. (allowing for signs). A058393 would then effectively provide the table for nonpositive n. - Henry Bottomley, Jun 25 2001

Programs

  • Maple
    read transforms; 1/(1-y-x*y-x^2); SERIES2(%,x,y,12); SERIES2TOLIST(%,x,y,12);
  • Mathematica
    t[n_, k_] := Sum[ (-1)^(n-j)*Binomial[j, k], {j, 0, n}]; Flatten[ Table[t[n, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Oct 20 2011, after Paul Barry *)
  • PARI
    T(n, k) = sum(j=0, n, (-1)^(n - j)*binomial(j, k));
    for(n=0, 12, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 11 2017
    
  • Python
    from sympy import binomial
    def T(n, k): return sum((-1)**(n - j)*binomial(j, k) for j in range(n + 1))
    for n in range(13): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 11 2017
  • Sage
    def A059260_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            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, n-k+1) for k in (1..n)]
    for n in (1..9): print(A059260_row(n)) # Peter Luschny, Mar 16 2016
    

Formula

G.f.: 1/(1-y-x*y-x^2) = 1 + y + x^2 + xy + y^2 + 2x^2y + 2xy^2 + y^3 + ...
E.g.f: (exp(-t)+(x+1)*exp((x+1)*t))/(x+2). - Tom Copeland, Mar 19 2014
O.g.f. (n-th row): ((-1)^n+(x+1)^(n+1))/(x+2). - Tom Copeland, Mar 19 2014
T(i, 0) = 1 if i is even or 0 if i is odd, T(0, i) = 1 and otherwise T(i, j) = T(i-1, j) + T(i, j-1); also T(i, j) = Sum_{m=j..i+j} (-1)^(i+j+m)*binomial(m, j). - Robert FERREOL, May 17 2002
T(i, j) ~ (i+j)/(2*i+j)*binomial(i+j, j); more precisely, abs(T(i, j)/binomial(i+j, j) - (i+j)/(2*i+j) )<=1/(4*(i+j)-2); the proof is by induction on i+j using the formula 2*T(i, j) = binomial(i+j, j)+T(i, j-1). - Claude Morin, May 21 2002
T(n, k) = Sum_{j=0..n} (-1)^(n-j)binomial(j, k). - Paul Barry, Aug 25 2004
T(n, k) = Sum_{j=0..n-k} binomial(n-j, j)*binomial(j, n-k-j). - Paul Barry, Jul 25 2005
Equals A097807 * A007318. - Gary W. Adamson, Feb 21 2007
Equals A128173 * A007318 as infinite lower triangular matrices. - Gary W. Adamson, Feb 17 2007
Equals A130595*A097805*A007318 = (inverse Pascal matrix)*(padded Pascal matrix)*(Pascal matrix) = A130595*A200139. Inverse is A097808 = A130595*(padded A130595)*A007318. - Tom Copeland, Nov 14 2016
T(i, j) = binomial(i+j, j)-T(i-1, j). - Laszlo Major, Apr 11 2017
Recurrence for row polynomials (with row indexing starting at n = 1): R(n,x) = x*R(n-1,x) + (x + 1)*R(n-2,x) with R(1,x) = 1 and R(2,x) = x. - Peter Bala, Feb 07 2024
From Miquel A. Fiol, Sep 30 2024: (Start)
The triangle can be seen as a slice of a 3-dimensional table that links it to well-known sequences as follows.
The j-th column of the triangle, T(i,j) for i >= j, equals A(n,c1,c2) = Sum_{k=0..floor(n/2)} binomial(c1+2*k-1,2*k)*binomial(c2+n-2*k-1,n-2*k) when c1=1, c2=j, and n=i-j.
This gives T(i,j) = Sum_{k=0..floor((i-j)/2)} binomial(i-2*k-1, j-1). For other values of (c1,c2), see the example below. (End)

