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%I #29 Oct 23 2018 05:49:48
%S 1,1,-1,1,0,-1,1,1,-1,-1,1,2,0,-2,-1,1,3,2,-2,-3,-1,1,4,5,0,-5,-4,-1,
%T 1,5,9,5,-5,-9,-5,-1,1,6,14,14,0,-14,-14,-6,-1,1,7,20,28,14,-14,-28,
%U -20,-7,-1,1,8,27,48,42,0,-42,-48,-27,-8,-1
%N Triangle T(n,k) of differences of pairs of consecutive terms of triangle A071919.
%C Row sums are 1,0,0,0,0,0, ... with g.f. 1 = (1-x)^0(1-2x)^0
%C (1,-1)-Pascal triangle; mirror image of triangle A112467. - _Philippe Deléham_, Nov 07 2006
%C Triangle T(n,k), read by rows, given by (1,0,0,0,0,0,0,0,0,...) DELTA (-1,2,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Nov 01 2011
%H T. M. Brown, <a href="https://arxiv.org/abs/1810.08235">On the unimodality of convolutions of sequences of binomial coefficients</a>, arXiv:1810.08235 [math.CO] (2018). See p. 8.
%H Pedro J. Miana, Hideyuki Ohtsuka, Natalia Romero, <a href="http://arxiv.org/abs/1602.04347">Sums of powers of Catalan triangle numbers</a>, arXiv:1602.04347 [math.NT], 2016.
%F T(n, k) = binomial(n, k) + 2*Sum{j=1...k} (-1)^j binomial(n, k-j).
%F Sum_{k=0..n} T(n, k)*x^k = (1-x)*(1+x)^(n-1), for n >= 1. - _Philippe Deléham_, Sep 05 2005
%F T(n,k) = T(n-1,k-1) + T(n-1,k) with T(n,0)=1, T(n,n)=-1 for n > 0. - _Philippe Deléham_, Nov 01 2011
%F T(n,k) =binomial(n-1,k) - binomial(n-1,k-1), for n > 0. T(n,k) = Sum_{i=-k..k} (-1)^i*binomial(n-1,k+i)*binomial(n+1,k-i), for n >= k. T(n,k)=0, for n < k. - _Mircea Merca_, Apr 28 2012
%F G.f.: (-1+2*x*y)/(-1+x*y+x). - _R. J. Mathar_, Aug 11 2015
%e Rows begin
%e 1;
%e 1, -1;
%e 1, 0, -1;
%e 1, 1, -1, -1;
%e 1, 2, 0, -2, -1;
%e 1, 3, 2, -2, -3, -1;
%e 1, 4, 5, 0, -5, -4, -1;
%e 1, 5, 9, 5, -5, -9, -5, -1;
%e 1, 6, 14, 14, 0, -14, -14, -6, -1;
%e 1, 7, 20, 28, 14, -14, -28, -20, -7, -1;
%e 1, 8, 27, 48, 42, 0, -42, -48, -27, -8, -1;
%p T(n,k):=piecewise(n=0,1,n>0,binomial(n-1,k)-binomial(n-1,k-1)) # _Mircea Merca_, Apr 28 2012
%Y Cf. A007318, A071919.
%Y Apart from initial term, same as A037012.
%K easy,sign,tabl
%O 0,12
%A _Paul Barry_, Feb 09 2003
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