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%I #8 Dec 07 2019 08:19:47
%S 1,12,15,18,25,45,75,125,1125,1875,5625,16875,140625,1171875
%N a(n) divides the number formed by concatenating the sum of the digits of a(n) with a(n), by a factor not previously used.
%C No more terms < 10^100. - _David Wasserman_, Mar 06 2008
%C 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
%e 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.
%Y Cf. A101170.
%K base,nonn,full,fini
%O 1,2
%A Chuck Seggelin (seqfan(AT)plastereddragon.com), Dec 03 2004
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