Extensions

Formula corrected by Philippe Deléham, Jan 11 2014

A035317 Pascal-like triangle associated with A000670.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 4, 2, 1, 4, 7, 6, 3, 1, 5, 11, 13, 9, 3, 1, 6, 16, 24, 22, 12, 4, 1, 7, 22, 40, 46, 34, 16, 4, 1, 8, 29, 62, 86, 80, 50, 20, 5, 1, 9, 37, 91, 148, 166, 130, 70, 25, 5, 1, 10, 46, 128, 239, 314, 296, 200, 95, 30, 6, 1, 11, 56, 174, 367, 553, 610, 496, 295, 125
Offset: 0

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Author

Keywords

Comments

From Johannes W. Meijer, Jul 20 2011: (Start)
The triangle sums, see A180662 for their definitions, link this "Races with Ties" triangle with several sequences, see the crossrefs. Observe that the Kn4 sums lead to the golden rectangle numbers A001654 and that the Fi1 and Fi2 sums lead to the Jacobsthal sequence A001045.
The series expansion of G(x, y) = 1/((y*x-1)*(y*x+1)*((y+1)*x-1)) as function of x leads to this sequence, see the second Maple program. (End)
T(2n,k) = the number of hatted frog arrangements with k frogs on the 2xn grid. See the linked paper "Frogs, hats and common subsequences". - Chris Cox, Apr 12 2024

Examples

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

Crossrefs

Row sums are A000975, diagonal sums are A080239.
Central terms are A014300.
Similar to the triangles A059259, A080242, A108561, A112555.
Cf. A059260.
Triangle sums (see the comments): A000975 (Row1), A059841 (Row2), A080239 (Kn11), A052952 (Kn21), A129696 (Kn22), A001906 (Kn3), A001654 (Kn4), A001045 (Fi1, Fi2), A023435 (Ca2), Gi2 (A193146), A190525 (Ze2), A193147 (Ze3), A181532 (Ze4). - Johannes W. Meijer, Jul 20 2011
Cf. A181971.

Programs

  • Haskell
    a035317 n k = a035317_tabl !! n !! k
    a035317_row n = a035317_tabl !! n
    a035317_tabl = map snd $ iterate f (0, [1]) where
       f (i, row) = (1 - i, zipWith (+) ([0] ++ row) (row ++ [i]))
    -- Reinhard Zumkeller, Jul 09 2012
    
  • Maple
    A035317 := proc(n,k): add((-1)^(i+k) * binomial(i+n-k+1, i), i=0..k) end: seq(seq(A035317(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Jul 20 2011
    A035317 := proc(n,k): coeff(coeftayl(1/((y*x-1)*(y*x+1)*((y+1)*x-1)), x=0, n), y, k) end: seq(seq(A035317(n,k), k=0..n), n=0..10); # Johannes W. Meijer, Jul 20 2011
  • Mathematica
    t[n_, k_] := (-1)^k*(((-1)^k*(n+2)!*Hypergeometric2F1[1, n+3, k+2, -1])/((k+1)!*(n-k+1)!) + 2^(k-n-2)); Flatten[ Table[ t[n, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, Dec 14 2011, after Johannes W. Meijer *)
  • PARI
    {T(n,k)=if(n==k,(n+2)\2,if(k==0,1,if(n>k,T(n-1,k-1)+T(n-1,k))))}
    for(n=0,12,for(k=0,n,print1(T(n,k),","));print("")) \\ Paul D. Hanna, Jul 18 2012
    
  • Sage
    def A035317_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            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)^k*prec(n+2, k) for k in (1..n)]
    for n in (1..11): print(A035317_row(n)) # Peter Luschny, Mar 16 2016

Formula

T(n,k) = Sum_{j=0..floor(n/2)} binomial(n-2j, k-2j). - Paul Barry, Feb 11 2003
From Johannes W. Meijer, Jul 20 2011: (Start)
T(n, k) = Sum_{i=0..k}((-1)^(i+k) * binomial(i+n-k+1,i)). (Mendelson)
T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = floor(n/2) + 1. (Mendelson)
Sum_{k = 0..n}((-1)^k * (n-k+1)^n * T(n, k)) = A000670(n). (Mendelson)
T(n, n-k) = A128176(n, k); T(n+k, n-k) = A158909(n, k); T(2*n-k, k) = A092879(n, k). (End)
T(2*n+1,n) = A014301(n+1); T(2*n+1,n+1) = A026641(n+1). - Reinhard Zumkeller, Jul 19 2012

Extensions

More terms from James Sellers

A108561 Triangle read by rows: T(n,0)=1, T(n,n)=(-1)^n, T(n+1,k)=T(n,k-1)+T(n,k) for 0 < k < n.

