A333715 -id:A333715 - cp's OEIS frontend

A035607 Table a(d,m) of number of points of L1 norm m in cubic lattice Z^d, read by antidiagonals (d >= 1, m >= 0).

1, 1, 2, 1, 4, 2, 1, 6, 8, 2, 1, 8, 18, 12, 2, 1, 10, 32, 38, 16, 2, 1, 12, 50, 88, 66, 20, 2, 1, 14, 72, 170, 192, 102, 24, 2, 1, 16, 98, 292, 450, 360, 146, 28, 2, 1, 18, 128, 462, 912, 1002, 608, 198, 32, 2, 1, 20, 162, 688, 1666, 2364, 1970, 952, 258, 36, 2, 1, 22, 200, 978, 2816

Offset: 0

N. J. A. Sloane

Table also gives coordination sequences of same lattices.
Rows sums are given by A001333. Rising and falling diagonals are the tribonacci numbers A000213, A001590. - Paul Barry, Feb 13 2003
a(d,m) also gives the number of ways to choose m squares from a 2 X (d-1) grid so that no two squares in the selection are (horizontally or vertically) adjacent. - Jacob A. Siehler, May 13 2006
Mirror image of triangle A113413. - Philippe Deléham, Oct 15 2006
The Ca1 sums lead to A126116 and the Ca2 sums lead to A070550, see A180662 for the definitions of these triangle sums. - Johannes W. Meijer, Aug 05 2011
A035607 is jointly generated with the Delannoy triangle A008288 as an array of coefficients of polynomials v(n,x):  initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) and v(n,x) = 2*x*u(n-1,x) + v(n-1,x).  See the Mathematica section. - Clark Kimberling, Mar 05 2012
Also, the polynomial v(n,x) above is x + (x + 1)*f(n-1,x), where f(0,x) = 1. - Clark Kimberling, Oct 24 2014
Rows also give the coefficients of the independence polynomial of the n-ladder graph. - Eric W. Weisstein, Dec 29 2017
Considering both sequences as square arrays (offset by one row), the rows of A035607 are the first differences of the rows of A008288, and the rows of A008288 are the partial sums of the rows of A035607. - Shel Kaphan, Feb 23 2023
Considering only points with nonnegative coordinates, the number of points at L1 distance = m in d dimensions is the same as the number of ways of putting m indistinguishable balls into d distinguishable urns, binomial(m+d-1, d-1). This is one facet of the cross-polytope. Allowing for + and - coordinates, there are binomial(d,i)*2^i facets containing points with up to i nonzero coordinates. Eliminating double counting of points with any coordinates = 0, there are Sum_{i=1..d} (-1)^(d-i)*binomial(m+i-1,i-1)*binomial(d,i)*2^i points at distance m in d dimensions. One may avoid the alternating sum by using binomial(m-1,i-1) to count only the points per facet with exactly i nonzero coordinates, avoiding any double counting, but the result is the same. - Shel Kaphan, Mar 04 2023

			From _Clark Kimberling_, Oct 24 2014: (Start)
As a triangle of coefficients in polynomials v(n,x) in Comments, the first 6 rows are
  1
  1   2
  1   4   2
  1   6   8   2
  1   8  18  12   2
  1  10  32  38  16   2
  ... (End)
From _Shel Kaphan_, Mar 04 2023: (Start)
For d=3, m=4:
There are binomial(3,1)*2^1 = 6 facets (vertices) of binomial(4+1-1,1-1) = 1 point with <= one nonzero coordinate.
There are binomial(3,2)*2^2 = 12 facets (edges) of binomial(4+2-1,2-1) = 5 points with <= two nonzero coordinates.
There are binomial(3,3)*2^3 = 8 facets (faces) of binomial(4+3-1,3-1) = 15 points with <= three nonzero coordinates.
a(3,4) = 8*15 - 12*5 + 6*1 = 120 - 60 + 6 = 66. (End)

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Bela Bajnok, Additive Combinatorics: A Menu of Research Problems, arXiv:1705.07444 [math.NT], May 2017. See Sect. 2.3.
J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 392.
R. J. Mathar, Bivariate Generating Functions for Non-attacking Wazirs on Rectangular Boards (2024), Table 2.
Emanuele Munarini, Combinatorial properties of the antichains of a garland, Integers, 9 (2009), 353-374.
Joan Serra-Sagrista, Enumeration of lattice points in l_1 norm, Inf. Proc. Lett. 76 (1-2) (2000) 39-44.
J. Siehler, Adjacency-free selections from a 2xN grid (Mathematica notebook) [broken link]
Jacob A. Siehler, Selections Without Adjacency on a Rectangular Grid, arXiv:1409.3869 [math.CO], 2014.
Eric Weisstein's World of Mathematics, Independence Polynomial
Eric Weisstein's World of Mathematics, Ladder Graph

