%I
%S 1,1,1,1,0,1,2,0,3,1,5,0,6,0,1,16,0,20,0,5,1,61,0,75,0,15,0,1,272,0,
%T 336,0,70,0,7,1,1385,0,1708,0,350,0,28,0,1,7936,0,9792,0,2016,0,168,0,
%U 9,1,50521,0,62325,0,12810,0,1050,0,45,0,1,353792,0,436480,0,89760,0,7392,0
%N Matrix inverse of A162169.
%C First column appears to be A000111. Third column is A162171. Row sums minus A000035 appears to be A062272.
%e Table begins:
%e .1
%e .1...1
%e .1...0...1
%e .2...0...3...1
%e .5...0...6...0...1
%e 16...0..20...0...5...1
%e 61...0..75...0..15...0...1
%o (PARI) T(n, k) = if (k % 2, binomial(n1, k1) * (1)^floor((n+k1)/2), if (n==k, 1 , 0));
%o tabl(nn) = {m = matrix(nn, nn, n, k, if (n>=k, T(n,k), 0)); m = m^(1); for (n=1, nn, for (k=1, n, print1(m[n,k], ", ");); print(););} \\ _Michel Marcus_, Jun 17 2015
%Y Cf. A000111, A162171, A062272.
%K nonn,tabl
%O 1,7
%A _Mats Granvik_, Jun 27 2009
