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Triangle T(n,k) read by rows given by [2,1,2,1,2,1,2,1,2,1,2,1,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938 .
1

%I #27 May 04 2019 03:46:45

%S 1,2,1,6,5,1,22,23,8,1,90,107,49,11,1,394,509,276,84,14,1,1806,2473,

%T 1505,556,128,17,1,8558,12235,8100,3429,974,181,20,1,41586,61463,

%U 43393,20355,6713,1557,243,23,1

%N Triangle T(n,k) read by rows given by [2,1,2,1,2,1,2,1,2,1,2,1,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938 .

%C Riordan array ((1-x-sqrt(1-6x+x^2))/(2x), (1-3x-sqrt(1-6x+x^2))/(4x)).

%C Inverse of Riordan array (1/(1+2x),x/(1+3x+2x^2)) (a signed version of A124237). _Paul Barry_, Apr 28 2009:

%C Peart and Woodson give a factorization of this array in the Riordan group as (1/(1 - 3*x), x/(1 - 3*x)) * (C(2*x^2), x*C(2*x^2)) * (1/(1 + x), x), where C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + ... is the o.g.f. of the Catalan numbers A000108. - _Peter Bala_, Aug 07 2014

%H Paul Barry, <a href="http://arxiv.org/abs/1311.2292">Laurent Biorthogonal Polynomials and Riordan Arrays</a>, arXiv preprint arXiv:1311.2292 [math.CA], 2013.

%H P. Peart and L. Woodson, <a href="http://www.fq.math.ca/Scanned/31-2/peart.pdf">Triple factorization of some Riordan matrices</a>, The Fib. Quart., Vol. 31 No. 2, May 1993.

%H Sheng-Liang Yang, Yan-Ni Dong, and Tian-Xiao He, <a href="https://doi.org/10.1016/j.disc.2017.07.006">Some matrix identities on colored Motzkin paths</a>, Discrete Mathematics 340.12 (2017): 3081-3091. See p. 3087.

%F 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)+3*T(n-1,k)+2*T(n-1,k+1) for k>=1 .

%F G.f.: 1/(1-xy-2x-x^2(2+y)/(1-3x-2x^2/(1-3x-2x^2/(1-3x-2x^2/(1- ... (continued fraction). - _Paul Barry_, Apr 28 2009

%F Sum_{k, k>=0} T(m,k)*T(n,k)*2^k = T(m+n,0) = A006318(m+n). - _Philippe Deléham_, Jan 24 2010

%F T(n,k) = S(n,n-k) - 2*S(n, n-k-2), where S(n,k) = Sum_{j = 0..k} binomial(n-1,k-j)*binomial(n,j)*2^j. - _Peter Bala_, Feb 20 2018

%e From _Paul Barry_, Apr 28 2009: (Start)

%e Triangle begins

%e 1,

%e 2, 1,

%e 6, 5, 1,

%e 22, 23, 8, 1,

%e 90, 107, 49, 11, 1,

%e 394, 509, 276, 84, 14, 1,

%e 1806, 2473, 1505, 556, 128, 17, 1

%e Production matrix begins

%e 2, 1,

%e 2, 3, 1,

%e 0, 2, 3, 1,

%e 0, 0, 2, 3, 1,

%e 0, 0, 0, 2, 3, 1,

%e 0, 0, 0, 0, 2, 3, 1,

%e 0, 0, 0, 0, 0, 2, 3, 1; (End)

%p S := proc (n, k)

%p add(binomial(n-1, k-j)*binomial(n, j)*2^j, j = 0..k);

%p end proc:

%p for n from 0 to 10 do

%p seq(S(n, n-k)-2*S(n, n-k-2), k = 0..n)

%p end do; # _Peter Bala_, Feb 20 2018

%t T[n_, 0] := Hypergeometric2F1[-n, n + 1, 2, -1]; T[n_, k_] := Binomial[-1 + n, -k + n] Hypergeometric2F1[k - n, -n, k, 2] - 2 Binomial[-1 + n, -2 - k + n] Hypergeometric2F1[2 + k - n, -n, 2 + k, 2]; Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* _Peter Luschny_, Feb 20 2018 *)

%Y Cf. A006318, A000012, A016789.

%K nonn,tabl,easy

%O 0,2

%A _Philippe Deléham_, Oct 27 2007