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%I #76 Jul 23 2020 19:21:16
%S 1,2,1,6,4,1,20,15,6,1,70,56,28,8,1,252,210,120,45,10,1,924,792,495,
%T 220,66,12,1,3432,3003,2002,1001,364,91,14,1,12870,11440,8008,4368,
%U 1820,560,120,16,1,48620,43758,31824,18564,8568,3060,816,153,18,1,184756,167960
%N Triangle T(n,k), read by rows, defined by T(n,k) = binomial(2*n,n-k).
%C Right-hand side of even-numbered rows of Pascal's triangle.
%C Row sums are A032443. Reverse of A062344. Right-hand side of A034870. Binomial transform of trinomial triangle A094531.
%C Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = 2*T(n-1,0) + 2*T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k >= 1. - _Philippe Deléham_, Mar 14 2007
%C Central coefficients T(2n,n) are binomial(4n,n) (A005810).
%C The A- and Z-sequence for this Riordan triangle is [1,2,1] and [2,2], respectively. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. See also the _Philippe Deléham_ comment above. - _Wolfdieter Lang_, Nov 22 2012
%H Indranil Ghosh, <a href="/A094527/b094527.txt">Rows 0..100 of triangle, flattened</a>
%H Paul Barry, <a href="http://dx.doi.org/10.1155/2013/657806">On the Connection Coefficients of the Chebyshev-Boubaker polynomials</a>, The Scientific World Journal, Volume 2013 (2013), Article ID 657806, 10 pages.
%H Paul Barry, <a href="https://arxiv.org/abs/1912.01124">A Note on Riordan Arrays with Catalan Halves</a>, arXiv:1912.01124 [math.CO], 2019.
%H Paul Barry, <a href="https://arxiv.org/abs/2004.04577">On a Central Transform of Integer Sequences</a>, arXiv:2004.04577 [math.CO], 2020.
%H Johann Cigler, <a href="https://arxiv.org/abs/1611.05252">Some elementary observations on Narayana polynomials and related topics</a>, arXiv:1611.05252 [math.CO], 2016. See p. 19.
%H A. Luzón, D. Merlini, M. A. Morón, R. Sprugnoli, <a href="http://dx.doi.org/10.1016/j.dam.2014.03.005">Complementary Riordan arrays</a>, Discrete Applied Mathematics, 172 (2014) 75-87.
%H Asamoah Nkwanta and Earl R. Barnes, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Nkwanta/nkwanta2.html">Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind</a>, Journal of Integer Sequences, Article 12.3.3, 2012. - From _N. J. A. Sloane_, Sep 16 2012
%H P. Peart and W.-J. Woan, <a href="http://dx.doi.org/10.1016/S0166-218X(99)00166-3">A divisibility property for a subgroup of Riordan matrices</a>, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263
%H T. M. Richardson, <a href="http://arxiv.org/abs/1405.6315">The Reciprocal Pascal Matrix</a>, arXiv preprint arXiv:1405.6315 [math.CO], 2014.
%F Riordan array (1/sqrt(1-4x)), (1-2x-sqrt(1-4x))/(2x)). Column k has e.g.f. exp(2x)Bessel_I(k, 2x). - _Paul Barry_, Jul 14 2005
%F Product of Riordan arrays (1/(1-x), x/(1-x)) (Pascal's triangle, A007318) and (1/sqrt(1-2x-3x^2), (1-x-sqrt(1-2x-3x^2))/(2x)) (A094531). Inverse is A110162. - _Paul Barry_, Jul 14 2005
%F T(n,k) = Sum_{j=0..n} C(n,j)*C(n,j-k). - _Paul Barry_, Mar 07 2006
%F T(n,k) = Sum_{h>=k} A039599(n,h). Sum_{k=0..n} T(n,k) = A032443(n). - _Philippe Deléham_, May 01 2006
%F Sum_{k=0..n} T(n,k)^2 = A036910(n). - _Philippe Deléham_, May 07 2006
%F Sum_{k=0..n} T(n,k)*(-1)^k = A088218(n). - _Philippe Deléham_, Mar 14 2007
%F From _Wolfdieter Lang_, Nov 22 2012: (Start)
%F The o.g.f. for the row polynomials P(n,x) := Sum_{k=0..n} T(n,k)*x^k is G(z,x) = (-x + (1+x)*z + x*z*c(z))/(sqrt(1-4*z)*((1+x)^2*z -x)) with c the o.g.f. of A000108 (Catalan). This follows from the Riordan property.
%F The o.g.f. for column no. k is (c(x)-1)^k/sqrt(1-4*x) (from the Riordan property). (End)
%F From _Peter Bala_, Jun 29 2015: (Start)
%F Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = ( 1 - 2*x - sqrt(1 - 4*x) )/(2*x) and so belongs to the hitting time subgroup of the Riordan group (see Peart and Woan, Example 5.1).
%F T(n,k) = [x^(n-k)] f(x)^n with f(x) = (1 + x)^2. In general the (n,k)th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)
%F From _Peter Bala_, Jul 21 2015: (Start)
%F n-th row polynomial R(n,t) = [x^n] ( (1 + (1 + t)*x)^2/(1 + t*x) )^n.
%F exp ( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (2 + t)*x + (5 + 4*t + t^2)*x^2 + ... is the o.g.f. for A039598. (End)
%e The triangle T(n,k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10
%e 0: 1
%e 1: 2 1
%e 2: 6 4 1
%e 3: 20 15 6 1
%e 4: 70 56 28 8 1
%e 5: 252 210 120 45 10 1
%e 6: 924 792 495 220 66 12 1
%e 7: 3432 3003 2002 1001 364 91 14 1
%e 8: 12870 11440 8008 4368 1820 560 120 16 1
%e 9: 48620 43758 31824 18564 8568 3060 816 153 18 1
%e 10: 184756 167960 125970 77520 38760 15504 4845 1140 190 20 1
%e ... Reformatted ad extended by _Wolfdieter Lang_, Nov 22 2012
%e From _Paul Barry_, Sep 07 2009: (Start)
%e Production array is
%e 2, 1,
%e 2, 2, 1,
%e 0, 1, 2, 1,
%e 0, 0, 1, 2, 1,
%e 0, 0, 0, 1, 2, 1,
%e 0, 0, 0, 0, 1, 2, 1,
%e 0, 0, 0, 0, 0, 1, 2, 1 (End)
%e From _Wolfdieter Lang_, Nov 22 2012: (Start)
%e Recurrence from the Riordan A-sequence [1,2,1]: T(4,1) = 56 = 1*T(3,0) + 2*T(3,1) + 1*T(3,2) = 1*20 + 2*15 + 1*6.
%e Recurrence from the Riordan Z-sequence [2,2]: T(7,0) = 3432 = 2*T(6,0) + 2*T(6,1) = 2*924 + 2*792. See the _Philippe Deléham_ comment above. (End)
%p A094527 := proc(n,k)
%p binomial(2*n,n-k) ;
%p end proc: # _R. J. Mathar_, Jun 04 2013
%t T[n_, k_] := Binomial[2*n, n - k];
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Nov 14 2017 *)
%Y Cf. A007318, A032443, A039599, A039598.
%K easy,nonn,tabl
%O 0,2
%A _Paul Barry_, May 07 2004
%E Entry revised by _N. J. A. Sloane_, Mar 23 2007
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