|
| |
|
|
A061554
|
|
Square table read by antidiagonals: a(n,k)=C(n+k,[k/2]).
|
|
39
| |
|
|
1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 6, 4, 4, 1, 1, 10, 10, 5, 5, 1, 1, 20, 15, 15, 6, 6, 1, 1, 35, 35, 21, 21, 7, 7, 1, 1, 70, 56, 56, 28, 28, 8, 8, 1, 1, 126, 126, 84, 84, 36, 36, 9, 9, 1, 1, 252, 210, 210, 120, 120, 45, 45, 10, 10, 1, 1, 462, 462, 330, 330, 165, 165, 55, 55, 11, 11, 1, 1
(list; table; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,4
|
|
|
COMMENTS
| Equivalently, a triangle read by rows, where the rows are obtained by sorting the elements of the rows of Pascal's triangle (A007318) into descending order. - Philippe DELEHAM, May 21 2005
Equivalently, as a triangle read by rows, this is T(n,k)=binomial(n,floor((n-k)/2)); column k then has e.g.f. Bessel_I(k,2x)+Bessel_I(k+1,2x). - Paul Barry (pbarry(AT)wit.ie), Feb 28 2006
Antidiagonal sums are A037952(n+1) = C(n+1,[n/2]). Matrix inverse is the row reversal of triangle A066170. Eigensequence is A125094(n) = Sum_{k=0..n-1} A125093(n-1,k)*A125094(k). - Paul D. Hanna (pauldhanna(AT)juno.com), Nov 20 2006
Riordan array (1/(1-x-x^2*c(x^2)),x*c(x^2)); where c(x)=g.f.for Catalan numbers A000108 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Mar 17 2007
Triangle T(n,k), 0<=k<=n, read by rows given by : T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+T(n-1,k+1) for k>=1 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Mar 27 2007
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1 . Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0)-> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; ((1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Sep 25 2007
T(n,k) is the number of paths from (0,k) to some (n,m) which never dip below y=0, touch y=0 at least once and are made up only of the steps (1,1) and (1,-1). This can be proved using the recurrence supplied by DELEHAM. [From Gerald McGarvey (gerald.mcgarvey(AT)comcast.net), Oct 15 2008]
Triangle read by rows = partial sums of A053121 terms starting from the right. [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 24 2008]
As a subset of the "family of triangles" (Deleham comment of Sep 25 2007), beginning with a signed variant of A061554, M = (-1,0) = (1; -1, 1; 2, -1, 1; -3, 3, -1, 1;...) successive binomial transforms of M yield (0,1) - A089942; (1,2) - A039599; (2,3) - A124733; (3,4) - A124574; (4,5) - A126331;...such that the binomial transform of the triangle generated from (n,n+1) = the triangle generated from (n+1,n+2). Similarly, another subset beginning with A053121 - (0,0), and taking successive binomial transforms yields (1,1) - A064189; (2,2) - A039598; (3,3) - A091965,... By rows, the triangle generated from (n,n) can be obtained by taking pairwise sums from the (n-1,n) triangle starting from the right. For example, row 2 of (1,2) - A039599 = (2, 3, 1); and taking pairwise sums from the right we obtain (5, 4, 1) = row 2 of (2,2) - A039598. - Gary W. Adamson, Aug 04 2011
The triangle by rows (n) with alternating signs (+-+...) from the top as a set of simultaneous equations solves for diagonal lengths of odd N (N = 2n+1) regular polygons. The constants in each case are powers of c = 2*Cos 2Pi/N. By way of example, the first 3 rows relate to the heptagon and the simultaneous equations are (1,0,0) = 1; (-1,1,0) = c = 1.24697...; and (2,-1,1) = c^2. The answers are 1, 2.24697..., and 1.801...; the 3 distinct diagonal lengths of the heptagon with edge = 1. - Gary W. Adamson, Sep 07 2011.
|
|
|
FORMULA
| As a triangle: T(n,k) = C(n,m) where m = floor[(n+1)/2 - (-1)^(n-k)*(k+1)/2].
a(0, k)=C(k, [k/2])=A001405(k); for n>0 T(n, k)=T(n+1, k-2)+T(n-1, k).
n-th row = M^n * V, where M = the infinite tridiagonal matrix with all 1's in the super and subdiagonals and (1,0,0,0...) in the main diagonal. V = the infinite vector [1,0,0,0...]. Example: (3,3,1,1,0,0,0...) = M^3 * V. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Nov 04 2006
Sum_{k, 0<=k<=n}T(m,k)*T(n,k)=T(m+n,0)=A001405(m+n). - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Feb 26 2007
Sum{k, 0<=k<=n}T(n,k)=2^n . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Mar 27 2007
Sum_{k, 0<=k<=n} T(n,k)*x^k = A127361(n), A126869(n), A001405(n), A000079(n), A127358(n), A127359(n), A127360(n) for x = -2, -1, 0, 1, 2, 3, 4 respectively. [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Dec 04 2009]
|
|
|
EXAMPLE
| Triangle version begins:
1;
1, 1;
2, 1, 1;
3, 3, 1, 1;
6, 4, 4, 1, 1;
10, 10, 5, 5, 1, 1;
20, 15, 15, 6, 6, 1, 1;
35, 35, 21, 21, 7, 7, 1, 1;
70, 56, 56, 28, 28, 8, 8, 1, 1;
126, 126, 84, 84, 36, 36, 9, 9, 1, 1;
252, 210, 210, 120, 120, 45, 45, 10, 10, 1, 1;
462, 462, 330, 330, 165, 165, 55, 55, 11, 11, 1, 1; ...
Matrix inverse begins:
1;
-1, 1;
-1, -1, 1;
1, -2, -1, 1;
1, 2, -3, -1, 1;
-1, 3, 3, -4, -1, 1;
-1, -3, 6, 4, -5, -1, 1;
1, -4, -6, 10, 5, -6, -1, 1;
1, 4, -10, -10, 15, 6, -7, -1, 1; ...
Contribution from Paul Barry (pbarry(AT)wit.ie), May 21 2009: (Start)
Production matrix is
.1, 1,
.1, 0, 1,
.0, 1, 0, 1,
.0, 0, 1, 0, 1,
.0, 0, 0, 1, 0, 1,
.0, 0, 0, 0, 1, 0, 1,
.0, 0, 0, 0, 0, 1, 0, 1 (End)
|
|
|
MATHEMATICA
| t[n_, k_] = Binomial[n, Floor[(n+1)/2 - (-1)^(n-k)*(k+1)/2]]; Flatten[Table[t[n, k], {n, 0, 11}, {k, 0, n}]] (* From Jean-François Alcover, May 31 2011 *)
|
|
|
PROG
| (PARI) T(n, k)=binomial(n, (n+1)\2-(-1)^(n-k)*((k+1)\2))
|
|
|
CROSSREFS
| Rows are A001405, A037952, A037955, A037951, A037956, A037953, A037957 etc. Columns are truncated pairs of A000012, A000027, A000217, A000292, A000332, A000389, A000579, etc. Main diagonal is alternate values of A051036.
Cf. A007318, A107430, A125094, A037952, A066170.
Sequence in context: A194672 A034364 A090011 * A088326 A181039 A124975
Adjacent sequences: A061551 A061552 A061553 * A061555 A061556 A061557
|
|
|
KEYWORD
| nonn,tabl
|
|
|
AUTHOR
| Henry Bottomley (se16(AT)btinternet.com), May 17 2001
|
|
|
EXTENSIONS
| Entry revised by N. J. A. Sloane (njas(AT)research.att.com), Nov 22 2006
|
| |
|
|
