A001779 -id:A001779 - cp's OEIS frontend

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), ...

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

N. J. A. Sloane, Jan 23 2001

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)

			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

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
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.
MathOverflow, Cyclotomic Polynomials in Combinatorics
Mark Norfleet, Characterization of second-order strong divisibility sequences of polynomials, The Fibonacci Quarterly, 43(2) (2005), 166-169.

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
Cf. A026641, A014300.
Cf. A007318, A097805, A097808, A130595, A200139, A062135.

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

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 *)

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

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

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

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)

Formula corrected by Philippe Deléham, Jan 11 2014

A001769 Expansion of 1/((1+x)*(1-x)^7).

Original entry on oeis.org

1, 6, 22, 62, 148, 314, 610, 1106, 1897, 3108, 4900, 7476, 11088, 16044, 22716, 31548, 43065, 57882, 76714, 100386, 129844, 166166, 210574, 264446, 329329, 406952, 499240, 608328, 736576, 886584, 1061208, 1263576, 1497105, 1765518, 2072862, 2423526, 2822260, 3274194, 3784858

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 4, 17.
Index entries for linear recurrences with constant coefficients, signature (6,-14,14,0,-14,14,-6,1).

Cf. A002620, A002623, A001752, A001753 (first differences), A158454 (signed column k=3), A001779 (partial sums), A169794 (binomial transf.).

[(4*n^6+96*n^5+910*n^4+4320*n^3+10696*n^2+12864*n+5715)/5760+(-1)^n/128: n in [0..40]]; // Vincenzo Librandi, Aug 15 2011

CoefficientList[Series[1/((1+x)(1-x)^7),{x,0,30}],x] (* or *) LinearRecurrence[ {6,-14,14,0,-14,14,-6,1},{1,6,22,62,148,314,610,1106},40] (* Harvey P. Dale, May 24 2015 *)

a(n)=(4*n^6+96*n^5+910*n^4+4320*n^3+10696*n^2+12864*n)\/5760+1 \\ Charles R Greathouse IV, Apr 17 2012

From Paul Barry, Jul 01 2003: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k)*C(k+6, 6).
a(n) = (4*n^6 +96*n^5 +910*n^4 +4320*n^3 +10696*n^2 +12864*n+5715)/5760+(-1)^n/128. (End)
Boas-Buck recurrence: a(n) = (1/n)*Sum_{p=0..n-1} (7 + (-1)^(n-p))*a(p), n >= 1, a(0) = 1. See the Boas-Buck comment in A046521 (here for the unsigned column k = 3 with offset 0).
a(n)+a(n+1) = A000579(n+7). - R. J. Mathar, Jan 06 2021

A112465 Riordan array (1/(1+x), x/(1-x)).

Original entry on oeis.org

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

Offset: 0

Paul Barry, Sep 06 2005

Inverse is A112466. Note that C(n,k) = Sum_{j = 0..n-k} C(j+k-1, j).

			Triangle starts
   1;
  -1, 1;
   1, 0, 1;
  -1, 1, 1,  1;
   1, 0, 2,  2,  1;
  -1, 1, 2,  4,  3,  1;
   1, 0, 3,  6,  7,  4,  1;
  -1, 1, 3,  9, 13, 11,  5, 1;
   1, 0, 4, 12, 22, 24, 16, 6, 1;
Production matrix begins
  -1, 1;
   0, 1, 1;
   0, 0, 1, 1;
   0, 0, 0, 1, 1;
   0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 0, 0, 1, 1; - _Paul Barry_, Apr 08 2011

