A122988 Number of possible arrangements of the last three digits of x^n for all x>0 (leading zeros omitted).

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

Robert G. Wilson v, Table of n, a(n) for n = 0..1000

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

Cf. A122986, A122987, A000578, A006716, A027676.

Sequence in context: A038459 A072990 A112023 * A069536 A126820 A033421

Adjacent sequences: A122985 A122986 A122987 * A122989 A122990 A122991

Keyword

base,nonn

Author

Sergio Pimentel, Sep 22 2006

Extensions

Edited and extended by Robert G. Wilson v, Sep 27 2006

Status

approved