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%I #19 Feb 13 2021 16:04:28
%S 1,-1,1,0,-1,1,1,-1,-1,1,0,1,-1,-1,1,0,1,0,-1,-1,1,-1,1,1,0,-1,-1,1,
%T -1,0,2,0,0,-1,-1,1,0,-1,0,2,0,0,-1,-1,1,0,-2,1,1,1,0,0,-1,-1,1,1,-2,
%U -1,1,1,1,0,0,-1,-1,1,1,-1,-2,0,2,0,1,0,0,-1,-1,1
%N Matrix inverse of partition triangle A008284.
%C Since A008284 has only ones in its first column, the sum of terms for any row n > 1 is 0. - _François Marques_, Feb 09 2021
%F T(n,n-k) = A010815(k) for k <= n/2. - _François Marques_, Feb 09 2021
%e Triangle begins:
%e 1;
%e -1,1;
%e 0,-1,1;
%e 1,-1,-1,1;
%e ...
%o (PARI) tp(n, k) = if (n<1, 0, if (k<1, 0, if (k == n, 1, if (k > n, 0, tp(n-1, k-1) + tp(n-k, k)))));
%o tabl(nn) = {mtp = matrix(nn, nn, n, k, tp(n, k)); mtpi = mtp^(-1); for (n = 1, nn, for (k = 1, n, print1(mtpi[n, k], ", ");); print(););} \\ _Michel Marcus_, Mar 04 2014
%Y Cf. A008284, A038497, A039800-A039810, A010815.
%K sign,tabl
%O 1,31
%A _Christian G. Bower_, Feb 15 1999
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