1,1

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.

No more terms < 10^29. - David Wasserman (wasserma(AT)spawar.navy.mil), Aug 26 2005

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

Cf. A069860, A088797, A088798, A088799, A088800.

Adjacent sequences: A088865 A088866 A088867 this_sequence A088869 A088870 A088871

Sequence in context: A019547 A067673 A045253 this_sequence A044236 A044617 A090095

base,nonn

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