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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), ...
25
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
0,8
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
LINKS
Roland Bacher, Chebyshev polynomials, quadratic surds and a variation of Pascal's triangle, arXiv:1509.09054 [math.CO], 2015.
Joseph Briggs, Alex Parker, Coy Schwieder, and Chris Wells, Frogs, hats and common subsequences, arXiv:2404.07285 [math.CO], 2024. See p. 28.
Robert Coquereaux and Éric Ragoucy, Currents on Grassmann algebras, J. of Geometry and Physics, 1995, Vol 15, pp 333-352.
Robert Coquereaux and Éric Ragoucy, Currents on Grassmann algebras, arXiv:hep-th/9310147, 1993.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. II. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See p. 8.
R. H. Hammack and G. D. Smith, Cycle bases of reduced powers of graphs, Ars Math. Contemp. 12 (2017) 183-203.
Christian Kassel, A Künneth formula for the cyclic cohomology of Z2-graded algebras, Math. Ann. 275 (1986) 683.
Ana Filipa Loureiro and Pascal Maroni, Polynomial sequences associated with the classical linear functionals, Numerical Algorithms, June 2012, Volume 60, Issue 2, pp 297-314. - From N. J. A. Sloane, Oct 12 2012
Ana Filipa Loureiro and Pascal Maroni, Polynomial sequences associated with the classical linear functionals, preprint, Centro de Matemática da Universidade do Porto.
Mark Norfleet, Characterization of second-order strong divisibility sequences of polynomials, The Fibonacci Quarterly, 43(2) (2005), 166-169.
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)
EXAMPLE
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):
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
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 *)
PROG
(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
(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
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
Sequence in context: A238453 A376991 A066287 * A239473 A135229 A257543
KEYWORD
nonn,tabl,nice
AUTHOR
N. J. A. Sloane, Jan 23 2001
EXTENSIONS
Formula corrected by Philippe Deléham, Jan 11 2014
STATUS
approved