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A074013
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Number of elements of GF(5^n) with trace 1 and subtrace 4.
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7
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0, 1, 6, 20, 125, 650, 3025, 15750, 78000, 390625, 1952500, 9765625, 48831250, 244125000
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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 }.
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PROG
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(Sage)
def a(n):
ans = 0
for x in GF(5^n):
if x.charpoly().coefficients(sparse=False)[-3:-1]==[4, 1]: ans += 1
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CROSSREFS
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KEYWORD
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nonn,more,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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