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Matrix inverse of number triangle A046854.
6

%I #14 Jan 22 2020 03:32:28

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

%T 5,-5,-9,9,5,-5,-1,1,0,14,-14,-14,14,6,-6,-1,1,-14,14,28,-28,-20,20,7,

%U -7,-1,1,0,-42,42,48,-48,-27,27

%N Matrix inverse of number triangle A046854.

%C First column is A105523; second column is A106181.

%C Triangle T(n,k), 0 <= k <= n, read by rows given by [ -1, 1, -1, 1, -1, 1, -1, 1, -1, 1,...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - _Philippe Deléham_, Sep 29 2006

%C A124448*A007318 as infinite lower triangular matrices. - _Philippe Deléham_, Oct 16 2007

%F Riordan array (1-y, y) where y=-(1-sqrt(1+4x^2))/(2x).

%F Sum_{k=0..n} abs(T(n,k)) = A063886(n). - _Philippe Deléham_, Oct 06 2006

%F 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,k-1) - T(n-1,k+1) for k >= 1. - _Philippe Deléham_, Oct 27 2007

%F T(2n,0) = A000007(n); T(2n+2,2k+2) = -T(2n+2,2k+1) = (-1)^(n-k)*A039598(n,k); T(2n+1,2k+1) = -T(2n+1,2k) = (-1)^(n-k)*A039599(n,k). - _Philippe Deléham_, Oct 29 2007

%F Sum_{k>=0} T(m,k)*T(n,k)*(-1)^k = T(m+n,0) = A105523(m+n). - _Philippe Deléham_, Jan 24 2010

%e Triangle begins

%e 1;

%e -1, 1;

%e 0, -1, 1;

%e 1, -1, -1, 1;

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

%e -2, 2, 3, -3, -1, 1;

%e 0, -5, 5, 4, -4, -1, 1;

%Y Cf. A000108.

%K easy,sign,tabl

%O 0,12

%A _Paul Barry_, Apr 24 2005