Other versions: A113413, A119800, A122542, A266213.
Cf. A008288, which has g.f. 1/(1-x-x*y-x^2*y).
Cf. A078057 (row sums), A050146 (central terms).
Cf. A050146.

a035607 n k = a035607_tabl !! n !! k
a035607_row n = a035607_tabl !! n
a035607_tabl = map fst $ iterate
   (\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
                      zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 2])
-- Reinhard Zumkeller, Jul 20 2013

A035607 := proc(d,m) local j: add(binomial(floor((d-1+j)/2),d-m-1)*binomial(d-m-1, floor((d-1-j)/2)),j=0..d-1) end: seq(seq(A035607(d,m),m=0..d-1),d=1..11); # d=dimension, m=norm # Johannes W. Meijer, Aug 05 2011

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A008288 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A035607 *)
(* Clark Kimberling, Mar 09 2012 *)
Reverse /@ CoefficientList[CoefficientList[Series[(1 + x)/(1 - x - x y - x^2 y), {x, 0, 10}], x], y] // Flatten (* Eric W. Weisstein, Dec 29 2017 *)

T(n, k) = if (k==0, 1, sum(i=0, k-1, binomial(n-k,i+1)*binomial(k-1,i)*2^(i+1)));
tabl(nn) = for (n=1, nn, for (k=0, n-1, print1(T(n, k), ", ")); print); \\ as a triangle; Michel Marcus, Feb 27 2018

def A035607_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return 1
        if k==0: return 0
        return prec(n-1,k-1)+2*sum(prec(n-i,k-1) for i in (2..n-k+1))
    return [prec(n, n-k) for k in (0..n-1)]
for n in (1..10): print(A035607_row(n)) # Peter Luschny, Mar 16 2016

From Johannes W. Meijer, Aug 05 2011: (Start)
f(d,m) = Sum_{j=0..d-1} binomial(floor((d-1+j)/2), d-m-1)*binomial(d-m-1, floor((d-1-j)/2)), d >= 1 and 0 <= m <= d-1.
f(d,m) = f(d-1,m-1) + f(d-1,m) + f(d-2,m-1) (d >= 3 and 1 <= m <= d-1) with f(d,0) = 1 (d >= 1) and f(d,d-1) = 2 (d>=2). (End)
From Roger Cuculière, Apr 10 2006: (Start)
The generating function G(x,y) of this double sequence is the sum of a(n,p)*x^n*y^p, n=1..oo, p=0..oo, which is G(x,y) = x*(1+y)/(1-x-y-x*y).
The horizontal generating function H_n(y), which generates the rows of the table: (1, 2, 2, 2, 2, ...), (1, 4, 8, 12, 16, ...), (1, 6, 18, 38, 66, ...), is the sum of a(n,p)*y^p, p=0..oo, for each fixed n. This is H_n(y) = ((1+y)^n)/((1-y)^n).
The vertical generating function V_p(x), which generates the columns of the table: (1, 1, 1, 1, 1, ...), (2, 4, 6, 8, 10, ...), (2, 8, 18, 32, 50, ...), is the sum of a(n,p)*x^n, n=1..oo, for each fixed p. This is V_p(x) = 2*((1+x)^(p-1))/((1-x)^(p+1)) for p >= 1 and V_0(x) = x/(1-x). (End)
G.f.: (1+x)/(1-x-x*y-x^2*y). - Vladeta Jovovic, Apr 02 2002 (But see previous lines!)
T(2*n,n) = A050146(n+1). - Reinhard Zumkeller, Jul 20 2013
Seen as a triangle read by rows: T(n,0) = 1, for n > 1: T(n,n-1) = 2, T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-2,k-1), 0 < k < n. - Reinhard Zumkeller, Jul 20 2013
Seen as a triangle T(n,k) with 0 <= k < n read by rows: T(n,0)=1 for n > 0 and T(n,k) = Sum_{i=0..k-1} binomial(n-k,i+1)*binomial(k-1,i)*2^(i+1) for k > 0. - Werner Schulte, Feb 22 2018
With p >= 1 and q >= 0, as a square array a(p,q) = T(p+q-1,q) = 2*p*Hypergeometric2F1[1-p, 1-q, 2, 2] for q >= 1. Consequently, a(p,q) = a(q,p)*p/q. - Shel Kaphan, Feb 14 2023
For n >= 1, T(2*n,n) = A002003(n), T(3*n,2*n) = A103885(n) and T(4*n,3*n) = A333715(n). - Peter Bala, Jun 15 2023

More terms from David W. Wilson

Maple program corrected and information added by Johannes W. Meijer, Aug 05 2011

A144097 The 4-Schroeder numbers: a(n) = number of lattice paths (Schroeder paths) from (0,0) to (3n,n) with unit steps N=(0,1), E=(1,0) and D=(1,1) staying weakly above the line y = 3x.

