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 1-4 of 4 results.

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

Views

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

A058395 Square array read by antidiagonals. Based on triangular numbers (A000217) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 3, 1, 1, 0, 3, 2, 1, 6, 3, 4, 3, 1, 0, 6, 6, 6, 4, 1, 10, 6, 9, 10, 9, 5, 1, 0, 10, 12, 15, 16, 13, 6, 1, 15, 10, 16, 21, 25, 25, 18, 7, 1, 0, 15, 20, 28, 36, 41, 38, 24, 8, 1, 21, 15, 25, 36, 49, 61, 66, 56, 31, 9, 1, 0, 21, 30, 45, 64, 85, 102, 104, 80, 39, 10, 1, 28, 21, 36, 55, 81, 113, 146, 168, 160, 111, 48, 11, 1
Offset: 0

Views

Author

Henry Bottomley, Nov 24 2000

Keywords

Comments

Changing the formula by replacing T(2n, 0) = T(n, 3) with T(2n, 0) = T(n, m) for some other value of m would change the generating function to the coefficient of x^n in expansion of (1 + x)^k / (1 - x^2)^m. This would produce A058393, A058394, A057884 (and effectively A007318).

Examples

			The array T(n, k) starts:
[0] 1, 0,  3,   0,   6,   0,  10,    0,   15,    0, ...
[1] 1, 1,  3,   3,   6,   6,  10,   10,   15,   15, ...
[2] 1, 2,  4,   6,   9,  12,  16,   20,   25,   30, ...
[3] 1, 3,  6,  10,  15,  21,  28,   36,   45,   55, ...
[4] 1, 4,  9,  16,  25,  36,  49,   64,   81,  100, ...
[5] 1, 5, 13,  25,  41,  61,  85,  113,  145,  181, ...
[6] 1, 6, 18,  38,  66, 102, 146,  198,  258,  326, ...
[7] 1, 7, 24,  56, 104, 168, 248,  344,  456,  584, ...
[8] 1, 8, 31,  80, 160, 272, 416,  592,  800, 1040, ...
[9] 1, 9, 39, 111, 240, 432, 688, 1008, 1392, 1840, ...
		

Crossrefs

Rows are A000217 with zeros, A008805, A002620, A000217, A000290, A001844, A005899.
Columns are A000012, A001477, A016028.
The triangle A055252 also appears in half of the array.

Programs

  • Maple
    gf := n -> (1 + x)^n / (1 - x^2)^3: ser := n -> series(gf(n), x, 20):
    seq(lprint([n], seq(coeff(ser(n), x, k), k = 0..9)), n = 0..9); # Peter Luschny, Apr 12 2023
  • Mathematica
    T[0, k_] := If[OddQ[k], 0, (k+2)(k+4)/8];
    T[n_, k_] := T[n, k] = If[k == 0, 1, T[n-1, k-1] + T[n-1, k]];
    Table[T[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Apr 13 2023 *)

Formula

T(n, k) = T(n-1, k-1) + T(n, k-1) with T(0, k) = 1, T(2*n, 0) = T(n, 3) and T(2*n + 1, 0) = 0. Coefficient of x^n in expansion of (1 + x)^k / (1 - x^2)^3.

A058394 A square array based on natural numbers (A000027) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 2, 1, 1, 0, 2, 2, 1, 3, 2, 3, 3, 1, 0, 3, 4, 5, 4, 1, 4, 3, 5, 7, 8, 5, 1, 0, 4, 6, 9, 12, 12, 6, 1, 5, 4, 7, 11, 16, 20, 17, 7, 1, 0, 5, 8, 13, 20, 28, 32, 23, 8, 1, 6, 5, 9, 15, 24, 36, 48, 49, 30, 9, 1, 0, 6, 10, 17, 28, 44, 64, 80, 72, 38, 10, 1, 7, 6, 11, 19, 32, 52, 80, 112, 129
Offset: 0

Views

Author

Henry Bottomley, Nov 24 2000

Keywords

Comments

Changing the formula by replacing T(2n,0)=T(n,2) by T(2n,0)=T(n,m) for some other value of m, would make the generating function change to coefficient of x^n in expansion of (1+x)^k/(1-x^2)^m. This would produce A058393, A058395, A057884 (and effectively A007318).

Examples

			Rows are (1,0,2,0,3,0,4,...), (1,1,2,2,3,3,...), (1,2,3,4,5,6,...), (1,3,5,7,9,11,...), etc.
		

Crossrefs

Rows are A027656 (A000027 with zeros), A008619, A000027, A005408, A008574 etc. Columns are A000012, A001477, A022856 etc. Diagonals include A034007, A045891, A045623, A001792, A001787, A000337, A045618, A045889, A034009, A055250, A055251 etc. The triangle A055249 also appears in half of the array.

Formula

T(n, k)=T(n-1, k-1)+T(n, k-1) with T(0, k)=1, T(2n, 0)=T(n, 2) and T(2n+1, 0)=0. Coefficient of x^n in expansion of (1+x)^k/(1-x^2)^2.

A131255 A004070 * A000012(signed).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 1, 0, 2, 1, 0, 1, 1, 3, 1, 0, 1, 1, 3, 4, 1, 1, 0, 2, 2, 6, 5, 1, 0, 1, 1, 3, 5, 10, 6, 1, 0, 1, 1, 3, 5, 11, 15, 7, 1, 1, 0, 2, 2, 6, 10, 21, 21, 8, 1
Offset: 0

Views

Author

Gary W. Adamson, Jun 23 2007

Keywords

Comments

Row sums = A004695 starting (1, 1, 2, 4, 6, 10, 17, 27, 44, 72, ...). A058393 = A000012(signed) * A004070.

Examples

			First few rows of the triangle:
  1;
  0,  1;
  0,  1,  1;
  1,  0,  2,  1;
  0,  1,  1,  3,  1;
  0,  1,  1,  3,  4,  1;
  1,  0,  2,  2,  6,  5,  1;
  0,  1,  1,  3,  5, 10,  6,  1;
  0,  1,  1,  3,  5, 11, 15,  7,  1;
  1,  0,  2,  2,  6, 10,  2, 21,  8,  1;
  ...
		

Crossrefs

Formula

A004070 * A000012, where A000012 = (1; -1,1; 1,-1,1; ...).
Showing 1-4 of 4 results.