|
%I #35 Jun 07 2021 05:48:23
%S 1,2,4,10,32,112,424,1808,8320,40384,210944,1170688,6783616,41411840,
%T 265451008,1765520128,12227526656,88163295232,656548065280,
%U 5054719287296,40261285543936,330010835894272,2783003772452864,24166721466204160,215318925894909952,1966855934183800832
%N Number of n X n (0,1)-matrices A over the reals such that A^2 is the transpose of A.
%C From _Peter Luschny_, Jun 04 2021: (Start)
%C a(n) = n! * [x^n] exp(x*(x^2 + 6)/3).
%C a(n) = 2*a(n - 1) + (n^2 - 3*n + 2)*a(n - 3) for n >= 3.
%C a(n) = Sum_{k=0..n/3} (2^(n-3*k)*n!)/(3^k*k!*(n-3*k)!).
%C a(n) = 2^n*hypergeom([-n/3, (1-n)/3, (2-n)/3], [], -9/8).
%C [The above formulas, first stated as conjectures, were proved by mjqxxxx at Mathematics Stack Exchange, see link.] (End)
%H <a href="https://math.stackexchange.com/users/5546/mjqxxxx">mjqxxxx</a>, <a href="https://math.stackexchange.com/q/4164050">Proof of conjectured formulas</a>, Mathematics Stack Exchange.
%F a(n) = A336174(n) + A000079(n).
%e a(3) = A336174(3) + A000079(3) = 2 + 8 = 10.
%p a := n -> add((2^(n - 3*k)*n!)/(3^k*k!*(n - 3*k)!), k=0..n/3):
%p seq(a(n), n=0..25); # _Peter Luschny_, Jun 05 2021
%o (PARI) m(n, t) = matrix(n, n, i, j, (t>>(i*n+j-n-1))%2)
%o a(n) = sum(t = 0, 2^n^2-1, m(n, t)^2 == m(n, t)~)
%o for(n = 0, 9, print1(a(n), ", "))
%o (Python)
%o from itertools import product
%o from sympy import Matrix
%o def A336614(n):
%o c = 0
%o for d in product((0,1),repeat=n*n):
%o M = Matrix(d).reshape(n,n)
%o if M*M == M.T:
%o c += 1
%o return c # _Chai Wah Wu_, Sep 29 2020
%Y Row sums of A344912.
%Y Cf. A000079, A001471, A336174.
%K nonn
%O 0,2
%A _Torlach Rush_, Jul 27 2020
%E More terms from _Peter Luschny_, Jun 05 2021
| 