A111910 -id:A111910 - cp's OEIS frontend

A140136 Numerator coefficients for generators of lattice path enumeration square array A111910.

1, 1, 1, 1, 7, 7, 1, 1, 20, 75, 75, 20, 1, 1, 42, 364, 1001, 1001, 364, 42, 1, 1, 75, 1212, 6720, 15288, 15288, 6720, 1212, 75, 1, 1, 121, 3223, 30723, 127908, 255816, 255816, 127908, 30723, 3223, 121, 1, 1, 182, 7371, 109538, 737737, 2510508

Offset: 0

Paul Barry, May 09 2008

			Triangle begins:
  1;
  1,  1;
  1,  7,    7,     1;
  1, 20,   75,    75,    20,     1;
  1, 42,  364,  1001,  1001,   364,   42,    1;
  1, 75, 1212,  6720, 15288, 15288, 6720, 1212, 75, 1;
  ...

Michael De Vlieger, Table of n, a(n) for n = 0..10100 (rows 0..100, flattened)
G. Kreweras, Sur une classe de problèmes de dénombrement liés au treillis des partitions des entiers, Cahiers du B.U.R.O. 6 (1965), 9-107; see p. 93.
G. Kreweras and H. Niederhausen, Solution of an enumerative problem connected with lattice paths, European J. Combin. 2 (1981), 55-60; see p. 60.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 8.

Row sums are A006335.

Cf. A111910.

T[n_, k_] := ((k + n - 1)! (2 (k + n) - 3)! HypergeometricPFQ[{2 - 3 k, 1/2 - n, 1 - n, -n}, {1 - k - n, 3/2 - k - n, 2 - k - n}, 1])/(k! (2 k - 1)! n! (2 n - 1)!);
Join[{{1}}, Table[T[n, k], {k, 2, 8}, {n, 1, 2 k - 2}]] // Flatten (* Peter Luschny, Sep 04 2019 *)

(Sum_{k=0..n} T(n,k) * x^k) / (1-x)^(3*n+1) generates row n of A111910.

Triangle T(q,n), where T(n,q) = Sum_{j = 0..n} (-1)^j*C(3*q+1,j)*K(n-j,q) with K(p,q) = A111910(p,q).

A196148 Antidiagonal sums of square array A111910.

Original entry on oeis.org

1, 2, 7, 30, 146, 772, 4331, 25398, 154158, 961820, 6137734, 39909740, 263665252, 1765815560, 11966535091, 81937361702, 566185489878, 3944202596652, 27676632525362, 195481707009220, 1388890568962556

Offset: 0

Peter Bala, Oct 13 2011

Michael De Vlieger, Table of n, a(n) for n = 0..1000
Anthony James Wood, Nonequilibrium steady states from a random-walk perspective, Ph. D. Thesis, The University of Edinburgh (Scotland, UK 2019).
Anthony J. Wood, Richard A. Blythe, and Martin R. Evans, Renyi entropy of the totally asymmetric exclusion process, arXiv:1708.00303 [cond-mat.stat-mech], 2017.
Anthony J. Wood, Richard A. Blythe, and Martin R. Evans, Combinatorial mappings of exclusion processes, arXiv:1908.00942 [cond-mat.stat-mech], 2019.

Cf. A111910.

Cf. A174119.

[(&+[(n-j+1)*Binomial(n+1, j)*Binomial(2*n+4, 2*j+2)/((n+1)*(n+2)*(2*n+3)): j in [0..n]]): n in [0..25]]; // G. C. Greubel, Feb 11 2021

Table[Sum[(n+1)! * (2*n+1)! / ((n-k+1)! * (k+1)! * (2*n-2*k+1)! * (2*k+1)!), {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Dec 16 2017 *)
Table[HypergeometricPFQ[{-n, -n-1/2, -n-1}, {3/2, 2}, -1], {n,0,25}] (* G. C. Greubel, Feb 11 2021 *)

S(n,k) = (n+k+1)!*(2*n+2*k+1)!/((n+1)!*(k+1)!*(2*n+1)!*(2*k+1)!);
a(n) = sum(k = 0, n, S(n-k,k)); \\ Michel Marcus, Dec 16 2017

