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%I #15 Aug 30 2019 15:21:20
%S 0,1,12,448,61440,32505856,67645734912,558551906910208,
%T 18374686479671623680,2413129272746388704198656,
%U 1266412660188944021221804081152,2657157917355198038900481496478384128,22295300680659888126120304278929453214597120
%N The number of eigenvectors with eigenvalue 1 summed over all linear operators on the vector space GF(2)^n.
%C Generally, for any prime power q, the total number of eigenvectors corresponding to any element lambda in the field GF(q) summed over all operators on GF(q)^n is equal to (q^n-1)*q^(n^2-n).
%F a(n) = (2^n-1)*2^(n^2-n).
%t Map[Total,Table[Table[(q^(n - k) - 1) Product[(q^n - q^i)^2/(q^k - q^i), {i, 0,k - 1}] /. q -> 2, {k, 0, n}], {n, 0, 11}]]
%Y Cf. A286331.
%K nonn
%O 0,3
%A _Geoffrey Critzer_, Aug 28 2019
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