%I #64 Nov 14 2022 09:54:49

%S 1,0,1,0,-1,1,0,0,-2,1,0,0,1,-3,1,0,0,0,3,-4,1,0,0,0,-1,6,-5,1,0,0,0,

%T 0,-4,10,-6,1,0,0,0,0,1,-10,15,-7,1,0,0,0,0,0,5,-20,21,-8,1,0,0,0,0,0,

%U -1,15,-35,28,-9,1,0,0,0,0,0,0,-6,35,-56,36,-10,1,0,0,0,0,0,0,1,-21,70,-84,45,-11,1,0,0,0,0

%N Riordan array (1, x(1-x)).

%C Inverse is Riordan array (1, xc(x)) (A106566).

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

%C Modulo 2, this sequence gives A106344. - _Philippe Deléham_, Dec 18 2008

%C Coefficient array of the polynomials Chebyshev_U(n, sqrt(x)/2)*(sqrt(x))^n. - _Paul Barry_, Sep 28 2009

%H Paul Barry, <a href="http://arxiv.org/abs/1312.0583">Embedding structures associated with Riordan arrays and moment matrices</a>, arXiv preprint arXiv:1312.0583 [math.CO], 2013.

%H Tom Copeland, <a href="http://tcjpn.wordpress.com/2015/10/12/the-elliptic-lie-triad-kdv-and-ricattt-equations-infinigens-and-elliptic-genera/">Addendum to Elliptic Lie Triad</a>

%F Number triangle T(n, k) = (-1)^(n-k)*binomial(k, n-k).

%F T(n, k)*2^(n-k) = A110509(n, k); T(n, k)*3^(n-k) = A110517(n, k).

%F Sum_{k=0..n} T(n,k)*A000108(k)=1. - _Philippe Deléham_, Jun 11 2007

%F From _Philippe Deléham_, Oct 30 2008: (Start)

%F Sum_{k=0..n} T(n,k)*A144706(k) = A082505(n+1).

%F Sum_{k=0..n} T(n,k)*A002450(k) = A100335(n).

%F Sum_{k=0..n} T(n,k)*A001906(k) = A100334(n).

%F Sum_{k=0..n} T(n,k)*A015565(k) = A099322(n).

%F Sum_{k=0..n} T(n,k)*A003462(k) = A106233(n). (End)

%F Sum_{k=0..n} T(n,k)*x^(n-k) = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = -12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12 respectively. - _Philippe Deléham_, Oct 27 2008

%F Sum_{k=0..n} T(n,k)*x^k = A000007(n), A010892(n), A099087(n), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n) for x = 0,1,2,3,4,5,6,7,8,9,10 respectively. - _Philippe Deléham_, Oct 28 2008

%F G.f.: 1/(1-y*x+y*x^2). - _Philippe Deléham_, Dec 15 2011

%F T(n,k) = T(n-1,k-1) - T(n-2,k-1), T(n,0) = 0^n. - _Philippe Deléham_, Feb 15 2012

%F Sum_{k=0..n} T(n,k)*x^(n-k) = F(n+1,-x) where F(n,x)is the n-th Fibonacci polynomial in x defined in A011973. - _Philippe Deléham_, Feb 22 2013

%F Sum_{k=0..n} T(n,k)^2 = A051286(n). - _Philippe Deléham_, Feb 26 2013

%F Sum_{k=0..n} T(n,k)*T(n+1,k) = -A110320(n). - _Philippe Deléham_, Feb 26 2013

%F For T(0,0) = 0, the signed triangle below has the o.g.f. G(x,t) = [t*x(1-x)]/[1-t*x(1-x)] = L[t*Cinv(x)] where L(x) = x/(1-x) and Cinv(x)=x(1-x) with the inverses Linv(x) = x/(1+x) and C(x)= [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108, so the inverse o.g.f. is Ginv(x,t) = C[Linv(x)/t] = [1-sqrt[1-4*x/(t(1+x))]]/2 (cf. A124644 and A030528). - _Tom Copeland_, Jan 19 2016

%e Rows begin:

%e 1;

%e 0, 1;

%e 0, -1, 1;

%e 0, 0, -2, 1;

%e 0, 0, 1, -3, 1;

%e 0, 0, 0, 3, -4, 1;

%e 0, 0, 0, -1, 6, -5, 1;

%e 0, 0, 0, 0, -4, 10, -6, 1;

%e 0, 0, 0, 0, 1, -10, 15, -7, 1;

%e 0, 0, 0, 0, 0, 5, -20, 21, -8, 1;

%e 0, 0, 0, 0, 0, -1, 15, -35, 28, -9, 1;

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

%e Production array is

%e 0, 1,

%e 0, -1, 1,

%e 0, -1, -1, 1,

%e 0, -2, -1, -1, 1,

%e 0, -5, -2, -1, -1, 1,

%e 0, -14, -5, -2, -1, -1, 1,

%e 0, -42, -14, -5, -2, -1, -1, 1,

%e 0, -132, -42, -14, -5, -2, -1, -1, 1,

%e 0, -429, -132, -42, -14, -5, -2, -1, -1, 1 (End)

%t (* The function RiordanArray is defined in A256893. *)

%t RiordanArray[1&, #(1-#)&, 13] // Flatten (* _Jean-François Alcover_, Jul 16 2019 *)

%o (Magma) /* As triangle */ [[(-1)^(n-k)*Binomial(k, n-k): k in [0..n]]: n in [0.. 15]]; // _Vincenzo Librandi_, Jan 14 2016

%Y Cf. A026729 (unsigned version), A000108, A030528, A124644.

%K easy,sign,tabl

%O 0,9

%A _Philippe Deléham_, Aug 28 2005