Original entry on oeis.org

1, 2, 14, 134, 1482, 17818, 226214, 2984206, 40503890, 561957362, 7934063678, 113622696470, 1646501710362, 24098174350986, 355715715691350, 5289547733908510, 79163575684710818, 1191491384838325474

Offset: 0

Joachim Schroeder (schroderjd(AT)qwa.uovs.ac.za), Sep 10 2008

a(n) is also the number of lattice path from (0,0) to (4n,0) with unit steps (1,3), (2,2) and (1,-1) staying weakly above the x-axis.
Also, the number of planar rooted trees with n non-leaf vertices such that each non-leaf vertex has either 3 or 4 children. - Cameron Marcott, Sep 18 2013
a(n) equals the number of ordered complete 4-ary trees with 3*n + 1 leaves, where the internal vertices come in two colors and such that each vertex and its rightmost child have different colors. See Drake, Example 1.6.9. - Peter Bala, Apr 30 2023

			a(2)=14, because
  01: NNNENNNE,
  02: NNDNNNE,
  03: NNNENND,
  04: NNDNND,
  05: NNNDNNE,
  06: NNNDND,
  07: NNNNENNE,
  08: NNNNEND,
  09: NNNNDNE,
  10: NNNNDD,
  11: NNNNNENE,
  12: NNNNNED,
  13: NNNNNDE,
  14: NNNNNNEE
are all the paths from (0,0) to (2,6) with steps N,E and D weakly above y=3x.

Sheng-Liang Yang and Mei-yang Jiang, The m-Schröder paths and m-Schröder numbers, Disc. Math. (2021) Vol. 344, Issue 2, 112209. doi:10.1016/j.disc.2020.112209. See Table 1.

Alois P. Heinz, Table of n, a(n) for n = 0..800
D. Bevan, D. Levin, P. Nugent, J. Pantone, and L. Pudwell, Pattern avoidance in forests of binary shrubs, arXiv preprint arXiv:1510.08036 [math.CO], 2015.
B. Drake, An inversion theorem for labeled trees and some limits of areas under lattice paths (Example 1.6.9), A dissertation presented to the Faculty of the Graduate School of Arts and Sciences of Brandeis University.
Gi-Sang Cheon, S.-T. Jin, and L. W. Shapiro, A combinatorial equivalence relation for formal power series, Linear Algebra and its Applications, Available online 30 March 2015.
J. Schroeder, Generalized Schroeder numbers and the rotation principle, Journal of Integer Sequences 10(7).
R. A. Sulanke, A recurrence restricted by a diagonal condition: generalized Catalan arrays, Fibonacci Q., 27 (1989), 33-46.
Jun Yan, Lattice paths enumerations weighted by ascent lengths, arXiv:2501.01152 [math.CO], 2025. See p. 11.
Lin Yang, Yu-Yuan Zhang, and Sheng-Liang Yang, The halves of Delannoy matrix and Chung-Feller properties of the m-Schröder paths, Linear Alg. Appl. (2024).
Sheng-liang Yang and Mei-yang Jiang, Pattern avoiding problems on the hybrid d-trees, J. Lanzhou Univ. Tech., (China, 2023) Vol. 49, No. 2, 144-150. (in Mandarin; improperly cited as A144079)

Cf. A027307 (the case y=2x), A008288 (Delannoy numbers), A008412 (4-dimensional coordination numbers).
This appears to equal 2*A243675. - N. J. A. Sloane, Mar 28 2021
The sequences listed in Yang-Jiang's Table 1 appear to be A006318, A001003, A027307, A034015, A144097, A243675, A260332, A243676. - N. J. A. Sloane, Mar 28 2021

