A174125 -id:A174125

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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.

+10
6

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

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,8

LINKS

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

FORMULA

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)

EXAMPLE

Triangle begins as:

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;

MATHEMATICA

(* 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 *)

PROG

(Sage)

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

(Magma)

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

CROSSREFS

Cf. A174116, A174117, A174124, A174125.

Cf. A111910, A196148.

KEYWORD

nonn,tabl,easy

AUTHOR

Roger L. Bagula, Mar 08 2010

EXTENSIONS

Edited by G. C. Greubel, Feb 11 2021

STATUS

approved

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.

+10
5

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

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,8

LINKS

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

FORMULA

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)

EXAMPLE

Triangle begins as:

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;

MATHEMATICA

(* 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 *)

PROG

(Sage)

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

(Magma)

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

CROSSREFS

Cf. A174117, A174119, A174124, A174125.

Cf. A000108, A001263.

KEYWORD

nonn,tabl

AUTHOR

Roger L. Bagula, Mar 08 2010

EXTENSIONS

Edited by G. C. Greubel, Feb 11 2021

STATUS

approved

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.

+10
5

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

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,5

LINKS

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

FORMULA

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)

EXAMPLE

Triangle begins as:

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;

MATHEMATICA

(* 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 *)

PROG

(Sage)

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

(Magma)

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

CROSSREFS

Cf. A174116, A174119, A174124, A174125.

Cf. A000108, A001263.

KEYWORD

nonn,tabl

AUTHOR

Roger L. Bagula, Mar 08 2010

EXTENSIONS

Edited by G. C. Greubel, Feb 11 2021

STATUS

approved

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.

+10
5

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

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,5

COMMENTS

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

LINKS

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.

FORMULA

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)

EXAMPLE

Triangle begins as:

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;

MATHEMATICA

(* 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 *)

PROG

(Sage)

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

(Magma)

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

CROSSREFS

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

Cf. A000108, A174116, A174117, A174119.

KEYWORD

nonn,tabl

AUTHOR

Roger L. Bagula, Mar 09 2010

EXTENSIONS

Edited by G. C. Greubel, Feb 11 2021

STATUS

approved

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