OFFSET
1,2
COMMENTS
Lemma: The sequence is finite with all terms in the sequence having at most 22 digits. Proof: Let n be an m-digit natural number in the sequence for some m. Then 10^(m-1) <= n and n <= 9 + 9^2 + ... + 9^m = 9(9^m-1)/8 < (9^(m+1))/8. Thus 10^(m-1) < (9^(m+1))/8. Taking logarithms of both sides and solving yields m < 22.97. QED. The sequence listed, found by a computer program searching up to 10^22, is therefore complete. - Francis J. McDonnell, Apr 12 2012
LINKS
Francis J. McDonnell, Java program
EXAMPLE
2537 = 2^1 + 3^2 + 5^3 + 7^4 = 2 + 9 + 125 + 2401.
60409 = 0^1 + 0^2 + 4^3 + 6^4 + 9^5 = 0 + 0 + 64 + 1296 + 59049.
PROG
CROSSREFS
KEYWORD
nonn,base,fini,full
AUTHOR
Francis J. McDonnell, Mar 29 2012
EXTENSIONS
More terms added by Francis J. McDonnell, Apr 12 2012
Faster program used to obtain more terms included by Francis J. McDonnell, Apr 16 2012
STATUS
approved