A344912 -id:A344912 - cp's OEIS frontend

A336614 Number of n X n (0,1)-matrices A over the reals such that A^2 is the transpose of A.

1, 2, 4, 10, 32, 112, 424, 1808, 8320, 40384, 210944, 1170688, 6783616, 41411840, 265451008, 1765520128, 12227526656, 88163295232, 656548065280, 5054719287296, 40261285543936, 330010835894272, 2783003772452864, 24166721466204160, 215318925894909952, 1966855934183800832

Offset: 0

Torlach Rush, Jul 27 2020

nonn

From Peter Luschny, Jun 04 2021: (Start)
a(n) = n! * [x^n] exp(x*(x^2 + 6)/3).
a(n) = 2*a(n - 1) + (n^2 - 3*n + 2)*a(n - 3) for n >= 3.
a(n) = Sum_{k=0..n/3} (2^(n-3*k)*n!)/(3^k*k!*(n-3*k)!).
a(n) = 2^n*hypergeom([-n/3, (1-n)/3, (2-n)/3], [], -9/8).
[The above formulas, first stated as conjectures, were proved by mjqxxxx at Mathematics Stack Exchange, see link.] (End)

			a(3) = A336174(3) + A000079(3) = 2 + 8 = 10.

mjqxxxx, Proof of conjectured formulas, Mathematics Stack Exchange.

Row sums of A344912.

Cf. A000079, A001471, A336174.

a := n -> add((2^(n - 3*k)*n!)/(3^k*k!*(n - 3*k)!), k=0..n/3):
seq(a(n), n=0..25); # Peter Luschny, Jun 05 2021

m(n, t) = matrix(n, n, i, j, (t>>(i*n+j-n-1))%2)
a(n) = sum(t = 0, 2^n^2-1, m(n, t)^2 == m(n, t)~)
for(n = 0, 9, print1(a(n), ", "))

from itertools import product
from sympy import Matrix
def A336614(n):
    c = 0
    for d in product((0,1),repeat=n*n):
        M = Matrix(d).reshape(n,n)
        if M*M == M.T:
            c += 1
    return c # Chai Wah Wu, Sep 29 2020

a(n) = A336174(n) + A000079(n).

More terms from Peter Luschny, Jun 05 2021

A336174 Number of non-symmetric binary n X n matrices M over the reals such that M^2 is the transpose of M.

Original entry on oeis.org

0, 0, 0, 2, 16, 80, 360, 1680, 8064, 39872, 209920, 1168640, 6779520, 41403648, 265434624, 1765487360, 12227461120, 88163164160, 656547803136, 5054718763008, 40261284495360, 330010833797120, 2783003768258560, 24166721457815552, 215318925878132736, 1966855934150246400

Offset: 0

Torlach Rush, Jul 10 2020

nonn

We classify the (0,1) n X n matrices M_n by k, the number of 1's.
Let [T(n,k), n >= 0, k=0..n], be the lower triangular matrix where T(n,k) is the number of M^2 matrices equal to the transpose of M for n and k. Then:
T(n,n) = A001471(n).
Column sequences k=3..7 (without leading 0's) are:
T(n,3) = A001471(3) * A000292(n+1).
T(n,4) = A001471(4) * A000332(n+4).
T(n,5) = A001471(5) * A000389(n+5).
T(n,6) = A001471(6) * A000579(n+6).
T(n,7) = A001471(7) * A000580(n+7).
Row sums of T(n,k) generate known terms of this sequence and the next term a(10) evaluates to 209920 (see conjectured formula below).

			a(3) = 2 because [0,1,0]    [0,1,0]    [0,0,1]
                 [0,0,1]  * [0,0,1]  = [1,0,0]
                 [1,0,0]    [1,0,0]    [0,1,0],
             and [0,0,1]    [0,0,1]    [0,1,0]
                 [1,0,0]  * [1,0,0]  = [0,0,1]
                 [0,1,0]    [0,1,0]    [1,0,0].

Cf. A000292, A000332, A000389, A000579, A000580, A001471, A344912.

a := n -> 2^n*(add(n!/(24^k*k!*(n-3*k)!), k=0..n/3) - 1): seq(a(n), n=0..25);
# Alternative:
gf := exp(x*(x^2+6)/3) - exp(2*x): ser := series(gf,x,32):
seq(n!*coeff(ser,x,n), n = 0..25); # Peter Luschny, Jun 05 2021

m(n, t) = matrix(n, n, i, j, (t>>(i*n+j-n-1))%2)
a(n) = sum(t = 0, 2^n^2-1, m(n, t)^2 == m(n, t)~) - 2^n
for(n = 0, 9, print1(a(n), ", "))

a(n) = Sum_{k=0..n} A001471(k) * binomial(n, k). [Previously conjectured, for a proof see the link in A344912.]
From Peter Luschny, Jun 05 2021: (Start)
a(n) = 2^n*(add(n!/(24^k * k! * (n - 3*k)!), k=0..n/3) - 1).
a(n) = 2^n*(hypergeom([-n/3, (1 - n)/3, (2 - n)/3], [], -9/8) - 1).
a(n) = [x^n] exp(x*(x^2 + 6)/3) - exp(2*x). (End)
D-finite with recurrence (-n+3)*a(n) +4*(n-2)*a(n-1) +4*(-n+1)*a(n-2) +(n-1)*(n-2)*(n-3)*a(n-3) -2*(n-1)*(n-2)*(n-3)*a(n-4)=0. - R. J. Mathar, Jul 27 2022

More terms from Peter Luschny, Jun 05 2021

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A336614 Number of n X n (0,1)-matrices A over the reals such that A^2 is the transpose of A.

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