A322556 The number of eigenvectors with eigenvalue 1 summed over all linear operators on the vector space GF(2)^n.

0, 1, 12, 448, 61440, 32505856, 67645734912, 558551906910208, 18374686479671623680, 2413129272746388704198656, 1266412660188944021221804081152, 2657157917355198038900481496478384128, 22295300680659888126120304278929453214597120

Offset: 0

Author

Geoffrey Critzer, Aug 28 2019

Keywords

• nonn

Comments

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).

Crossrefs

Cf. A286331.

Programs

Mathematica

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}]]

Formula

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