|
%I #2 Feb 27 2009 03:00:00
%S 16,43,86,3923,7846,1320532,14869252,37031789,74063578,1770748607,
%T 3541497214,7082994428,50541770557,1040474831558,1076026363388,
%U 2080949663116,2746369613531,3376243036861,5492739227062,6529386313577
%N Numbers n which are divisors of the number formed by concatenating (n-1), (n-2), (n-3) and (n-4) in that order.
%C Each member of this sequence appears to also be a factor of the number formed by concatenating (n+1), (n+2), (n+3) and (n+4) in that order. When evaluating concat((n+1),(n+2),(n+3),(n+4)) - concat((n-1),(n-2),(n-3),(n-4)) for members larger than 86 the difference appears to always be a number of the form 2(0)...4(0)...6(0)...8 with the same number of zeros following the 2, 4 and 6. The member will be a factor of this number. Further terms for the 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 2(0)...4(0)...6(0)...8. 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 likely candidates and should be tested (note that apart from 16, candidates which are divisible by 8 appear to never be members). For example let d = 2000000000000000400000000000000060000000000000008. z=15. The divisors of d are numerous, but only one is z+1 (16) digits long: 7547657634163187. Testing this candidate confirms that it is also a member of this sequence.
%e a(3)=86 because 86 is a factor of 85848382.
%Y Cf. A069860, A088797, A088798, A088799.
%K base,nonn
%O 1,1
%A Chuck Seggelin (barkeep(AT)plastereddragon.com), Oct 20 2003
|