Schr:=proc(n,m,l)(n-l*m+1)/m*sum(2^v*binomial(m,v)*binomial(n,v-1),v=1..m) end proc; where n=3m and l=3, also
Schr:=proc(n,m,l)(n-l*m+1)/(n+1)*sum(2^v*binomial(m-1,v-1)*binomial(n+1,v),v=0..m) end proc; where n=3m and l=3, also
Schr:=proc(n,m,l)(n-l*m+1)/m*sum(binomial(m,v)*binomial(n+v,m-1),v=0..m) end proc; where n=3m and l=3, also
Schr:=proc(n,m,l)(n-l*m+1)/(n+1)*sum(binomial(n+1,v)*binomial(m-1+v,n),v=0..n+1) end proc; where n=3m and l=3.
# alternative Maple program:
a:= proc(n) option remember; `if`(n<2, n+1,
      ((15610*n^5 -67123*n^4 +106824*n^3 -77633*n^2
       +25514*n-3000)*a(n-1) -(3*(n-2))*(3*n-4)*
       (3*n-5)*(35*n^2-28*n+5)*a(n-2)) / ((3*(3*n-1))
       *(3*n+1)*n*(35*n^2-98*n+68)))
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, May 26 2015

d[n_, k_] := Binomial[n+k, k] Hypergeometric2F1[-k, -n, -n-k, -1]; a[0] = 1; a[n_] = d[3n, n] - 3d[3n+1, n-1] - 2d[3n, n-1]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Feb 25 2017 *)

{a(n) = sum(k=0, n, binomial(n, k) * binomial(3*n+k+1, n)/(3*n+k+1))} \\ Seiichi Manyama, Jul 25 2020

{a(n) = if(n==0, 1, sum(k=1, n, 2^k*binomial(n, k) * binomial(3*n, k-1)/n))} \\ Seiichi Manyama, Jul 25 2020

G.f. A(z) satisfies A(z) = 1 + z(A(z)^3 + A(z)^4) a(n)= S_{3n+1}(n) - 3S_n(3n + 1), where S_a(b) are coordination numbers, i.e., the number of points in the a-dimensional cubic lattice Z^a having distance b in the L_1 norm.
Also a(n) = D(3n,n) - 3D(3n + 1,n-1) - 2D(3n,n-1), where D(a,b) are the Delannoy numbers, i.e., the number of paths with N, E and D steps from (0,0) to (a,b).
D-finite with recurrence 3*n*(3*n-1)*(3*n+1)*(35*n^2-98*n+68) *a(n) +(-15610*n^5+67123*n^4-106824*n^3+77633*n^2-25514*n+3000)*a(n-1) +3*(n-2) *(3*n-4) *(3*n-5) *(35*n^2-28*n+5) *a(n-2)=0. - R. J. Mathar, Sep 06 2016
From Seiichi Manyama, Jul 25 2020: (Start)
a(n) = Sum_{k=0..n} binomial(n,k) * binomial(3*n+k+1, n)/(3*n+k+1).
a(n) = (1/n) * Sum_{k=1..n} 2^k * binomial(n,k) * binomial(3*n,k-1) for n > 0. (End)
a(n) ~ sqrt(12160 + 3853*sqrt(10)) * 3^(3*n - 9/2) / (2*sqrt(5*Pi) * n^(3/2) * (223 - 70*sqrt(10))^(n - 1/2)). - Vaclav Kotesovec, Jul 31 2021
Series reversion of x*(1 - x^3)/(1 + x^3) = x + 2*x^4 + 14*x^7 + 134*x^10 + ... = Sum_{n >= 0} a(n)*x^(3*n+1). - Peter Bala, Apr 30 2023
From Peter Bala, Jun 16 2023: (Start)
The g.f. A(x) = 1 + 2*x + 14*x^2 + 134*x^3 + ... satisfies A(x)^3 = (1/x) * the series reversion of ((1 - x)/(1 + x))^3.
Define b(n) = [x^(3*n)] ( (1 + x)/(1 - x) )^n = (1/3) * [x^n] ((1 + x)/(1 - x))^(3*n) = A333715(n). Then A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ).
a(n) = 2*hypergeom([1 - n, -3*n], [2], 2) for n >= 1. (End)
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 2^(n-k) * binomial(n,k) * binomial(4*n-k,n-1-k) for n > 0. - Seiichi Manyama, Aug 09 2023

A351858 a(n) = [x^n] (1 + x + x^2)^(3n)/(1 + x)^(2n).

