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Number of {0,1} n X n matrices with no rows or columns in which all entries are the same.
2

%I #26 Mar 26 2017 13:54:32

%S 1,0,2,102,22874,17633670,46959933962,451575174961302,

%T 16271255119687320314,2253375946574190518740230,

%U 1219041140314101911449662059402,2601922592659455476330065914740044182,22040870572750372076278589658097827953983034

%N Number of {0,1} n X n matrices with no rows or columns in which all entries are the same.

%C Every row and column must contain both a 0 and a 1 .

%C a(n) is the number of relations on n labeled points such that for every point x there exists y,z,t,u such that xRy, zRx, not(xRt), and not(uRx).

%F a(n) = 2*Sum_{k=0..n} ((-1)^(n+k)*binomial(n,k)*(2^k-1)^n) + 2^(n^2) + 2*(2^n-2)^n - 4*(2^n-1)^n.

%F a(n) = 2*A048291(n) + 2^(n^2) + 2*(2^n-2)^n - 4*(2^n-1)^n.

%e For n=2 the a(2)=2 matrices are

%e 0 1

%e 1 0

%e and

%e 1 0

%e 0 1

%p seq(2*sum((-1)^(n+k)*binomial(n,k)*(2^k-1)^n,k=0..n)+2^(n^2)+2*(2^n-2)^n-4*(2^n-1)^n,n=0..10)

%t Table[If[n==0, 1, 2 Sum[(-1)^(n + k) * Binomial[n, k] * (2^k - 1)^n, {k, 0,n}] + 2^(n^2) + 2*(2^n - 2)^n - 4*(2^n - 1)^n], {n, 0, 12}] (* _Indranil Ghosh_, Mar 12 2017 *)

%o (PARI) for(n=0, 12, print1(2*sum(k=0, n, (-1)^(n + k) * binomial(n, k) * (2^k - 1)^n) + 2^(n^2) + 2*(2^n - 2)^n - 4*(2^n - 1)^n,", ")) \\ _Indranil Ghosh_, Mar 12 2017

%o (Python)

%o import math

%o f = math.factorial

%o def C(n,r): return f(n)/f(r)/f(n - r)

%o def A(n):

%o ....s=0

%o ....for k in range(0, n+1):

%o ........s+=(-1)**(n + k) * C(n, k) * (2**k -1)**n

%o ....return 2*s + 2**(n**2) + 2*(2**n - 2)**n - 4*(2**n - 1)**n # _Indranil Ghosh_, Mar 12 2017

%Y Cf. A048291.

%Y Diagonal of A283654.

%K nonn

%O 0,3

%A _Robert FERREOL_, Mar 12 2017

%E a(11)-a(12) from _Indranil Ghosh_, Mar 12 2017