[seqfan] Hidden divisors of [a(n) and a(n+1) concatenated]

[Eric Angelini]

Hello SeqFans,
 
... let's start a monotonically increasing S with 1,2:
 
S= 1,2,...
 
... if we concatenate 1 and 2 we get '12' which has
1,2,3,4,6,12 as divisors being different. What would
be the smallest integer now (to prolong S), which, 
concatenated with 2, has 5 in it's divisors? (This 5
is the "next hidden divisor" we are looking for.)
We try 3 -- but [23] has no 5 among it's divisors --,
we try 4 -- but [24] has no 5 among it's divisors --,
we try 5 -- and get a hit as [25] is divisible by 5:
 
S= 1,2,5,...
 
[12] produces "hidden div." 1,2,3,4,6,12
[25] produces "hidden div." 1,5,25
 
The next hidden divisor is 7 (we already have 6 thanks
to [12]); which a(4) would fit S?
 
This is 6 -- as 56 is divisible by 7:
 
S= 1,2,5,6,...
 
We must always keep track of the hidden divisors (HD)
we have found so far, in order not to search twice
for divisors we already have:
 
[12] produces HD = 1,2,3,4,6,12
[25] produces HD = 1,5,25
[56] produces HD = 1,2,4,7,8,14,28,56
 
The next HD we want is thus 9; a(5) is 12:
 
S= 1,2,5,6,12,...
 
 [12] produces HD = 1,2,3,4,6,12
 [25] produces HD = 1,5,25
 [56] produces HD = 1,2,4,7,8,14,28,56
[612] produces HD = 1,2,3,4,6,9,12,17,18,34,36,51,68,102,153,204,306
 
... etc. I've computed (by hand) a few more terms,
which I'm sure are wrong at some point (this was
late night yesterday):
 
S= 1,2,5,6,12,20,24,31,35,36,54,72,...
 
Could someone check and compute a few more terms?
 
---
 
A similar seq. dealing with HDs would be T:
 
T= 1,0,2,0,6,...
 
Where we concatenate _every integer_ printed so far
to produce the HD:
 
     [1] produces HD = 1
    [10] produces HD = 1,2,5,10
   [102] produces HD = 1,3,6,17,34,102
  [1020] produces HD = 1,2,3,4,5,6,10,12,15,17,20,30,34,51,60,68,85,102,170,204,255,340,510
 
... and to get the next HD (7), I guess we must 
prolong T with '6'...
 
Lots of fun with HDs!
Best,
É.