

A088868


Numbers n which are divisors of the number formed by concatenating (n4), (n3), (n2) and (n1) in that order.


6




OFFSET

1,1


COMMENTS

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((n4),(n3),(n2),(n1)) 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.
No more terms < 10^29.  David Wasserman, Aug 26 2005


LINKS

Table of n, a(n) for n=1..9.


EXAMPLE

a(2)=109 because 109 is a factor of 105106107108.


CROSSREFS

Cf. A069860, A088797, A088798, A088799, A088800.
Sequence in context: A067673 A250653 A045253 * A044236 A044617 A216169
Adjacent sequences: A088865 A088866 A088867 * A088869 A088870 A088871


KEYWORD

base,nonn


AUTHOR

Chuck Seggelin (barkeep(AT)plastereddragon.com), Oct 20 2003


STATUS

approved



