OFFSET
0,3
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).
FORMULA
a(n) = (2^n-1)*2^(n^2-n).
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}]]
CROSSREFS
KEYWORD
nonn
AUTHOR
Geoffrey Critzer, Aug 28 2019
STATUS
approved