[seqfan] Hidden divisors of [a(n) and a(n+1) concatenated]
Eric Angelini
Eric.Angelini at kntv.be
Sun Feb 21 18:33:46 CET 2010
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.
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