Original entry on oeis.org

1, 1, -1, 1, 0, 1, 1, 1, 1, -1, 1, 2, 2, 0, 1, 1, 3, 4, 2, 1, -1, 1, 4, 7, 6, 3, 0, 1, 1, 5, 11, 13, 9, 3, 1, -1, 1, 6, 16, 24, 22, 12, 4, 0, 1, 1, 7, 22, 40, 46, 34, 16, 4, 1, -1, 1, 8, 29, 62, 86, 80, 50, 20, 5, 0, 1, 1, 9, 37, 91, 148, 166, 130, 70, 25, 5, 1, -1, 1, 10, 46, 128, 239, 314, 296, 200, 95, 30, 6, 0, 1, 1, 11, 56, 174, 367
Offset: 0

Views

Author

Reinhard Zumkeller, Jun 10 2005

Keywords

Comments

Sum_{k=0..n} T(n,k) = A078008(n);
Sum_{k=0..n} abs(T(n,k)) = A052953(n-1) for n > 0;
T(n,1) = n - 2 for n > 1;
T(n,2) = A000124(n-3) for n > 2;
T(n,3) = A003600(n-4) for n > 4;
T(n,n-6) = A001753(n-6) for n > 6;
T(n,n-5) = A001752(n-5) for n > 5;
T(n,n-4) = A002623(n-4) for n > 4;
T(n,n-3) = A002620(n-1) for n > 3;
T(n,n-2) = A008619(n-2) for n > 2;
T(n,n-1) = n mod 2 for n > 0;
T(2*n,n) = A072547(n+1).
Sum_{k=0..n} T(n,k)*x^k = A232015(n), A078008(n), A000012(n), A040000(n), A001045(n+2), A140725(n+1) for x = 2, 1, 0, -1, -2, -3 respectively. - Philippe Deléham, Nov 17 2013, Nov 19 2013
(1,a^n) Pascal triangle with a = -1. - Philippe Deléham, Dec 27 2013
T(n,k) = A112465(n,n-k). - Reinhard Zumkeller, Jan 03 2014

Examples

			From _Philippe Deléham_, Nov 17 2013: (Start)
Triangle begins:
  1;
  1, -1;
  1,  0,  1;
  1,  1,  1, -1;
  1,  2,  2,  0,  1;
  1,  3,  4,  2,  1, -1;
  1,  4,  7,  6,  3,  0,  1; (End)
		

Crossrefs

Cf. A007318 (a=1), A008949(a=2), A164844(a=10).
Similar to the triangles A035317, A059259, A080242, A112555.
Cf. A072547 (central terms).

Programs

  • GAP
    Flat(List([0..13],n->List([0..n],k->Sum([0..k],i->Binomial(n,i)*(-2)^(k-i))))); # Muniru A Asiru, Feb 19 2018
  • Haskell
    a108561 n k = a108561_tabl !! n !! k
    a108561_row n = a108561_tabl !! n
    a108561_tabl = map reverse a112465_tabl
    -- Reinhard Zumkeller, Jan 03 2014
    
  • Maple
    A108561 := (n, k) -> add(binomial(n, i)*(-2)^(k-i), i = 0..k):
    seq(seq(A108561(n,k), k = 0..n), n = 0..12); # Peter Bala, Feb 18 2018
  • Mathematica
    Clear[t]; t[n_, 0] = 1; t[n_, n_] := t[n, n] = (-1)^Mod[n, 2]; t[n_, k_] := t[n, k] = t[n-1, k] + t[n-1, k-1]; Table[t[n, k], {n, 0, 13}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)
  • Sage
    def A108561_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            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)^k*prec(n, k) for k in (1..n-1)]+[(-1)^(n+1)]
    for n in (1..12): print(A108561_row(n)) # Peter Luschny, Mar 16 2016
    

