OFFSET
1,2
COMMENTS
No more terms < 10^100. - David Wasserman, Mar 06 2008
All terms are of the form t=d*b^k where b=2 or 5 and d divides s = sum of digits of t. Let m be the decimal length of t. Assume that k >= 140. Then we have m >= 42 and thus log_10(m) < m/25. Since d < 10*m, we have m <= 1 + log_10(d*b^k) < 2 + log_10(m) + k*log_10(b) < 2 + (m/25) + 0.7*k, implying that m < 2.1 + 0.73*k. On the other hand, b^k must divide s*10^m, implying that k < log_10(s)/log_10(2) + m < (1 + log_10(m))/log_10(2) + m < 3.4 + 1.14*m < 5.8 + 0.84*k and hence k < 37. This contradiction proves that no term has k >= 140, i.e., sequence is finite. - Max Alekseyev, May 08 2009
EXAMPLE
a(3) = 15 because the digit sum of 15 is 6 and 615/15 = 41. Any future number which divides its digit sum concatenated with itself by exactly 41 (such as 150, 225, 375, etc.) will be excluded.
CROSSREFS
KEYWORD
base,nonn,full,fini
AUTHOR
Chuck Seggelin (seqfan(AT)plastereddragon.com), Dec 03 2004
STATUS
approved