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%I #74 Jan 25 2024 11:50:25
%S 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,
%T 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,
%U 210,210,120,120,45,45,10,10,1,1,462,462,330,330,165,165,55,55,11,11,1,1
%N Square table read by antidiagonals: a(n,k) = binomial(n+k, floor(k/2)).
%C 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
%C 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
%C 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_, Nov 20 2006
%C 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
%C 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
%C 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
%C 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
%C Triangle read by rows = partial sums of A053121 terms starting from the right. - _Gary W. Adamson_, Oct 24 2008
%C 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
%C 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
%H G. C. Greubel, <a href="/A061554/b061554.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%H Reed Acton, T. Kyle Petersen, Blake Shirman, and Bridget Eileen Tenner, <a href="https://arxiv.org/abs/2401.11680">The clairvoyant maître d'</a>, arXiv:2401.11680 [math.CO], 2024. See p. 15.
%H Johann Cigler, <a href="http://arxiv.org/abs/1501.04750">Some remarks and conjectures related to lattice paths in strips along the x-axis</a>, arXiv:1501.04750 [math.CO], 2015.
%H Aoife Hennessy, <a href="http://repository.wit.ie/1693/1/AoifeThesis.pdf">A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths</a>, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
%H Yidong Sun and Luping Ma, <a href="https://doi.org/10.1016/j.ejc.2014.01.004">Minors of a class of Riordan arrays related to weighted partial Motzkin paths</a>. Eur. J. Comb. 39, 157-169 (2014), Table 2.2.
%F As a triangle: T(n,k) = binomial(n,m) where m = floor((n+1)/2 - (-1)^(n-k)*(k+1)/2).
%F 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).
%F 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
%F Sum_{k=0..n} T(m,k)*T(n,k) = T(m+n,0) = A001405(m+n). - _Philippe Deléham_, Feb 26 2007
%F Sum_{k=0..n} T(n,k)=2^n. - _Philippe Deléham_, Mar 27 2007
%F 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
%e The array starts:
%e 1, 1, 2, 3, 6, 10, 20, 35, 70, 126, ...
%e 1, 1, 3, 4, 10, 15, 35, 56, 126, 210, ...
%e 1, 1, 4, 5, 15, 21, 56, 84, 210, 330, ...
%e 1, 1, 5, 6, 21, 28, 84, 120, 330, 495, ...
%e 1, 1, 6, 7, 28, 36, 120, 165, 495, 715, ...
%e 1, 1, 7, 8, 36, 45, 165, 220, 715, 1001, ...
%e 1, 1, 8, 9, 45, 55, 220, 286, 1001, 1365, ...
%e 1, 1, 9, 10, 55, 66, 286, 364, 1365, 1820, ...
%e 1, 1, 10, 11, 66, 78, 364, 455, 1820, 2380, ...
%e 1, 1, 11, 12, 78, 91, 455, 560, 2380, 3060, ...
%e Triangle (antidiagonal) version begins:
%e 1;
%e 1, 1;
%e 2, 1, 1;
%e 3, 3, 1, 1;
%e 6, 4, 4, 1, 1;
%e 10, 10, 5, 5, 1, 1;
%e 20, 15, 15, 6, 6, 1, 1;
%e 35, 35, 21, 21, 7, 7, 1, 1;
%e 70, 56, 56, 28, 28, 8, 8, 1, 1;
%e 126, 126, 84, 84, 36, 36, 9, 9, 1, 1;
%e 252, 210, 210, 120, 120, 45, 45, 10, 10, 1, 1;
%e 462, 462, 330, 330, 165, 165, 55, 55, 11, 11, 1, 1; ...
%e Matrix inverse begins:
%e 1;
%e -1, 1;
%e -1, -1, 1;
%e 1, -2, -1, 1;
%e 1, 2, -3, -1, 1;
%e -1, 3, 3, -4, -1, 1;
%e -1, -3, 6, 4, -5, -1, 1;
%e 1, -4, -6, 10, 5, -6, -1, 1;
%e 1, 4, -10, -10, 15, 6, -7, -1, 1; ...
%e From _Paul Barry_, May 21 2009: (Start)
%e Production matrix is
%e 1, 1,
%e 1, 0, 1,
%e 0, 1, 0, 1,
%e 0, 0, 1, 0, 1,
%e 0, 0, 0, 1, 0, 1,
%e 0, 0, 0, 0, 1, 0, 1,
%e 0, 0, 0, 0, 0, 1, 0, 1 (End)
%p T := proc(n, k) option remember;
%p if n = k then 1 elif k < 0 or n < 0 or k > n then 0
%p 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:
%p for n from 0 to 9 do seq(T(n, k), k = 0..n) od; # _Peter Luschny_, May 25 2021
%t 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 *)
%o (PARI) T(n,k)=binomial(n,(n+1)\2-(-1)^(n-k)*((k+1)\2))
%Y 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.
%Y Cf. A007318, A107430, A125094, A037952, A066170.
%K nonn,tabl
%O 0,4
%A _Henry Bottomley_, May 17 2001
%E Entry revised by _N. J. A. Sloane_, Nov 22 2006
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