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A122988
Number of possible arrangements of the last three digits of x^n for all x>0 (leading zeros omitted).
3
1, 1000, 159, 505, 52, 105, 102, 505, 52, 505, 22, 505, 52, 505, 102, 105, 52, 505, 102, 505, 12, 505, 102, 505, 52, 25, 102, 505, 52, 505, 22, 505, 52, 505, 102, 105, 52, 505, 102, 505, 12, 505, 102, 505, 52, 105, 102, 505, 52, 505, 6, 505, 52, 505, 102, 105, 52
OFFSET
0,2
COMMENTS
Only possible values are {1, 4, 6, 12, 22, 25, 52, 102, 105, 159, 505, 1000}. - Robert G. Wilson v, Sep 27 2006.
LINKS
Eric Weisstein's World of Mathematics, Cubic Number.
Eric Weisstein's World of Mathematics, Square Number.
FORMULA
a(n)=1 for n=0 only,
a(n)=4 for n=100*k, k>=1,
a(n)=6 for n=100*k-50, k>=1,
a(n)=12 for n=20*k, k>=1 except if k == 0 (mod 5),
a(n)=22 for n=20*k-10, k>=1 except if k == 3 (mod 5),
a(n)=25 for n=50*k-25, k>=1,
a(n)=52 for n=4*k, k>=1 except if k == 0 (mod 5),
a(n)=102 for n=4*k-2, k>=2 except if k == 3 (mod 5),
a(n)=105 for n=10*k-5, k>=1 except if k == 3 (mod 5),
a(n)=159 for n=2 only,
a(n)=505 for n=2*k-1, k>=2 except if k == 3 (mod 5) and
a(n)=1000 for n=1 only.
EXAMPLE
a(0) = 1 because the last three digits of x^0 are always 001 (just one possibility).
a(100)=4 because the last three digits of x^100 can be 000, 001, 376 or 625 (which is four possibilities).
MATHEMATICA
f[n_] := Length@ Union@ PowerMod[ Range@1000, n, 1000]; Table[ f@n, {n, 0, 56}] (* Robert G. Wilson v *)
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Sergio Pimentel, Sep 22 2006
EXTENSIONS
Edited and extended by Robert G. Wilson v, Sep 27 2006
STATUS
approved