Unsolved? problem

[Richard Guy]

PS.  It's well-known to those who well know
it that    1/999998999999 = 
0.000000 000001 000001 000002 000003 000005
   000008 000013 000021 000034 000055 000089
   000144 000233 000377 000610 000987 001597
   002584 004181 006765 010946 017711 028657
   046368 075025 121393 196418 317811 514229
   8320413462711783125245837028962....    R.

On Thu, 28 Apr 2005, msubbara wrote:

> Dear Richard:
> It is indeed a good idea to collect Murray Klamkin problems. I am particularly
> fond of one his problems-but it did not appear in Math journals, but in the
> book by L.A.Graham : Ingenious Mathematical problems and methods, Dover.
> Murray's problem,which appears as #72 in the book,reads: find the smallest
> number such that if the last digit is removed and placed at the  beginning to
> become the  first digit, this new number is nine times the original one.
> The solution  given in the book itself is the astronomical fijure
>
> 10,112,359,550,561,797,752,808,988,764,044,943,820,224,719.
> What would be the solution if we do similar operation with the last two digits
> and ask '99 times the original number", or replace the last three digits
> similarly to   get 999 times the original number etc.
> Subbarao
>
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>> ===== Original Message From Richard Guy <rkg at cpsc.ucalgary.ca> =====
>> I'm collecting Murray Klamkin
>> problems and solutions and am
>> currently going thru Math Mag.
>>
>> I came across Problem 886,
>> Math Mag 48(1975) 57--58
>> [nothing to do with Murray]
>> which isn't properly stated
>> but should read as in OEIS
>> A003508 :
>>
>> a(n) = a(n-1) + 1 + sum of
>> distinct prime factors of
>> a(n-1) that are < a(n-1).
>>
>> This leads to
>> 1,2,3,4,7,8,11,12,18,24,30,41,42,55,...
>>
>> The original problem asked that
>> if you start elsewhere, e.g.,
>>
>> 5,6,12, ...  or
>> 9,13,14,24, ... or
>> 10,18, ... or
>> 15,24, ...
>>
>> do you always merge with the
>> original sequence?  Evidently
>>
>> 91,112,122,186,... takes a little
>> while.
>>
>> Has anyone ... Can anyone prove
>> Charles Trigg's guess ?    R.
>
>