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a(n) = (2^(n^2) - 1)/(1 - 1/2^n).
2

%I #13 Apr 27 2020 06:31:46

%S 2,20,584,69904,34636832,69810262080,567382630219904,

%T 18519084246547628288,2422583247133816584929792,

%U 1268889750375080065623288448000,2659754699919401766201267083003561984,22306191045953951743035482794815064402563072

%N a(n) = (2^(n^2) - 1)/(1 - 1/2^n).

%C Sum of the geometric progression of ratio 2^n.

%C Number of all partial binary matrices with rows of length n: A partial binary matrix has 1<=k<=n rows of length n. The number of different partial matrices with k rows is 2^(k*n). a(n) is the sum for k between 1 and n.

%H Andrew Howroyd, <a href="/A275779/b275779.txt">Table of n, a(n) for n = 1..50</a>

%F a(n) = Sum_{k=1..n} 2^(k*n).

%t Table[(2^(n^2) - 1)/(1 - 1/2^n), {n, 1, 10}]

%o (PARI) a(n) = {(2^(n^2) - 1)/(1 - 1/2^n)} \\ _Andrew Howroyd_, Apr 26 2020

%Y Cf. A128889 (accepting the null matrix and excluding the full n*n matrices)

%Y Cf. A096131, A057524.

%K nonn,easy

%O 1,1

%A _Olivier GĂ©rard_, Aug 08 2016

%E Terms a(11) and beyond from _Andrew Howroyd_, Apr 26 2020