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%I #11 Dec 21 2020 18:14:36
%S 1,-1,1,-1,0,1,-3,1,1,1,-12,4,5,2,1,-60,20,25,11,3,1,-360,120,150,66,
%T 19,4,1,-2520,840,1050,462,133,29,5,1,-20160,6720,8400,3696,1064,232,
%U 41,6,1,-181440,60480,75600,33264,9576,2088,369,55,7,1,-1814400,604800,756000,332640,95760,20880,3690,550,71,8,1
%N Matrix inverse of triangle A176270, read by rows.
%F T(n,n) = 1 for n >= 0; T(n,n-1) = n - 2 for n > 0; T(n,n-2) = n^2 - 3*n + 1 for n > 1; T(n,k) = (k^2 + k - 1) * n! / (k+2)! for 0 <= k <= n-2.
%F T(n,k) = n * T(n-1,k) for 0 <= k < n-2.
%F T(n,k) = T(k+2,k) * n! / (k+2)! for 0 <= k <= n-2.
%F Row sums are A000007(n) for n >= 0.
%e The triangle T(n,k) for 0 <= k <= n starts:
%e n\k : 0 1 2 3 4 5 6 7 8 9
%e =============================================================
%e 0 : 1
%e 1 : -1 1
%e 2 : -1 0 1
%e 3 : -3 1 1 1
%e 4 : -12 4 5 2 1
%e 5 : -60 20 25 11 3 1
%e 6 : -360 120 150 66 19 4 1
%e 7 : -2520 840 1050 462 133 29 5 1
%e 8 : -20160 6720 8400 3696 1064 232 41 6 1
%e 9 : -181440 60480 75600 33264 9576 2088 369 55 7 1
%e etc.
%o (PARI) for(n=0,10,for(k=0,n,if(k==n,print(" 1"),if(k==n-1,print1(n-2,", "),print1((k^2+k-1)*n!/(k+2)!,", ")))))
%o (PARI) 1/matrix(10, 10, n, k, n--; k--; if (n>=k, 1 + k*(k-n))) \\ _Michel Marcus_, Nov 11 2020
%Y Cf. A000007, A176270.
%K sign,easy,tabl
%O 0,7
%A _Werner Schulte_, Nov 10 2020
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