[hypergeometric([-n, -n-1/2, -n-1], [3/2, 2], -1).simplify_hypergeometric() for n in (0..25)] # G. C. Greubel, Feb 11 2021

a(n) = Sum_{k = 0..n} S(n-k,k) where S(n,k) = (n+k+1)!*(2*n+2*k+1)!/((n+1)!*(k+1)!*(2*n+1)!*(2*k+1)!).
From Vaclav Kotesovec, Dec 16 2017: (Start)
a(n) ~ 2^(3*n+3) / (sqrt(3*Pi) * n^(5/2)).
Recurrence: (n+2)*(2*n+3)*a(n) = 2*(7*n^2 + 7*n + 1)*a(n-1) + 8*(n-1)*(2*n-1)*a(n-2). (End)
a(n) = hypergeometric3F2([-n, -n-1/2, -n-1], [3/2, 2], -1). - G. C. Greubel, Feb 11 2021
Let E(x) = Sum_{n >= 0} x^n/((n+1)!*(2*n+1)!). Then E(x)^2 = 1 + 2*x/(2!*3!) + 7*x^2/(3!*5!) + 30*x^3/(4!*7!) + ... + a(n)*x^n/((n+1)!*(2*n+1)!) + ... is a generating function for the sequence. - Peter Bala, Sep 20 2021

A174119 Triangle T(n, k) = ((n-k)/6)binomial(n-1, k-1)binomial(2n, 2k) with T(n, 0) = T(n, n) = 1, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 5, 5, 1, 1, 14, 70, 14, 1, 1, 30, 420, 420, 30, 1, 1, 55, 1650, 4620, 1650, 55, 1, 1, 91, 5005, 30030, 30030, 5005, 91, 1, 1, 140, 12740, 140140, 300300, 140140, 12740, 140, 1, 1, 204, 28560, 519792, 2042040, 2042040, 519792, 28560, 204, 1, 1, 285, 58140, 1627920, 10581480, 19399380, 10581480, 1627920, 58140, 285, 1

Offset: 0

Roger L. Bagula, Mar 08 2010

			Triangle begins as:
  1;
  1,   1;
  1,   1,     1;
  1,   5,     5,       1;
  1,  14,    70,      14,        1;
  1,  30,   420,     420,       30,        1;
  1,  55,  1650,    4620,     1650,       55,        1;
  1,  91,  5005,   30030,    30030,     5005,       91,       1;
  1, 140, 12740,  140140,   300300,   140140,    12740,     140,     1;
  1, 204, 28560,  519792,  2042040,  2042040,   519792,   28560,   204,   1;
  1, 285, 58140, 1627920, 10581480, 19399380, 10581480, 1627920, 58140, 285, 1;

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A174116, A174117, A174124, A174125.

Cf. A111910, A196148.

T:= func< n,k | k eq 0 or k eq n select 1 else ((n-k)/6)*Binomial(n-1, k-1)*Binomial(2*n, 2*k) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 11 2021

(* First program *)
c[n_]:= If[n<2, 1, Product[j*(j-1)*(2*j-1)/6, {j, 2, n}]];
T[n_, k_]:= c[n]/(c[k]*c[n-k]);
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten
(* Second program *)
T[n_, k_]:= If[k==0 || k==n, 1, ((n-k)/6)*Binomial[n-1, k-1]*Binomial[2*n, 2*k]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Feb 11 2021 *)

def T(n,k): return 1 if (k==0 or k==n) else ((n-k)/6)*binomial(n-1, k-1)*binomial(2*n, 2*k)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 11 2021

T(n, k) = c(n)/(c(k)*c(n-k)) where c(n) = Product_{j=2..n} j*(j-1)*(2*j-1)/6 for n > 2 otherwise 1.
From G. C. Greubel, Feb 11 2021: (Start)
T(n, k) = ((n-k)/6)*binomial(n-1, k-1)*binomial(2*n, 2*k) with T(n, 0) = T(n, n) =1.
Sum_{k=0..n} T(n, k) = (n*(n-1)*(2*n-1)/6)*HypergeometricPFQ[{1-n, 3/2-n, 2-n}, {3/2, 2}, -1] + 2 - [n=0] (n*(n-1)*(2*n-1)/6)*A196148[n-2] + 2 - [n=0]. (End)

Edited by G. C. Greubel, Feb 11 2021

A091044 One half of odd-numbered entries of even-numbered rows of Pascal's triangle A007318.