Formula

G.f.: (1-y*x)/(1-x-(y+y^2)*x). - Philippe Deléham, Nov 17 2013
T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 17 2013
From Peter Bala, Feb 18 2018: (Start)
T(n,k) = Sum_{i = 0..k} binomial(n,i)*(-2)^(k-i), 0 <= k <= n.
The n-th row polynomial is the n-th degree Taylor polynomial of the rational function (1 + x)^n/(1 + 2*x) about 0. For example, for n = 4, (1 + x)^4/(1 + 2*x) = 1 + 2*x + 2*x^2 + x^4 + O(x^5). (End)

Extensions

Definition corrected by Philippe Deléham, Dec 26 2013

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

Views

Author

Emeric Deutsch, Oct 16 2006

Keywords

Comments

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)

Examples

			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;
		

References

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

Crossrefs

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

Programs

  • Magma
    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
    
  • Maple
    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
  • Mathematica
    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 *)
  • SageMath
    @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

Formula

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

A157897 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-3,k-3) + delta(n,0)*delta(k,0), T(n,k<0) = T(n

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 0, 1, 1, 3, 1, 2, 0, 1, 4, 3, 3, 2, 0, 1, 5, 6, 5, 6, 0, 1, 1, 6, 10, 9, 12, 3, 3, 0, 1, 7, 15, 16, 21, 12, 6, 3, 0, 1, 8, 21, 27, 35, 30, 14, 12, 0, 1, 1, 9, 28, 43, 57, 61, 35, 30, 6, 4, 0, 1, 10, 36, 65, 91, 111, 81, 65, 30, 10, 4, 0, 1, 11, 45, 94, 142, 189, 169, 135, 90, 30, 20, 0, 1
Offset: 0

Views

Author

Gary W. Adamson, Mar 08 2009

Keywords

Comments

T(n, k) is the number of tilings of an n-board that use k (1/2, 1)-fences and n-k squares. A (1/2, 1)-fence is a tile composed of two pieces of width 1/2 separated by a gap of width 1. (Result proved in paper by K. Edwards - see the links section.) - Michael A. Allen, Apr 28 2019
T(n, k) is the (n, n-k)-th entry in the (1/(1-x^3), x*(1+x)/(1-x^3)) Riordan array. - Michael A. Allen, Mar 11 2021

Examples

			First few rows of the triangle are:
  1;
  1,  0;
  1,  1,  0;
  1,  2,  0,  1;
  1,  3,  1,  2,  0;
  1,  4,  3,  3,  2,  0;
  1,  5,  6,  5,  6,  0,  1;
  1,  6, 10,  9, 12,  3,  3,  0;
  1,  7, 15, 16, 21, 12,  6,  3,  0;
  1,  8, 21, 27, 35, 30, 14, 12,  0,  1;
  ...
T(9,3) = 27 = T(8,3) + T(7,2) + T(6,0) = 16 + 10 + 1.
		

Crossrefs

Cf. A000073 (row sums), A006498, A120415.
Other triangles related to tiling using fences: A059259, A123521, A335964.

Programs

  • Magma
    function T(n,k) // T = A157897
      if k lt 0 or k gt n then return 0;
      elif k eq 0 then return 1;
      else return T(n-1, k) + T(n-2, k-1) + T(n-3, k-3);
      end if; return T;
    end function;
    [T(n,k): k in [0..n], n in [0..14]]; // G. C. Greubel, Sep 01 2022
    
  • Mathematica
    T[n_,k_]:= If[nMichael A. Allen, Apr 28 2019 *)
  • SageMath
    def T(n,k): # T = A157897
        if (k<0 or k>n): return 0
        elif (k==0): return 1
        else: return T(n-1, k) + T(n-2, k-1) + T(n-3, k-3)
    flatten([[T(n,k) for k in (0..n)] for n in (0..14)]) # G. C. Greubel, Sep 01 2022

