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A300819
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Consider the concatenation of a(n) and a(n+1) as the integer A; consider the concatenation of the last digit of a(n) and the first digit of a(n+1) as the integer B. Now A is divisible by B.
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1
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1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 28, 71, 17, 111, 24, 19, 110, 12, 18, 48, 72, 50, 13, 68, 73, 16, 38, 151, 14, 88, 74, 39, 77, 106, 95, 76, 26, 91, 20, 15, 37, 92, 40, 22, 31, 33, 82, 34, 78, 165, 36, 112, 84, 64, 80, 44, 55, 131, 30, 60, 66, 93, 86, 94, 136, 62, 118, 179, 216, 123, 42, 32, 109, 204
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OFFSET
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1,2
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COMMENTS
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The sequence starts with a(1) = 1 and is always extended with the smallest integer not yet present in the sequence that does not lead to a contradiction.
Leading zeros must be erased from the divisor: the pair [10,11] is thus accepted because 1011 is divisible by 1.
Lars Blomberg (in a private mail):
...The integer 21 is never used because:
...Let the previous number be x*10 + y, the new number is 21.
...Then x*1000 + y*100 + 21 (which is odd) must be divisible by y*10+2
...which is even, not possible.
...The same goes for 23, 25, 27, 29 and 41, 43, 45, 47, 49 and many more.
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LINKS
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EXAMPLE
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The concatenation [12] is divisible by 12, of course. A similar situation applies for the pairs [23], [34], [45], etc. When it comes to [910] we see that 910 is divisible by 91 (concatenation of 9 and 1, which are the last and first digits of the pair [9,10]). As we consider that [1011] is divisible by 1 (see the comment section), we accept that the integer 11 comes immediately after 10. But 12 cannot follow now, as [1112] is not divisible by 11. We have then to wait until 28 shows up as [1128] is divisible by 12 (quotient = 94). Etc.
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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