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A074013
Number of elements of GF(5^n) with trace 1 and subtrace 4.
7
0, 1, 6, 20, 125, 650, 3025, 15750, 78000, 390625, 1952500, 9765625, 48831250, 244125000
OFFSET
1,3
COMMENTS
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.
EXAMPLE
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 }.
PROG
(Sage)
def a(n):
ans = 0
for x in GF(5^n):
if x.charpoly().coefficients(sparse=False)[-3:-1]==[4, 1]: ans += 1
return ans # Robin Visser, Apr 26 2024
KEYWORD
nonn,more
AUTHOR
Frank Ruskey and Nate Kube, Aug 19 2002
EXTENSIONS
a(8)-a(14) from Robin Visser, Apr 26 2024
STATUS
approved