Formula

T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-3,k-3) + delta(n,0)*delta(k,0), T(n,k<0) = T(n
Sum_{k=0..n} T(n, k) = A000073(n+2). - Reinhard Zumkeller, Jun 25 2009
From G. C. Greubel, Sep 01 2022: (Start)
T(n, k) = T(n-1, k) + T(n-2, k-1) + T(n-3, k-3), with T(n, 0) = 1.
T(n, n) = A079978(n).
T(n, n-1) = A087508(n), n >= 1.
T(n, 1) = A001477(n-1).
T(n, 2) = A161680(n-2).
Sum_{k=0..floor(n/2)} T(n-k, k) = A120415(n). (End)

Extensions

Name clarified by Michael A. Allen, Apr 28 2019
Definition improved by Michael A. Allen, Mar 11 2021

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

Author

Michael A. Allen, Jul 01 2020

Keywords

Comments

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

Examples

			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;
  ...
		

Crossrefs

Other triangles related to tiling using fences: A059259, A123521, A157897, A158909.
Cf. A006498 (row sums), A011973, A348445.

Programs

  • Mathematica
    T[n_,k_]:=If[n
    				
  • PARI
    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

Formula

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.

A081297 Array T(k,n), read by antidiagonals: T(k,n) = ((k+1)^(n+1)-(-k)^(n+1))/(2k+1).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 7, 5, 1, 1, 1, 13, 13, 11, 1, 1, 1, 21, 25, 55, 21, 1, 1, 1, 31, 41, 181, 133, 43, 1, 1, 1, 43, 61, 461, 481, 463, 85, 1, 1, 1, 57, 85, 991, 1281, 2653, 1261, 171, 1, 1, 1, 73, 113, 1891, 2821, 10501, 8425, 4039, 341, 1, 1, 1, 91, 145, 3305
Offset: 0

Author

Paul Barry, Mar 17 2003

Keywords

Comments

Square array of solutions of a family of recurrences.
Rows of the array give solutions to the recurrences a(n)=a(n-1)+k(k-1)a(n-2), a(0)=a(1)=1.
Subarray of array in A072024. - Philippe Deléham, Nov 24 2013

Examples

			Rows begin
  1, 1,  1,  1,   1,    1, ...
  1, 1,  3,  5,  11,   21, ...
  1, 1,  7, 13,  55,  133, ...
  1, 1, 13, 25, 181,  481, ...
  1, 1, 21, 41, 461, 1281, ...
		

Crossrefs

Columns include A002061, A001844, A072025.
Diagonals include A081298, A081299, A081300, A081301, A081302.

Programs

  • Mathematica
    T[n_, k_]:=((n + 1)^(k + 1) - (-n)^(k + 1)) / (2n + 1); Flatten[Table[T[n - k, k], {n, 0, 10}, {k, 0, n}]] (* Indranil Ghosh, Mar 27 2017 *)
  • PARI
    for(k=0, 10, for(n=0, 9, print1(((k+1)^(n+1)-(-k)^(n+1))/(2*k+1), ", "); ); print(); ) \\ Andrew Howroyd, Mar 26 2017
    
  • Python
    def T(n, k): return ((n + 1)**(k + 1) - (-n)**(k + 1)) // (2*n + 1)
    for n in range(11):
        print([T(n - k, k) for k in range(n + 1)]) # Indranil Ghosh, Mar 27 2017

Formula

T(k, n) = ((k+1)^(n+1)-(-k)^(n+1))/(2k+1).
Rows of the array have g.f. 1/((1+kx)(1-(k+1)x)).

Extensions

Name clarified by Andrew Howroyd, Mar 27 2017

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

Author

Michael A. Allen, Dec 21 2021

Keywords

Comments

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).

Examples

			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;
		

Crossrefs

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.

Programs

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

Formula

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

Author

Michael A. Allen, Dec 22 2021

Keywords

Comments

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).

Examples

			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;
		

Crossrefs

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.

Programs

  • Mathematica
    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
    				

Formula

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.
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