Original entry on oeis.org

1, 1, 7, 19, 103, 376, 1825, 7547, 35175, 153838, 708132, 3181091, 14616481, 66582283, 306501377, 1407473269, 6497464679, 29991098982, 138844558150, 643215119214, 2985368996228, 13868212710623, 64508509024241, 300324344452479, 1399598738196897, 6527698842078501

Offset: 0

Peter Bala, Feb 27 2022

Given an integer sequence (g(n))n>=1, there exists a formal power series G(x), with rational coefficients, such that g(n) = [x^n] G(x)^n. The power series G(x) has integer coefficients iff the Gauss congruences g(n*p^r) == g(n*p^(r-1)) (mod p^r) hold for all primes p and positive integers n and r.
The central binomial coefficient binomial(2*n,n) = A000984(n) may be defined using the coefficient extraction operator as binomial(2*n,n) = [x^n] ((1 + x)^2)^n and hence the Gauss congruences hold for A000984. Moreover, it is known that the stronger supercongruences A000984(n*p^r) == A000984(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. See Meštrović, equation 39.
We define an infinite family of sequences as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. In particular, G_1(x) = (1 + x)^2 and the sequence u_1 is the sequence of central binomial coefficients. The present sequence is the case k = 2. See A351859 for the case k = 3.
Conjecture: for k >= 2, each sequence u_k satisfies the same supercongruences as the central binomial coefficients.
More generally, if r is a positive integer and s an integer then the sequence defined by  u_k(r,s;n) = [x^(r*n)] G_k(x)^(s*n) may satisfy the same supercongruences.

			Examples of supercongruences:
a(5) - a(1) = 376 - 1 = 3*(5^3) == 0 (mod 5^3)
a(2*7)- a(2) = 306501377 - 7 = 2*5*(7^3)*193*463 == 0 (mod 7^3)
A(5^2) - a(5) = 6527698842078501 - 376 = (5^6)*17*107*229671647 == 0 (mod 5^6)

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A103885, A333090, A333564, A333565, A333579, A333715, A351856, A351857, A351859, A372211, A372212, A372213.

seq(add(add((-1)^(n-k-j)*binomial(n,k)*binomial(3*n,j)* binomial(4*n-2*j-k-1,n-k-j), j = 0..n-k), k = 0..n), n = 0..25);

A351858[n_]:=Sum[(-1)^(n-k-j)Binomial[n,k]Binomial[3n,j]Binomial[4n-2j-k-1,n-k-j],{k,0,n},{j,0,n-k}];Array[A351858,25,0] (* Paolo Xausa, Oct 04 2023 *)
a[n_]:=SeriesCoefficient[(1 + x + x^2)^(3*n)/(1 + x)^(2*n),{x,0,n}]; Array[a,26,0] (* Stefano Spezia, Apr 30 2024 *)

a(n) = Sum_{k = 0..n} Sum_{j = 0..n-k} (-1)^(n-k-j)*C(n,k)*C(3*n,j)*C(4*n-2*j-k-1,n-k-j).
Conjecture: a(n) = Sum_{k = 0..floor(n/2)} C(3*n,k)*C(n-k,k).
The o.g.f. A(x) = 1 + x + 7*x^2 + 19*x^3 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2)^3/(1 + x)^2) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 + x)^2/(1 + x + x^2)^3 ) = 1 + x + 4*x^2 + 10*x^3 + 40*x^4 + 133*x^5 + 536*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).
a(n) ~ sqrt(2/9 + 2*sqrt(53/47)*cos(arccos(1259*sqrt(47/53)/1696)/3)/9) * (2*sqrt(164581)*cos(arccos(-90631279/(1316648*sqrt(164581)))/3)/81 - 293/81)^n / sqrt(Pi*n). - Vaclav Kotesovec, Jun 05 2022

A336521 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) is the coefficient of x^(k*n) in expansion of ( (1 + x)/(1 - x) )^n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 8, 1, 1, 2, 16, 38, 1, 1, 2, 24, 146, 192, 1, 1, 2, 32, 326, 1408, 1002, 1, 1, 2, 40, 578, 4672, 14002, 5336, 1, 1, 2, 48, 902, 11008, 69002, 142000, 28814, 1, 1, 2, 56, 1298, 21440, 216002, 1038984, 1459810, 157184, 1, 1, 2, 64, 1766, 36992, 525002, 4320608, 15856206, 15158272, 864146, 1

Offset: 0

Seiichi Manyama, Jul 24 2020

			Square array begins:
  1,    1,     1,     1,      1,      1, ...
  1,    2,     2,     2,      2,      2, ...
  1,    8,    16,    24,     32,     40, ...
  1,   38,   146,   326,    578,    902, ...
  1,  192,  1408,  4672,  11008,  21440, ...
  1, 1002, 14002, 69002, 216002, 525002, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Column k=0-3 give A000012, A123164, A103885, A333715.
Main diagonal gives A336522.
Cf. A122542, A266213, A336534.

