OFFSET
1,1
COMMENTS
With leading zeros, the initial terms are 02, 05, 06.
To compute the sequence, it is sufficient to consider the residue mod 100 of powers of numbers < 100 until the same value is reached for the second time. - M. F. Hasler, Dec 13 2018
EXAMPLE
9 (09!) not in the list because the perfect power 2209 = 47^2 ends with 09.
MAPLE
s:={$(0..99)}: for b from 0 to 99 do for p from 2 to 101 do s:=s minus {b^p mod 100}: od: od: op(s); # Nathaniel Johnston, Jun 22 2011
MATHEMATICA
S=Range[2, 99]; Do[n=1; T={}; While[T != (T = Union[T, {PowerMod[k, ++n, 100]}]), S=Complement[S, T]], {k, 2, 99}]; S (* Amiram Eldar, Dec 13 2018 after M. F. Hasler's pari code *)
PROG
(PARI) S=[2..99]; for(k=2, 99, my(m=Mod(k, 100), n=1, T=[]); while(T!=T=setunion(T, [m^n+=1]), ); S=setminus(S, lift(T))); S \\ Slightly shorter. - M. F. Hasler, Dec 13 2018
(PARI) S=0; for(k=2, 99, my(m=Mod(k, 100), n=1, T=0); while(T<T=bitor(T, 2^lift(m^n+=1)), ); S=bitor(S, T)); vecextract([0..99], 2^100-S-1) \\ Slightly faster. - M. F. Hasler, Dec 13 2018
CROSSREFS
KEYWORD
fini,full,easy,nonn,base
AUTHOR
Zak Seidov, Oct 14 2002
EXTENSIONS
Edited and confirmed by Nathaniel Johnston, Jun 22 2011
STATUS
approved