OFFSET
0,4
COMMENTS
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:
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).
FORMULA
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
EXAMPLE
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].
MAPLE
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
PROG
(PARI) 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), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Torlach Rush, Jul 10 2020
EXTENSIONS
More terms from Peter Luschny, Jun 05 2021
STATUS
approved