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Roland Bacher, Chebyshev polynomials, quadratic surds and a variation of Pascal's triangle, arXiv:1509.09054 [math.CO], 2015.
E. Deutsch, L. Ferrari and S. Rinaldi, Production Matrices, Advances in Mathematics, 34 (2005) pp. 101-122.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A059260, A108561, A112466, A112468.
Columns: A033999(n) (k=0), A000035(n) (k=1), A004526(n) (k=2), A002620(n-1) (k=3), A002623(n-4) (k=4), A001752(n-5) (k=5), A001753(n-6) (k=6), A001769(n-7) (k=7), A001779(n-8) (k=8), A001780(n-9) (k=9), A001781(n-10) (k=10), A001786(n-11) (k=11), A001808(n-12) (k=12).
Diagonals: A000012(n) (k=n), A023443(n) (k=n-1), A152947(n-1) (k=n-2), A283551(n-3) (k=n-3).
Main diagonal: A072547.
Sums: A078008 (row), A078024 (diagonal), A092220 (signed diagonal), A280560 (signed row).

a112465 n k = a112465_tabl !! n !! k
a112465_row n = a112465_tabl !! n
a112465_tabl = iterate f [1] where
   f xs'@(x:xs) = zipWith (+) ([-x] ++ xs ++ [0]) ([0] ++ xs')
-- Reinhard Zumkeller, Jan 03 2014

A112465:= func< n,k | (-1)^(n+k)*(&+[(-1)^j*Binomial(j+k-1,j): j in [0..n-k]]) >;
[A112465(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Apr 18 2025

T[n_, k_]:= Sum[Binomial[j+k-1, j]*(-1)^(n-k-j), {j, 0, n-k}];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Jul 23 2018 *)

def A112465(n,k): return (-1)^(n+k)*sum((-1)^j*binomial(j+k-1,j) for j in range(n-k+1))
print(flatten([[A112465(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Apr 18 2025

Number triangle T(n, k) = Sum_{j=0..n-k} (-1)^(n-k-j)*C(j+k-1, j).
T(2*n, n) = A072547(n) (main diagonal). - Paul Barry, Apr 08 2011
From Reinhard Zumkeller, Jan 03 2014: (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k), 0 < k < n, with T(n, 0) = (-1)^n and T(n, n) = 1.
T(n, k) = A108561(n, n-k). (End)
T(n, k) = T(n-1, k-1) + T(n-2, k) + T(n-2, k-1), T(0, 0) = 1, T(1, 0) = -1, T(1, 1) = 1, T(n, k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 11 2014
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(-1 + x + x^2/2! + x^3/3!) = -1 + 2*x^2/2! + 6*x^3/3! + 13*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014

A001780 Expansion of 1/((1+x)(1-x)^9).

Original entry on oeis.org

1, 8, 37, 128, 367, 920, 2083, 4352, 8518, 15792, 27966, 47616, 78354, 125136, 194634, 295680, 439791, 641784, 920491, 1299584, 1808521, 2483624, 3369301, 4519424, 5998876, 7885280, 10270924, 13264896

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 4, 17.
Index entries for linear recurrences with constant coefficients, signature (8,-27,48,-42,0,42,-48,27,-8,1).

Cf. A001769, A158454 (signed column k=4), A001779 (first differences), A169796 (binomial trans.).

Vec(1/(1+x)/(1-x)^9+O(x^99)) \\ Charles R Greathouse IV, Apr 18 2012

a(n) = 431*n/168 + (-1)^n/512 + 391*n^3/288 + 26011*n^2/10080 + 797*n^4/1920 + 11*n^5/144 + n^6/120 + n^7/2016 + n^8/80640 + 511/512. - R. J. Mathar, Mar 15 2011
Boas-Buck recurrence: a(n) = (1/n)*Sum_{p=0..n-1} (9 + (-1)^(n-p))*a(p), n >= 1, a(0) = 1. See the Boas-Buck comment in A046521 (here for the unsigned column k = 4 with offset 0). - Wolfdieter Lang, Aug 10 2017
a(n)+a(n+1) = A000581(n+9) . - R. J. Mathar, Jan 06 2021

A128494 Coefficient table for sums of Chebyshev's S-Polynomials.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, -1, 1, 1, 1, -1, -2, 1, 1, 1, 2, -2, -3, 1, 1, 0, 2, 4, -3, -4, 1, 1, 0, -2, 4, 7, -4, -5, 1, 1, 1, -2, -6, 7, 11, -5, -6, 1, 1, 1, 3, -6, -13, 11, 16, -6, -7, 1, 1, 0, 3, 9, -13, -24, 16, 22, -7, -8, 1, 1, 0, -3, 9, 22, -24, -40, 22, 29, -8, -9, 1, 1, 1, -3, -12, 22, 46, -40, -62, 29, 37, -9, -10, 1, 1, 1, 4, -12

Offset: 0

Wolfdieter Lang, Apr 04 2007

See A049310 for the coefficient table of Chebyshev's S(n,x)=U(n,x/2) polynomials.