Original entry on oeis.org

1, 2, 2, 3, 10, 3, 4, 28, 28, 4, 5, 60, 126, 60, 5, 6, 110, 396, 396, 110, 6, 7, 182, 1001, 1716, 1001, 182, 7, 8, 280, 2184, 5720, 5720, 2184, 280, 8, 9, 408, 4284, 15912, 24310, 15912, 4284, 408, 9, 10, 570, 7752, 38760, 83980, 83980, 38760, 7752, 570, 10, 11

Offset: 1

Wolfdieter Lang, Jan 23 2004

The odd-numbered columns of this triangle can be reduced: see triangle A091043.
The odd-numbered rows coincide with the ones of the reduced triangle A091043.
binomial(2*n,2*m+1) is even for n >= m + 1 >= 1, hence every T(n,m) is a positive integer.
The GCD (greatest common divisor) of the entries of each odd-numbered row n=2*k+1, k>=0, is 1.
The GCD of the entries of the even-numbered row n=2*k, k>=1, is A006519(n) (highest power of 2 in n=2*k).

			Triangle begins:
  [1];
  [2,2];
  [3,10,3];
  [4,28,28,4];
  [5,60,126,60,5];
  [6,110,396,396,110,6];
  ...
n = 6 = 2*3: gcd(6,110,396) = 2 = A006519(6);
n = 5: gcd(5,60,126) = 1 = A006519(5).

Indranil Ghosh, Rows 1..125, flattened
Matthew Blair, Rigoberto Flórez, and Antara Mukherjee, Honeycombs in the Pascal triangle and beyond, arXiv:2203.13205 [math.HO], 2022. See p. 4.
Kevin Buchin, Man-Kwun Chiu, Stefan Felsner, Günter Rote, and André Schulz, The Number of Convex Polyominoes with Given Height and Width, arXiv:1903.01095 [math.CO], 2019.
Hernan de Alba, W. Carballosa, J. Leaños, and L. M. Rivera, Independence and matching numbers of some token graphs, arXiv preprint arXiv:1606.06370 [math.CO], 2016.
Wolfdieter Lang, First 9 rows.

Cf. A000302 (row sums), A001109, A006519, A007318, A053124, A091042, A091043, A111910.

Flatten[Table[Binomial[2n,2m+1]/2,{n,1,11},{m,0,n-1}]] (* Indranil Ghosh, Feb 22 2017 *)

{A(i, j) = binomial(2*i + 2*j - 2, 2*i - 1) / 2}; /* Michael Somos, Oct 15 2017 */

T(n, m)= binomial(2*n, 2*m+1)/2, n >= m + 1 >= 1, else 0.
  Put a(n) = n!*(n+1/2)!/(1/2)!. T(n+1,k) = (n+1)*a(n)/(a(k)*a(n-k)).
  T(n-1,k-1)*T(n,k+1)*T(n+1,k) = T(n-1,k)*T(n,k-1)*T(n+1,k+1). Cf. A111910. - Peter Bala, Oct 13 2011
From Peter Bala, Jul 29 2013: (Start)
O.g.f.: 1/(1 - 2*t*(x + 1) + t^2*(x - 1)^2)= 1 + (2 + 2*x)*t + (3 + 10*x + 3*x^2)*t^2 + ....
The n-th row polynomial R(n,x) = 1/(4*sqrt(x))*( (1 + sqrt(x))^(2*n) - (sqrt(x) - 1)^(2*n) ) and has n-1 real zeros given by the formula -cot^2(k*Pi/(2*n)) for k = 1,2,...,n-1. Cf A091042.
The row polynomial R(n,x) satisfies (x - 1)^n*R(n,x/(x - 1)) = U(n,2*x - 1), the n-th row polynomial of A053124.
Row sums A000302. Sum {k = 0..n-1} 2^k*T(n,k) = A001109(n). (End)
From Werner Schulte, Jan 13 2017: (Start)
(1) T(n,m) = T(n-1,m) + T(n-1,m-1)*(2*n-1-m)/m for 0 < m < n-1 with T(n,0) = n and T(n,n) = 0;
(2) T(n,m) = 2*T(n-1,m) + 2*T(n-1,m-1) - T(n-2,m) + 2*T(n-2,m-1) - T(n-2,m-2) for 0 < m < n-1 with T(n,0) = T(n,n-1) = n and T(n,m) = 0 if m < 0 or m >= n;
(3) The row polynomials p(n,x) = Sum_{m=0..n-1} T(n,m)*x^m satisfy the recurrence equation p(n+2,x) = (2+2*x)*p(n+1,x) - (x-1)^2*p(n,x) for n >= 1 with initial values p(1,x) = 1 and p(2,x) = 2+2*x.
(End)
G.f.: x*y /(1 - 2*(x+y) + (x-y)^2) with the entries regarded as an infinite square array A(i, j) read by antidiagonals. - Michael Somos, Oct 15 2017

