|
|
A080247
|
|
Formal inverse of triangle A080246. Unsigned version of A080245.
|
|
8
|
|
|
1, 2, 1, 6, 4, 1, 22, 16, 6, 1, 90, 68, 30, 8, 1, 394, 304, 146, 48, 10, 1, 1806, 1412, 714, 264, 70, 12, 1, 8558, 6752, 3534, 1408, 430, 96, 14, 1, 41586, 33028, 17718, 7432, 2490, 652, 126, 16, 1, 206098
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
T(n,k) is the number of dissections of a convex (n+3)-gon by nonintersecting diagonals with exactly k diagonals emanating from a fixed vertex. Example: T(2,1)=4 because the dissections of the convex pentagon ABCDE having exactly one diagonal emanating from the vertex A are: {AC}, {AD}, {AC,EC} and {AD,BD}. - Emeric Deutsch, May 31 2004
|
|
LINKS
|
|
|
FORMULA
|
Essentially same triangle as triangle T(n,k), n > 0 and k > 0, read by rows; given by [0, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...] DELTA A000007 where DELTA is Deléham's operator defined in A084938.
T(n, k) = T(n-1, k-1) + 2*Sum_{j>=0} T(n-1, k+j) with T(0, 0) = 1 and T(n, k)=0 if k < 0. - Philippe Deléham, Jan 19 2004
T(n, k) = (k+1)*Sum_{j=0..n-k} (binomial(n+1, k+j+1)*binomial(n+j, j))/(n+1). - Emeric Deutsch, May 31 2004
Recurrence: T(0,0)=1; T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n,k+1). - David Callan, Jul 03 2006
T(n, k) = binomial(n, k)*hypergeom([k - n, n + 1], [k + 2], -1). - Peter Luschny, Jan 08 2018
T(n,k) = (k+1)/(n+1)*Sum_{m=0..n-k} 2^m*binomial(n+1,m)*binomial(n-k-1,n-k-m). - Vladimir Kruchinin, Jan 10 2022
|
|
EXAMPLE
|
Triangle starts:
[0] 1
[1] 2, 1
[2] 6, 4, 1
[3] 22, 16, 6, 1
[4] 90, 68, 30, 8, 1
[5] 394, 304, 146, 48, 10, 1
[6] 1806, 1412, 714, 264, 70, 12, 1
...
n-th row = top row of M^n, M = the following infinite square production matrix:
2, 1, 0, 0, 0, ...
2, 2, 1, 0, 0, ...
2, 2, 2, 1, 0, ...
2, 2, 2, 2, 1, ...
... (End)
|
|
MAPLE
|
A080247:=(n, k)->(k+1)*add(binomial(n+1, k+j+1)*binomial(n+j, j), j=0..n-k)/(n+1):
seq(seq(A080247(n, k), k=0..n), n=0..9);
|
|
MATHEMATICA
|
Clear[w] w[n_, k_] /; k < 0 || k > n := 0 w[0, 0]=1 ; w[n_, k_] /; 0 <= k <= n && !n == k == 0 := w[n, k] = w[n-1, k-1] + w[n-1, k] + w[n, k+1] Table[w[n, k], {n, 0, 10}, {k, 0, n}] (* David Callan, Jul 03 2006 *)
T[n_, k_] := Binomial[n, k] Hypergeometric2F1[k - n, n + 1, k + 2, -1];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Peter Luschny, Jan 08 2018 *)
|
|
PROG
|
(Sage)
@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, k+i-1) for i in (2..n-k+1))
return [(-1)^(n-k)*prec(n, k) for k in (1..n)]
(Maxima)
T(n, k):=((k+1)*sum(2^m*binomial(n+1, m)*binomial(n-k-1, n-k-m), m, 0, n-k))/(n+1); /* Vladimir Kruchinin, Jan 10 2022 */
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|