T[n_, 0] := 1; T[n_, k_] := Sum[Binomial[n, j] * Binomial[k*n + j - 1, n - 1], {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Jul 24 2020 *)

T(n,k) = (1/k) * [x^n] ( (1 + x)/(1 - x) )^(k*n) for k > 0 and n > 0.
T(n,k) = Sum_{j=0..n} binomial(n,j) * binomial(k*n+j-1,n-1).
T(n,k) = (1/k) * Sum_{j=0..n} binomial(k*n,n-j) * binomial(k*n+j-1,j) for k > 0 and n > 0.
T(n,k) = Sum_{j=1..n} 2^j * binomial(n,j) * binomial(k*n-1,j-1) for n > 0.
T(n,k) = binomial(k*n-1, n-1)*hypergeom([-n, k*n], [1+(k-1)*n], -1) for k > 0. - Stefano Spezia, Aug 09 2025

A333563 a(n) = [x^n] G(x)^n, where G(x) is the o.g.f. of A079489.

Original entry on oeis.org

1, 3, 53, 1056, 22181, 480003, 10588508, 236720424, 5344683429, 121590541641, 2782821611053, 64001191118956, 1477895865330092, 34243264651422596, 795729752353810824, 18537154747116799056, 432781371485493257637, 10123439350286679005973

Offset: 0

Peter Bala, Apr 12 2020

It can be shown that a(n) satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for prime p and positive integers n and k.
The o.g.f. G(x) of A079489 is given by G(x) = c(sqrt(x))*c(-sqrt(x)), where c(x) = ( (1 - sqrt(1 - 4*x))/(2*x) ) is the o.g.f. of the Catalan numbers A000108. It is known that the sequence b(n) := [x^n] c(x)^n  = 1/3*binomial(3*n,n) satisfies the supercongruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k - see Meštrović, equation 39. We conjecture that the present sequence  satisfies the same congruences. Some examples are given below.
More generally, if r > 0 and s are integers then the sequence a(r,s;n) := [x^(r*n)] G(x)^(s*n) may also satisfy the above congruences.

			Examples of congruences:
a(17) - a(1) = 10123439350286679005973 - 3 = 2(3^3)*5*(17^3)* 7631634401766047  == 0 ( mod 17^3 ).
a(3*5) - a(3) = 18537154747116799056 - 1056 = (2^4)*3*(5^3)*13* 237655830091241 == 0 ( mod 5^3 ).
a(5^2) - a(5) = 952866706104433648666617525245628 - 480003 = 3*(5^7)*17* 3642302759*65659247842693913 == 0 ( mod 5^6 ).

R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Cf. A000108, A079489, A333564, A333715.

c:= x -> (1/2)*(1-sqrt(1-4*x))/x:
G:= x -> c(sqrt(x))*c(-sqrt(x)):
H:= series(G(x)^n, x, 26):
seq(coeff(H, x, n), n = 0..25);

Join[{1}, Table[n^2 * Sum[(-1)^k/((n + 2*k)*(5*n - 2*k))*Binomial[n + 2*k, k] * Binomial[5*n - 2*k, 2*n - k], {k, 0, 2*n}], {n, 1, 20}]] (* Vaclav Kotesovec, Apr 20 2020 *)
Join[{1}, Table[Binomial[5*n, 2*n] * HypergeometricPFQ[{1/2 + n/2, -3*n, -2*n, n/2}, {1/2 - 5*n/2, 1 - 5*n/2, 1 + n}, -1]/5, {n, 1, 20}]] (* Vaclav Kotesovec, May 16 2020 *)

a(n) = if (n==0, 1, n^2 * sum(k = 0, 2*n, (-1)^k/((n+2*k)*(5*n-2*k))*binomial(n+2*k,k)*binomial(5*n-2*k, 2*n-k))); \\ Michel Marcus, May 16 2020

a(n) = n^2 * Sum_{k = 0..2*n} (-1)^k/((n+2*k)*(5*n-2*k))*C(n+2*k,k)* C(5*n-2*k, 2*n-k) for n >= 1.
a(p) == 3 ( mod p^3) for prime p >= 3, follows from the above formula.
P-recursive: 6*n*(n - 1)*(2*n - 1)*(3*n - 1)*(3*n - 2)*(2117*n^4 - 12615*n^3 + 27976*n^2 - 27348*n + 9936)*a(n) =  -(n - 1)*(27269077*n^8 - 217031969*n^7 + 722440183*n^6 - 1304402267*n^5 + 1384804360*n^4 - 874884704*n^3 + 315932544*n^2 - 57998736*n + 3913920)*a(n-1)  + 48*(6*n - 7)*(6*n - 8)*(6*n - 9)*(6*n - 10)*(6*n - 11)*(2117*n^4 - 4147*n^3 + 2833*n^2 - 773*n + 66)*a(n-2) with a(1) = 3, a(2) = 53.
[We note that the sequence u(n) := n^2 * Sum_{k = 0..2*n} 1/((n+2*k)*(5*n-2*k))*C(n+2*k,k)*C(5*n-2*k, 2*n-k) = (1/3)*C(6*n,2*n) is known to satisfy the congruences u(n*p^k) == u(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k - see Meštrović, equation 39. If, in the binomial sum formulas for a(n) and u(n) given above, we restrict the summation range to k = 0..n then we conjecture that the resulting pair of sequences satisfy the same congruences.]
a(n) ~ sqrt(1/24 + 1/(8*sqrt(73))) * ((2117*sqrt(73) - 12881)/216)^n / sqrt(Pi*n). - Vaclav Kotesovec, Apr 20 2020
a(n) = (1/5)*binomial(5*n, 2*n)*hypergeom([1/2 + n/2, -3*n, -2*n, n/2], [1/2 - 5*n/2, 1 - 5*n/2, 1 + n], -1) for n >= 1. - Vaclav Kotesovec, May 16 2020