This is a 'repetition triangle' based on a signed version of triangle A059260: a(2*p,2*k) = a(2*p+1,2*k) = A059260(p+k,2*k)*(-1)^(p+k) and a(2*p+1,2*k+1) = a(2*p+2,2*k+1) = A059260(p+k+1,2*k+1)*(-1)^(p+k), k >= 0.

			The triangle a(n,m) begins:
  n\m  0   1   2   3   4   5   6   7   8   9  10
   0:  1
   1:  1   1
   2:  0   1   1
   3:  0  -1   1   1
   4:  1  -1  -2   1   1
   5:  1   2  -2  -3   1   1
   6:  0   2   4  -3  -4   1   1
   7:  0  -2   4   7  -4  -5   1   1
   8:  1  -2  -6   7  11  -5  -6   1   1
   9:  1   3  -6 -13  11  16  -6  -7   1   1
  10:  0   3   9 -13 -24  16  22  -7  -8   1   1
... reformatted by _Wolfdieter Lang_, Oct 16 2012
Row polynomial S(1;4,x) = 1 - x - 2*x^2 + x^3 + x^4 = Sum_{k=0..4} S(k,x).
S(4,y)*S(5,y)/y = 3 - 13*y^2 + 16*y^4 - 7*y^6 + y^8, with y=sqrt(2+x) this becomes S(1;4,x).
From _Wolfdieter Lang_, Oct 16 2012: (Start)
S(1;4,x) = (1 - (S(5,x) - S(4,x)))/(2-x) = (1-x)*(2-x)*(1+x)*(1-x-x^2)/(2-x) = (1-x)*(1+x)*(1-x-x^2).
S(5,x) - S(4,x) = R(11,sqrt(2+x))/sqrt(2+x) = -1 + 3*x + 3*x^2 - 4*x^3 - x^4 + x^5. (End)

Wolfdieter Lang, First 15 rows
Per Alexandersson, Luis Angel González-Serrano, and Egor A. Maximenko, Mario Alberto Moctezuma-Salazar, Symmetric polynomials in the symplectic alphabet and their expression via Dickson-Zhukovsky variables, arXiv:1912.12725 [math.CO], 2019.
Per Alexandersson, Luis Angel González-Serrano, Egor A. Maximenko, and Mario Alberto Moctezuma-Salazar, Symmetric Polynomials in the Symplectic Alphabet and the Change of Variables z_j = x_j + x_j^(-1), The Elec. J. of Combinatorics (2021) Vol. 28, No. 1, #P1.56.

Row sums (signed): A021823(n+2). Row sums (unsigned): A070550(n).
Cf. A128495 for S(2; n, x) coefficient table.
The column sequences (unsigned) are, for m=0..4: A021923, A002265, A008642, A128498, A128499.
For m >= 1 the column sequences (without leading zeros) are of the form a(m, 2*k) = a(m, 2*k+1) = ((-1)^k)*b(m, k) with the sequences b(m, k), given for m=1..11 by A008619, A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808.

