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%I #35 Dec 24 2018 21:28:42
%S 1,0,0,0,0,1,1,0,3,2,9,24,24,24,9,216,540,610,420,210,44,7570,18000,
%T 20175,13720,6300,1920,265,357435,829920,909741,617610,284235,91140,
%U 19005,1854,22040361,50223600,54295528,36663312,17072790,5679184,1337280,203952,14833
%N Triangle T(n,k) read by rows: the number of n X n {0,1} matrices with trace k where each row sum and each column sum is 2.
%H Alois P. Heinz, <a href="/A284989/b284989.txt">Rows n = 0..14, flattened</a>
%H <a href="/index/Mat#binmat">Index entries for sequences related to binary matrices</a>
%F Let z1..zn be n variables and s1 = Sum_{k=1..n} zk, s2 = Sum_{k=1..n} zk^2, s12 = (s1^2 - s2)/2, fk = t*(s12 - zk*(s1 - zk)) + zk*(s1 - zk) for k=1..n, P_n(t) = [(z1..zn)^2] Product_{k=1..n} fk. Then P_n(t) = Sum_{k=0..n} T(n,k)*t^(n-k), n >= 3. - _Gheorghe Coserea_, Dec 21 2018
%e 0: 1
%e 1: 0 0
%e 2: 0 0 1
%e 3: 1 0 3 2
%e 4: 9 24 24 24 9
%e 5: 216 540 610 420 210 44
%e 6: 7570 18000 20175 13720 6300 1920 265
%e 7: 357435 829920 909741 617610 284235 91140 19005 1854
%e 8: 22040361 50223600 54295528 36663312 17072790 5679184 1337280 203952 14833
%o (PARI)
%o P(n, t='t) = {
%o my(z=vector(n, k, eval(Str("z", k))),
%o s1=sum(k=1, #z, z[k]), s2=sum(k=1, #z, z[k]^2), s12=(s1^2 - s2)/2,
%o f=vector(n, k, t*(s12 - z[k]*(s1 - z[k])) + z[k]*(s1 - z[k])), g=1);
%o for (i=1, n, g *= f[i]; for(j=1, n, g=substpol(g, z[j]^3, 0)));
%o for (k=1, n, g=polcoef(g, 2, z[k]));
%o g;
%o };
%o seq(N) = concat([[1], [0, 0], [0, 0, 1]], apply(n->Vec(P(n)), [3..N]));
%o concat(seq(8)) \\ _Gheorghe Coserea_, Dec 21 2018
%Y Cf. A001499 (row sums), A000166 (diagonal), A007107 (column 0).
%Y Cf. A008290, A098825, A284990, A284991.
%K nonn,tabl
%O 0,9
%A _R. J. Mathar_, Apr 07 2017
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