Extending A062885

[John Layman]

I have just submitted the following extension and comments to A062885, where terms from a(13)=208
on are new.  Note the modification in %N (name, i.e. definition).

Subject: COMMENT FROM John W. Layman RE A062885

%I A062885
%S A062885
0,2,2,6,4,20,6,42,8,864,20,42086,420,208,42,420,64,8642086,864,642086,20,42,42086,6420864,864,
-1,208,864,420,8642,420
%N A062885 Smallest multiple of n with property that digits are even and each digit is two less
(mod 10) 
than the previous digit, if such a multiple exists, else -1.
%C A062885 It is conjectured that the next term, a(31), is -1.  Terms a(32) through a(44) are 64,
-1, 864
2086, 420, 864, -1, 642086, -1, -1, -1, 42, 86, 2086420864, where a(40)=-1 has been established and
other
 values of a(n)=-1 are only conjectured (no multiple up to (10^8)*n meets the necessary conditions
in tho
se cases).
%e A062885 For n=25, the conditions require that the desired multiple 25k have k even, i.e.
25k=25(2i)=50
i=(5i)(10).  Thus the final digit is 0, so the next-to-last digit must be 2, but this is impossible
since
 5i always ends in 0 or 5.  Thus a(25)=-1.
%O A062885 0
%K A062885 ,nonn,
%A A062885 John W. Layman (layman at math.vt.edu), Nov 01 2001
RH 
RA 128.173.42.69
RU 
RI 


An unexpected point of interest arose when encountering values of n for which a(n)=-1, that is,
when extensive calculation indicates that NO multiple of n has the desired property of it digits.
The interesting part is proving that no such multiple exists, i.e. that a(n) is INDEED -1.  The
proof for n=25 is given in the comment above and is also easy for n=40.  It  might be of some
interest to see proofs for n=31, 33, 37, 39, 41, etc.  Some of these may be easy but, who knows,
some may be rather challenging.

John Layman