A174119 -id:A174119 - cp's OEIS frontend

A174116 Triangle T(n, k) = (n/2)binomial(n-1, k-1)binomial(n-1, k) with T(n, 0) = T(n, n) = 1, read by rows.

1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 6, 18, 6, 1, 1, 10, 60, 60, 10, 1, 1, 15, 150, 300, 150, 15, 1, 1, 21, 315, 1050, 1050, 315, 21, 1, 1, 28, 588, 2940, 4900, 2940, 588, 28, 1, 1, 36, 1008, 7056, 17640, 17640, 7056, 1008, 36, 1, 1, 45, 1620, 15120, 52920, 79380, 52920, 15120, 1620, 45, 1

Offset: 0

Roger L. Bagula, Mar 08 2010

			Triangle begins as:
  1;
  1,  1;
  1,  1,    1;
  1,  3,    3,     1;
  1,  6,   18,     6,     1;
  1, 10,   60,    60,    10,     1;
  1, 15,  150,   300,   150,    15,     1;
  1, 21,  315,  1050,  1050,   315,    21,     1;
  1, 28,  588,  2940,  4900,  2940,   588,    28,    1;
  1, 36, 1008,  7056, 17640, 17640,  7056,  1008,   36,  1;
  1, 45, 1620, 15120, 52920, 79380, 52920, 15120, 1620, 45, 1;

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

Cf. A174117, A174119, A174124, A174125.

Cf. A000108, A001263.

T:= func< n,k | k eq 0 or k eq n select 1 else (n/2)*Binomial(n-1, k-1)*Binomial(n-1, 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[Binomial[j,2], {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/2)*Binomial[n-1, k-1]*Binomial[n-1, 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/2)*binomial(n-1, k-1)*binomial(n-1, k)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 11 2021

Let c(n) = Product_{j=2..n} binomial(j,2) for n > 1 otherwise 1 then the number triangle is given by T(n, k) = c(n)/(c(k)*c(n-k)).
From G. C. Greubel, Feb 11 2021: (Start)
T(n, k) = (n/2)*binomial(n-1, k-1)*binomial(n-1, k) with T(n, 0) = T(n, n) = 1.
T(n, k) = binomial(n-k+1, 2)*A001263(n, k) with T(n, 0) = T(n, n) = 1.
Sum_{k=0..n} T(n,k) = binomial(n, 2)*C_{n-1} + 2 - [n=0], where C_{n} are the Catalan numbers (A000108) and [] is the Iverson bracket. (End)

Edited by G. C. Greubel, Feb 11 2021

A174117 Triangle T(n, k) = (2k/(k+1))binomial(n-1, k)*binomial(n+1, k) with T(n, 0) = T(n, n) = 1, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 8, 8, 1, 1, 15, 40, 15, 1, 1, 24, 120, 120, 24, 1, 1, 35, 280, 525, 280, 35, 1, 1, 48, 560, 1680, 1680, 560, 48, 1, 1, 63, 1008, 4410, 7056, 4410, 1008, 63, 1, 1, 80, 1680, 10080, 23520, 23520, 10080, 1680, 80, 1, 1, 99, 2640, 20790, 66528, 97020, 66528, 20790, 2640, 99, 1

Offset: 0

Roger L. Bagula, Mar 08 2010

			Triangle begins as:
  1;
  1,  1;
  1,  3,    1;
  1,  8,    8,     1;
  1, 15,   40,    15,     1;
  1, 24,  120,   120,    24,     1;
  1, 35,  280,   525,   280,    35,     1;
  1, 48,  560,  1680,  1680,   560,    48,     1;
  1, 63, 1008,  4410,  7056,  4410,  1008,    63,    1;
  1, 80, 1680, 10080, 23520, 23520, 10080,  1680,   80,  1;
  1, 99, 2640, 20790, 66528, 97020, 66528, 20790, 2640, 99, 1;

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

Cf. A174116, A174119, A174124, A174125.

Cf. A000108, A001263.

T:= func< n,k | k eq 0 or k eq n select 1 else (2*k/(k+1))*Binomial(n-1, k)*Binomial(n+1, 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[i^2 -1, {i,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, (2*k/(k+1))*Binomial[n+1, k]*Binomial[n-1, 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 (2*k/(k+1))*binomial(n-1, k)*binomial(n+1, k)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 11 2021

Let c(n) = Product_{j=2..n} (j^2 - 1) for n > 1 otherwise 1 then the number triangle is given by T(n, k) = c(n)/(c(k)*c(n-k)).
From G. C. Greubel, Feb 11 2021: (Start)
T(n, k) = (2*k/(k+1))*binomial(n-1, k)*binomial(n+1, k) with T(n, 0) = T(n, n) = 1.
T(n, k) = 2*((n+1)*(n-k)/(k+1))*A001263(n, k).
Sum_{k=0..n} T(n, k) = (2/(n+2))*( (n^2-1)*C_{n} + 1), where C_{n} are the Catalan numbers (A000108). (End)

Edited by G. C. Greubel, Feb 11 2021

A174124 Triangle T(n, k, q) = (q+1)binomial(n, k)(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and q = 1, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 12, 12, 1, 1, 20, 40, 20, 1, 1, 30, 100, 100, 30, 1, 1, 42, 210, 350, 210, 42, 1, 1, 56, 392, 980, 980, 392, 56, 1, 1, 72, 672, 2352, 3528, 2352, 672, 72, 1, 1, 90, 1080, 5040, 10584, 10584, 5040, 1080, 90, 1, 1, 110, 1650, 9900, 27720, 38808, 27720, 9900, 1650, 110, 1

