A284989 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.

1, 0, 0, 0, 0, 1, 1, 0, 3, 2, 9, 24, 24, 24, 9, 216, 540, 610, 420, 210, 44, 7570, 18000, 20175, 13720, 6300, 1920, 265, 357435, 829920, 909741, 617610, 284235, 91140, 19005, 1854, 22040361, 50223600, 54295528, 36663312, 17072790, 5679184, 1337280, 203952, 14833

Offset: 0

Author

R. J. Mathar, Apr 07 2017

Keywords

• nonn
• tabl

Examples

			0:         1
1:         0        0
2:         0        0        1
3:         1        0        3        2
4:         9       24       24       24        9
5:       216      540      610      420      210      44
6:      7570    18000    20175    13720     6300    1920     265
7:    357435   829920   909741   617610   284235   91140   19005   1854
8:  22040361 50223600 54295528 36663312 17072790 5679184 1337280 203952 14833
		

Links

• Alois P. Heinz, Rows n = 0..14, flattened
• Index entries for sequences related to binary matrices

Crossrefs

Cf. A001499 (row sums), A000166 (diagonal), A007107 (column 0).

Cf. A008290, A098825, A284990, A284991.

Programs

PARI

P(n, t='t) = {
  my(z=vector(n, k, eval(Str("z", k))),
     s1=sum(k=1, #z, z[k]), s2=sum(k=1, #z, z[k]^2), s12=(s1^2 - s2)/2,
     f=vector(n, k, t*(s12 - z[k]*(s1 - z[k])) + z[k]*(s1 - z[k])), g=1);
  for (i=1, n, g *= f[i]; for(j=1, n, g=substpol(g, z[j]^3, 0)));
  for (k=1, n, g=polcoef(g, 2, z[k]));
  g;
};
seq(N) = concat([[1], [0, 0], [0, 0, 1]], apply(n->Vec(P(n)), [3..N]));
concat(seq(8)) \\ Gheorghe Coserea, Dec 21 2018

Formula

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