1,3

Same as the number of elements of GF(5^n) with trace 2 and subtrace 1. Same as the number of elements of GF(5^n) with trace 3 and subtrace 1. Same as the number of elements of GF(5^n) with trace 4 and subtrace 4.

F. Ruskey, Number of Elements of GF(5^n) with given trace and subtrace

a(3;3,1)=6. Let GF(5^3) be defined by the field extension GF(5)[x]/( 3+2b+3b^2+b^3 ). The six elements of GF(5^3) with trace 3 and subtrace 1 are { 2+b+b^2, 3+2b+b^2, 4+3b+2b^2, 3+2b+3b^2, 4+3b+4b^2, 4b+4b^2 }.

Cf. A074006, A074007, A074008, A074009, A074010, A074011, A074012.

Sequence in context: A117998 A099840 A074012 this_sequence A114959 A000386 A145221

Adjacent sequences: A074010 A074011 A074012 this_sequence A074014 A074015 A074016

easy,nonn

Frank Ruskey, Nate Kube (fruskey(AT)cs.uvic.ca), Aug 19 2002