Offset: 0

Roger L. Bagula, Mar 09 2010

Triangles of this class, depending upon q, are of the form T(n, k, q) = (q+1)*binomial(n, k)*(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and have the row sums Sum_{k=0..n} T(n, k, q) = q*(q+1)*C_{n+q}/binomial(n+2*q, q-1) - 2*q + q*[n=0], where C_{n} are the Catalan numbers (A000108) and [] is the Iverson bracket. - G. C. Greubel, Feb 11 2021

			Triangle begins as:
  1;
  1,   1;
  1,   6,    1;
  1,  12,   12,    1;
  1,  20,   40,   20,     1;
  1,  30,  100,  100,    30,     1;
  1,  42,  210,  350,   210,    42,     1;
  1,  56,  392,  980,   980,   392,    56,    1;
  1,  72,  672, 2352,  3528,  2352,   672,   72,    1;
  1,  90, 1080, 5040, 10584, 10584,  5040, 1080,   90,   1;
  1, 110, 1650, 9900, 27720, 38808, 27720, 9900, 1650, 110, 1;

Michael De Vlieger, Table of n, a(n) for n = 0..11475
Samuele Giraudo, Tree series and pattern avoidance in syntax trees, arXiv:1903.00677 [math.CO], 2019.

Cf. this sequence (q=1), A174125 (q=2).

Cf. A000108, A174116, A174117, A174119.

c:= func< n,q | n lt 2 select 1 else (&*[j*(j+q): j in [2..n]]) >;
T:= func< n,k,q | c(n, q)/(c(k, q)*c(n-k, q)) >;
[T(n,k,1): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 11 2021

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

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

Let c(n, q) = Product_{j=2..n} j*(j+q) for n > 2, otherwise 1, then the number triangle is given by T(n, k, q) = c(n, q)/(c(k, q)*c(n-k, q)) for q = 1.
From G. C. Greubel, Feb 11 2021: (Start)
T(n, k, q) = (q+1)*binomial(n, k)*(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and q = 1.
Sum_{k=0..n} T(n, k, 1) = 2*A000108(n+1) - 2 + [n=0]. (End)

Edited by G. C. Greubel, Feb 11 2021

A174125 Triangle T(n, k, q) = (q+1)binomial(n, k)(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and q = 2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 15, 15, 1, 1, 24, 45, 24, 1, 1, 35, 105, 105, 35, 1, 1, 48, 210, 336, 210, 48, 1, 1, 63, 378, 882, 882, 378, 63, 1, 1, 80, 630, 2016, 2940, 2016, 630, 80, 1, 1, 99, 990, 4158, 8316, 8316, 4158, 990, 99, 1, 1, 120, 1485, 7920, 20790, 28512, 20790, 7920, 1485, 120, 1

Offset: 0

Roger L. Bagula, Mar 09 2010

			Triangle begins as:
  1;
  1,   1;
  1,   8,    1;
  1,  15,   15,    1;
  1,  24,   45,   24,     1;
  1,  35,  105,  105,    35,     1;
  1,  48,  210,  336,   210,    48,     1;
  1,  63,  378,  882,   882,   378,    63,    1;
  1,  80,  630, 2016,  2940,  2016,   630,   80,    1;
  1,  99,  990, 4158,  8316,  8316,  4158,  990,   99,   1;
  1, 120, 1485, 7920, 20790, 28512, 20790, 7920, 1485, 120, 1;

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

Cf. A174124 (q=1), this sequence (q=2).

Cf. A000108, A174116, A174117, A174119.

c:= func< n,q | n lt 2 select 1 else (&*[j*(j+q): j in [2..n]]) >;
T:= func< n,k,q | c(n, q)/(c(k, q)*c(n-k, q)) >;
[T(n,k,2): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 11 2021

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

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

Let c(n, q) = Product_{j=2..n} j*(j+q) for n > 2, otherwise 1, then the number triangle is given by T(n, k, q) = c(n, q)/(c(k, q)*c(n-k, q)) for q = 2.
From G. C. Greubel, Feb 11 2021: (Start)
T(n, k, q) = (q+1)*binomial(n, k)*(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and q = 2.
Sum_{k=0..n} T(n, k, 1) = 6*A000108(n+2)/(n+4) - 4 + 2*[n=0]. (End)

Edited by G. C. Greubel, Feb 11 2021

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

cp's OEIS Frontend

A174116 Triangle T(n, k) = (n/2)*binomial(n-1, k-1)*binomial(n-1, k) with T(n, 0) = T(n, n) = 1, read by rows.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Sage

Formula

Extensions

A174117 Triangle T(n, k) = (2*k/(k+1))*binomial(n-1, k)*binomial(n+1, k) with T(n, 0) = T(n, n) = 1, read by rows.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Sage

Formula

Extensions

A174124 Triangle T(n, k, q) = (q+1)*binomial(n, k)*(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and q = 1, read by rows.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Sage

Formula

Extensions

A174125 Triangle T(n, k, q) = (q+1)*binomial(n, k)*(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and q = 2, read by rows.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Sage

Formula

Extensions

A196148 Antidiagonal sums of square array A111910.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

A174116 Triangle T(n, k) = (n/2)binomial(n-1, k-1)binomial(n-1, k) with T(n, 0) = T(n, n) = 1, read by rows.

A174117 Triangle T(n, k) = (2k/(k+1))binomial(n-1, k)*binomial(n+1, k) with T(n, 0) = T(n, n) = 1, read by rows.

A174124 Triangle T(n, k, q) = (q+1)binomial(n, k)(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and q = 1, read by rows.

A174125 Triangle T(n, k, q) = (q+1)binomial(n, k)(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and q = 2, read by rows.