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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A061554 Square table read by antidiagonals: a(n,k) = binomial(n+k, floor(k/2)).

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%I A061554 #77 Aug 17 2024 22:42:52
%S A061554 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 A061554 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 A061554 210,210,120,120,45,45,10,10,1,1,462,462,330,330,165,165,55,55,11,11,1,1
%N A061554 Square table read by antidiagonals: a(n,k) = binomial(n+k, floor(k/2)).
%C A061554 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 A061554 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 A061554 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 A061554 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 A061554 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 A061554 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 A061554 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 A061554 Triangle read by rows = partial sums of A053121 terms starting from the right. - _Gary W. Adamson_, Oct 24 2008
%C A061554 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 A061554 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 A061554 G. C. Greubel, <a href="/A061554/b061554.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%H A061554 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 A061554 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 A061554 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 A061554 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 A061554 As a triangle: T(n,k) = binomial(n,m) where m = floor((n+1)/2 - (-1)^(n-k)*(k+1)/2).
%F A061554 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 A061554 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 A061554 Sum_{k=0..n} T(m,k)*T(n,k) = T(m+n,0) = A001405(m+n). - _Philippe Deléham_, Feb 26 2007
%F A061554 Sum_{k=0..n} T(n,k)=2^n. - _Philippe Deléham_, Mar 27 2007
%F A061554 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 A061554 The array starts:
%e A061554    1, 1,  2,  3,  6, 10,  20,  35,   70,  126, ...
%e A061554    1, 1,  3,  4, 10, 15,  35,  56,  126,  210, ...
%e A061554    1, 1,  4,  5, 15, 21,  56,  84,  210,  330, ...
%e A061554    1, 1,  5,  6, 21, 28,  84, 120,  330,  495, ...
%e A061554    1, 1,  6,  7, 28, 36, 120, 165,  495,  715, ...
%e A061554    1, 1,  7,  8, 36, 45, 165, 220,  715, 1001, ...
%e A061554    1, 1,  8,  9, 45, 55, 220, 286, 1001, 1365, ...
%e A061554    1, 1,  9, 10, 55, 66, 286, 364, 1365, 1820, ...
%e A061554    1, 1, 10, 11, 66, 78, 364, 455, 1820, 2380, ...
%e A061554    1, 1, 11, 12, 78, 91, 455, 560, 2380, 3060, ...
%e A061554 Triangle (antidiagonal) version begins:
%e A061554     1;
%e A061554     1,   1;
%e A061554     2,   1,   1;
%e A061554     3,   3,   1,   1;
%e A061554     6,   4,   4,   1,   1;
%e A061554    10,  10,   5,   5,   1,   1;
%e A061554    20,  15,  15,   6,   6,   1,  1;
%e A061554    35,  35,  21,  21,   7,   7,  1,  1;
%e A061554    70,  56,  56,  28,  28,   8,  8,  1,  1;
%e A061554   126, 126,  84,  84,  36,  36,  9,  9,  1,  1;
%e A061554   252, 210, 210, 120, 120,  45, 45, 10, 10,  1, 1;
%e A061554   462, 462, 330, 330, 165, 165, 55, 55, 11, 11, 1, 1; ...
%e A061554 Matrix inverse begins:
%e A061554    1;
%e A061554   -1,  1;
%e A061554   -1, -1,   1;
%e A061554    1, -2,  -1,   1;
%e A061554    1,  2,  -3,  -1,  1;
%e A061554   -1,  3,   3,  -4, -1,  1;
%e A061554   -1, -3,   6,   4, -5, -1,  1;
%e A061554    1, -4,  -6,  10,  5, -6, -1,  1;
%e A061554    1,  4, -10, -10, 15,  6, -7, -1, 1; ...
%e A061554 From _Paul Barry_, May 21 2009: (Start)
%e A061554 Production matrix is
%e A061554   1, 1,
%e A061554   1, 0, 1,
%e A061554   0, 1, 0, 1,
%e A061554   0, 0, 1, 0, 1,
%e A061554   0, 0, 0, 1, 0, 1,
%e A061554   0, 0, 0, 0, 1, 0, 1,
%e A061554   0, 0, 0, 0, 0, 1, 0, 1 (End)
%p A061554 T := proc(n, k) option remember;
%p A061554 if n = k then 1 elif k < 0 or n < 0 or k > n then 0
%p A061554 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 A061554 for n from 0 to 9 do seq(T(n, k), k = 0..n) od; # _Peter Luschny_, May 25 2021
%t A061554 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 A061554 (PARI) T(n,k)=binomial(n,(n+1)\2-(-1)^(n-k)*((k+1)\2))
%Y A061554 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 A061554 Cf. A007318, A107430, A125094, A037952, A066170.
%K A061554 nonn,tabl
%O A061554 0,4
%A A061554 _Henry Bottomley_, May 17 2001
%E A061554 Entry revised by _N. J. A. Sloane_, Nov 22 2006