S(1;n,x) = Sum_{k=0..n} S(k,x) = Sum_{m=0..n} a(n,m)*x^m, n >= 0.
a(n,m) = [x^m](S(n,y)*S(n+1,y)/y) with y:=sqrt(2+x).
G.f. for column m: (x^m)/((1-x)*(1+x^2)^(m+1)), which shows that this is a lower diagonal matrix of the Riordan type, named (1/((1+x^2)*(1-x)), x/(1+x^2)).
From Wolfdieter Lang, Oct 16 2012: (Start)
a(n,m) = [x^m](1- (S(n+1,x) - S(n,x)))/(2-x). From the Binet - de Moivre formula for S and use of the geometric sum.
a(n,m) = [x^m](1- R(2*n+3,sqrt(2+x))/sqrt(2+x))/(2-x) with the monic integer T-polynomials R with coefficient triangle given in A127672. From the odd part of the bisection of the T-polynomials. (End)

A166189 Number of 3 X 3 X 3 triangular nonnegative integer arrays with all sums of an element and its neighbors <= n.

Original entry on oeis.org

1, 7, 29, 90, 232, 524, 1072, 2030, 3613, 6111, 9905, 15484, 23464, 34608, 49848, 70308, 97329, 132495, 177661, 234982, 306944, 396396, 506584, 641186, 804349, 1000727, 1235521, 1514520, 1844144, 2231488, 2684368, 3211368

Offset: 0

R. H. Hardin, Oct 09 2009

a(n) gives the number of hexagons that have vertices at the lattice points and sides on lattice lines of a triangular lattice with sides n+3. Note that the hexagons can be non-regular. This problem appeared as ConvexHexagons in Single Round Match 455 in TopCoder. - Dmitry Kamenetsky, Dec 17 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Jun-Ming Zhu, The number of convex pentagons and hexagons in an n-triangular net, arXiv:1012.4058 [math.CO], 2010; See H(n), formula 3, on page 4.
Index entries for linear recurrences with constant coefficients, signature (6,-14,14,0,-14,14,-6,1).

Cf. A176646, A011888 (first differences).

R:=PowerSeriesRing(Integers(), 50);
Coefficients(R!( (1-x^3)/((1-x^2)*(1-x)^7) )); // G. C. Greubel, Jul 02 2021

LinearRecurrence[{6,-14,14,0,-14,14,-6,1}, {1,7,29,90,232,524,1072,2030}, 51] (* G. C. Greubel, Jul 02 2021 *)

\\ using Zhu expressions
f(k) = (8*k^6 + 24*k^5 + 25*k^4 + 10*k^3 - 3*k^2 -4*k)/60;
g(k) = (8*k^6 - 5*k^4 - 3*k^2)/60;
a(n) = n+=3; if (n%2, f((n-1)/2), g(n/2)); \\ Michel Marcus, Jul 04 2021

def a(n): return (n+2)*(n+4)*(2*n^4 +24*n^3 +105*n^2 +198*n +120)/960 if (n%2==0) else (n+1)*(n+3)^2*(n+5)*(2*n*(n+6) +21)/960
[a(n) for n in (0..50)] # G. C. Greubel, Jul 02 2021

From G. C. Greubel, Jul 02 2021: (Start)
a(n) = (1/1920)*(4*n^6 +72*n^5 +530*n^4 +2040*n^3 +4296*n^2 +4608*n +1905 +15*(-1)^n).
a(2*n+1) = (1/15)*binomial(n+2, 2)*binomial(n+3, 2)*(8*n^2 + 32*n + 35).
a(2*n) = (1/30)*binomial(n+2, 2)*(8*n^4 + 48*n^3 + 105*n^2 + 99*n + 30).
G.f.: (1 - x^3)/((1-x^2)*(1-x)^7).
E.g.f.: (1/1920)*((1905 +2*x*(5775 +7665*x +3690*x^2 +755*x^3 +66*x^4 +2*x^5))*exp(x) + 15*exp(-x)). (End)
a(n) = A001779(n)-A001779(n-3). - R. J. Mathar, Jul 04 2021

cp's OEIS Frontend

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), ...

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A001769 Expansion of 1/((1+x)*(1-x)^7).

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A112465 Riordan array (1/(1+x), x/(1-x)).

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A001780 Expansion of 1/((1+x)(1-x)^9).

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A128494 Coefficient table for sums of Chebyshev's S-Polynomials.

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A166189 Number of 3 X 3 X 3 triangular nonnegative integer arrays with all sums of an element and its neighbors <= n.

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Magma

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