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A088868
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Numbers n which are divisors of the number formed by concatenating (n-4), (n-3), (n-2) and (n-1) in that order.
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6
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OFFSET
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1,1
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COMMENTS
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Each member of this sequence appears to also be a factor of the number formed by concatenating (n+4), (n+3), (n+2) and (n+1) in that order. When evaluating concat((n+4),(n+3),(n+2),(n+1)) - concat((n-4),(n-3),(n-2),(n-1)) for members of this sequence the difference appears to always be a number of the form 8(0)...6(0)...4(0)...2 with the same number of zeros following the 8, 6 and 4. The member will be a factor of this number. Terms for this sequence can be produced by factoring numbers of this form. Let z=the number of zeros in one of the segments of a number d of the form 8(0)...6(0)...4(0)...2. Find the divisors of d. All divisors which are not of length z+1 are not members of this sequence and those that are of length z+1 are possible candidates and should be tested. For example let d = 8000000000000000006000000000000000004000000000000000002. z=17. The divisors of d are numerous, but only two are z+1 (18) digits long: 138599925259848671 and 27719985051 9697342. Testing these candidates confirms that the first one is a member of this sequence.
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LINKS
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EXAMPLE
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a(2)=109 because 109 is a factor of 105106107108.
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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Chuck Seggelin (barkeep(AT)plastereddragon.com), Oct 20 2003
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STATUS
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approved
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