A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4n)/(1 + x + x^2)^(3n).

Original entry on oeis.org

1, 1, 3, 19, 67, 251, 1137, 4803, 20035, 87013, 377753, 1634469, 7134385, 31261114, 137121113, 603206144, 2660097603, 11749336328, 51981371895, 230336544210, 1021976441817, 4539784391763, 20188837618799, 89871081815631, 400427435522737, 1785639575031501

Offset: 0

Peter Bala, Mar 01 2022

This sequence is the third in an infinite family of sequences defined as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k, so that G_1(x) = (1 + x)^2, and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. See A000984, the sequence of central binomial coefficients, for the case k = 1 and A351858 for the case k = 2. The present sequence is the case k = 3.
Given a power series G(x) with integer coefficients it is known that the sequence (g(n))n>=1 defined by g(n) := [x^n] G(x)^n satisfies the Gauss congruences g(n*p^r) == g(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
Thus a(n) satisfies the Gauss congruences. Calculation suggests that, in fact, the stronger supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. These supercongruences are known to hold for the central binomial coefficients A000984(n) = [x^n] ((1 + x)^2)^n (Meštrović, equation 39).
More generally, if r is a positive integer and s an integer then the sequence defined by a(r,s;n) = [x^(r*n)] G_3(x)^(s*n) may satisfy the same supercongruences.

			Examples of supercongruences:
a(5) - a(1) = 251 - 1 = 2*(5^3) == 0 (mod 5^3)
a(2*7) - a(2) = 137121113 - 3 = 2*5*(7^4)*5711 == 0 (mod 7^4)
a(5^2) - a(5) = 1785639575031501 - 251 = 2*(5^6)*1373*3989*10433 == 0 (mod 5^6)

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Paolo Xausa, Table of n, a(n) for n = 0..425
R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A103885, A333090, A333564, A333565, A333579, A333715, A351856, A351857, A351858.

seq(add(add(add((-1)^j*binomial(4*n,n-2*i-j-k)*binomial(4*n,i)* binomial(3*n+j-1,j)*binomial(j,k), k = 0..j), j = 0..n), i = 0..n), n = 0..25);

A351859[n_] := Sum[(-1)^j*Binomial[4*n, n-2*i-j-k]*Binomial[4*n, i]*Binomial[3*n+j-1, j]*Binomial[j, k], {i, 0, n}, {j, 0, n}, {k, 0, j}];
Array[A351859, 25, 0] (* Paolo Xausa, May 30 2025 *)

a(n)=sum(i=0,n,sum(j=0,n,sum(k=0,j,(-1)^j*binomial(4*n,n-2*i-j-k)*binomial(4*n,i)*binomial(3*n+j-1,j)*binomial(j,k))));
vector(25,n,a(n-1)) \\ Paolo Xausa, May 04 2022

a(n) = Sum_{i = 0..n} Sum_{j = 0..n} Sum_{k = 0..j} (-1)^j* C(4n,n-2*i-j-k) *C(4n,i)*C(3n+j-1,j)*C(j,k).
The o.g.f. A(x) = 1 + x + 3*x^2 + 19*x^3 + 67*x^4 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2 + x^3)^4/(1 + x + x^2)^3) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 + x + x^2)^3/(1 + x + x^2 + x^3)^4 ) = 1 + x + 2*x^2 + 8*x^3 + 25*x^4 + 81*x^5 + 305*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).

A363418 Square array read by ascending antidiagonals: T(n,k) = [x^(n*k)] ((1 + x)/(1 - x))^k for n, k >= 1.