A111911 a(n) = (4n+1)!/( (2n+1)! * ((n+1)!)^2 ).

Original entry on oeis.org

1, 5, 84, 2145, 68068, 2469012, 98062800, 4159088505, 185392049700, 8592433629780, 410935420867920, 20167102448028900, 1011343194858833424, 51656474975499371600, 2680436673901084633920, 141007991981718802584105, 7507710828193055843153700

Offset: 0

Emeric Deutsch, Aug 19 2005

nonn

Main diagonal of the square array A111910, i.e., a(n) = A111910(n,n).

G. C. Greubel, Table of n, a(n) for n = 0..550
G. Kreweras and H. Niederhausen, Solution of an enumerative problem connected with lattice paths, European J. Combin., 2 (1981), 55-60.

Cf. A111910.

Cf. A000108.

[((4*n+1)/(n+1))*Catalan(n)*Catalan(2*n): n in [0..30]]; // G. C. Greubel, Feb 12 2021

a:=n->1/(2*n+1)!*(4*n+1)!/(n+1)!^2: seq(a(n),n=1..17);
ogf := -1/(4*x)-Int(x^(-3/2)*hypergeom([-1/4, 1/4],[1],64*x),x)/(8*x^(1/2));
series( eval(ogf, Int = proc(a,x) int(series(a,x=0,32),x) end), x=0, 30); # Mark van Hoeij, May 01 2013

Table[((4*n+1)/(n+1))*CatalanNumber[n]*CatalanNumber[2*n], {n,0,30}] (* G. C. Greubel, Feb 12 2021 *)

[((4*n+1)/(n+1))*catalan_number(n)*catalan_number(2*n) for n in (0..30)] # G. C. Greubel, Feb 12 2021

G.f.: expression with a 2F1 function and an anti-derivative, see Maple program below. - Mark van Hoeij, May 01 2013
a(n) ~ 2^(6*n + 1/2) / (Pi * n^3). - Vaclav Kotesovec, Dec 16 2017
D-finite with recurrence (2*n+1)*(n+1)^2*a(n) -4*(4*n+1)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Feb 08 2021
From G. C. Greubel, Feb 12 2021: (Start)
a(n) = binomial(4*n+1, 2*n+1)*binomial(2*n, n)/(n+1)^2.
a(n) = ((4*n+1)/(n+1))*C_{n}*C_{2*n}, where C_{n} are the Catalan numbers (A000108). (End)

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A140136 Numerator coefficients for generators of lattice path enumeration square array A111910.

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A196148 Antidiagonal sums of square array A111910.

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A174119 Triangle T(n, k) = ((n-k)/6)*binomial(n-1, k-1)*binomial(2*n, 2*k) with T(n, 0) = T(n, n) = 1, read by rows.

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A091044 One half of odd-numbered entries of even-numbered rows of Pascal's triangle A007318.

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A111911 a(n) = (4*n+1)!/( (2*n+1)! * ((n+1)!)^2 ).

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A174119 Triangle T(n, k) = ((n-k)/6)binomial(n-1, k-1)binomial(2n, 2k) with T(n, 0) = T(n, n) = 1, read by rows.

A111911 a(n) = (4n+1)!/( (2n+1)! * ((n+1)!)^2 ).