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A061554
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Square table read by antidiagonals: a(n,k) = binomial(n+k, floor(k/2)).
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45
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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
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OFFSET
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0,4
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COMMENTS
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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 Deléham, 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, Feb 28 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 Deléham, 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 Deléham, 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 Deléham, 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 Deléham. - Gerald McGarvey, Oct 15 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(2*Pi/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
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LINKS
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Reed Acton, T. Kyle Petersen, Blake Shirman, and Bridget Eileen Tenner, The clairvoyant maître d', arXiv:2401.11680 [math.CO], 2024. See p. 15.
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FORMULA
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As a triangle: T(n,k) = binomial(n,m) where m = floor((n+1)/2 - (-1)^(n-k)*(k+1)/2).
a(0, k) = binomial(k, floor(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, Nov 04 2006
Sum_{k=0..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. - Philippe Deléham, Dec 04 2009
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EXAMPLE
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The array starts:
1, 1, 2, 3, 6, 10, 20, 35, 70, 126, ...
1, 1, 3, 4, 10, 15, 35, 56, 126, 210, ...
1, 1, 4, 5, 15, 21, 56, 84, 210, 330, ...
1, 1, 5, 6, 21, 28, 84, 120, 330, 495, ...
1, 1, 6, 7, 28, 36, 120, 165, 495, 715, ...
1, 1, 7, 8, 36, 45, 165, 220, 715, 1001, ...
1, 1, 8, 9, 45, 55, 220, 286, 1001, 1365, ...
1, 1, 9, 10, 55, 66, 286, 364, 1365, 1820, ...
1, 1, 10, 11, 66, 78, 364, 455, 1820, 2380, ...
1, 1, 11, 12, 78, 91, 455, 560, 2380, 3060, ...
Triangle (antidiagonal) 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; ...
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)
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MAPLE
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T := proc(n, k) option remember;
if n = k then 1 elif k < 0 or n < 0 or k > n then 0
elif k = 0 then T(n-1, 0) + T(n-1, 1) else T(n-1, k-1) + T(n-1, k+1) fi end:
for n from 0 to 9 do seq(T(n, k), k = 0..n) od; # Peter Luschny, May 25 2021
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MATHEMATICA
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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}]] (* Jean-François Alcover, May 31 2011 *)
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PROG
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(PARI) T(n, k)=binomial(n, (n+1)\2-(-1)^(n-k)*((k+1)\2))
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CROSSREFS
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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.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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