Original entry on oeis.org

2, 2, 8, 2, 16, 38, 2, 24, 146, 192, 2, 32, 326, 1408, 1002, 2, 40, 578, 4672, 14002, 5336, 2, 48, 902, 11008, 69002, 142000, 28814, 2, 56, 1298, 21440, 216002, 1038984, 1459810, 157184, 2, 64, 1766, 36992, 525002, 4320608, 15856206, 15158272, 864146

Offset: 1

Peter Bala, Jun 12 2023

The n-th row sequence {T(n, k) : k >= 1} satisfies the Gauss congruences, that is, T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^r ) for all primes p and positive integers m and r.

We conjecture that each row sequence satisfies the stronger supercongruences T(n, m*p^r) == T(n, m*p^(r-1)) ( mod p^(3*r) ) for all primes p >= 5 and positive integers m and r.

			Square array begins
 n\k |  1   2     3      4        5          6           7
 - - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  1  |  2   8    38    192     1002       5336       28814   ...   (A002003)
  2  |  2  16   146   1408    14002     142000     1459810   ...   (A103885)
  3  |  2  24   326   4672    69002    1038984    15856206   ...   (A333715)
  4  |  2  32   578  11008   216002    4320608    87588482   ...
  5  |  2  40   902  21440   525002   13104184   331482062   ...
  6  |  2  48  1298  36992  1086002   32497680   985524066   ...
  7  |  2  56  1766  58688  2009002   70097384  2478629134   ...
  8  |  2  64  2306  87552  3424002  136485568  5513464322   ...

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 antidiagonals)

A002003 (row 1), A103885 (row 2), A333715 (row 3). Cf. A035607, A362724 - A362733, A363419.

# display as a square array
T := (n,k) -> add( binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j) , j = 0..k): for n from 1 to 10 do seq(T(n, k), k = 1..10) end do;
#alternative program
seq(print(seq(simplify(2*k*hypergeom([1 - k, 1 - n*k], [2], 2)), k = 1..10)), n = 1..10);
# display as a sequence
seq(seq(T(n+1-i, i), i = 1..n), n = 1..10);

T(n,k) = sum(j=0, k, binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j)) \\ Andrew Howroyd, Jan 05 2024

T(n,k) = Sum_{j = 0..k} binomial(k, j)*binomial((n + 1)*k - j - 1, n*k - j).
T(n,k) = 1/n * [x^k] ((1 + x)/(1 - x))*(n*k).
T(n,k) = (1/n)*Sum_{j = 0..k} binomial(n*k, j)*binomial((n + 1)*k - j - 1, k - j).
T(2*n,k) = [x^(n*k)] Chebyshev_T(k,(1 + x)/(1 - x)), where Chebyshev_T(n,x) denotes the n-th Chebyshev polynomial of the first kind. See A053120.
T(n,k) = Sum_{j = 1..k} (2^j)*binomial(k, j)*binomial(n*k - 1, n*k - j).
T(n,k) = (2*k) * hypergeom([1 - k, 1 - n*k], [2], 2).
Define E(n,x) = exp( Sum_{j >= 1} T(n,j)*x^j/j ). Then T(n+1,k) = [x^k] E(n,x)^k.
E(n,x) = (1/x) * the series reversion of x/E(n-1,x) for n >= 2.
E(n,x)^n = (1/x) * the series reversion of x*((1 - x)/(1 + x))^n.
E(m,x) appears to be the g.f. of the (m + 1)-Schroeder numbers. See A027307 (m = 2) and the cross references there.
The o.g.f. for row n is the diagonal of the bivariate rational function (1/n) * t*f(x)^n/(1 - t*f(x)^n), where f(x) = (1 + x)/(1 - x), and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.

cp's OEIS Frontend

A035607 Table a(d,m) of number of points of L1 norm m in cubic lattice Z^d, read by antidiagonals (d >= 1, m >= 0).

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A144097 The 4-Schroeder numbers: a(n) = number of lattice paths (Schroeder paths) from (0,0) to (3n,n) with unit steps N=(0,1), E=(1,0) and D=(1,1) staying weakly above the line y = 3x.

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A351858 a(n) = [x^n] (1 + x + x^2)^(3*n)/(1 + x)^(2*n).

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A336521 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) is the coefficient of x^(k*n) in expansion of ( (1 + x)/(1 - x) )^n.

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A333563 a(n) = [x^n] G(x)^n, where G(x) is the o.g.f. of A079489.

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A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4*n)/(1 + x + x^2)^(3*n).

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A351858 a(n) = [x^n] (1 + x + x^2)^(3n)/(1 + x)^(2n).

A351859 a(n) = [x^n] (1 + x + x^2 + x^3)^(4n)/(1 + x + x^2)^(3n).