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A101171
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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.
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1
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1, 12, 15, 18, 25, 45, 75, 125, 1125, 1875, 5625, 16875, 140625, 1171875
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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base,nonn,full,fini
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AUTHOR
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Chuck Seggelin (seqfan(AT)plastereddragon.com), Dec 03 2004
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STATUS
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approved
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