A088868 Numbers n which are divisors of the number formed by concatenating (n-4), (n-3), (n-2) and (n-1) in that order.

49, 109, 307121, 3837881, 415922011, 44886856951, 73071640562111, 11741452251865261, 138599925259848671

Offset: 1

Author

Chuck Seggelin, Oct 20 2003

Keywords

• base
• nonn

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((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, Aug 26 2005

Examples

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

Crossrefs

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