[seqfan] Re: A computational challenge from 1973

[Neil Sloane]

Thanks Alex and Olivier!  Just to get the ball rolling, I will create
a new entry for the "squares" problem that Alex solved: it will be
A258718 in a few hours.  I will attach a scan of the original problem
from Popular Computing.
Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com



On Wed, Jun 17, 2015 at 3:27 AM, Olivier Gerard
<olivier.gerard at gmail.com> wrote:
> On Wed, Jun 17, 2015 at 7:25 AM, Neil Sloane <njasloane at gmail.com> wrote:
>
>>
>> Suppose m is a k-digit number. Let c(m) = m/100^k.
>> Thus, if m is a 2-digit number, c(m)  = .00m
>> Example: m=16, c(m)=.0016 = 16/100^2.
>> If m=127, c(m)=.000127 = 127/10^6.
>>
>> HERE IS THE PROBLEM:
>> For a sequence a = a(1), a(2), a(3), ...,
>> define f(a) = Sum_{n >= 1} c(a(n)).
>>
>> For example, take a = 1 4 9 16 25 36 ... the nonzero squares
>> Then f(a) is the infinite sum of
>> .01
>> .04
>> .09
>> .0016
>> .0025
>> .0036
>> ...
>>
>>
> The principle is reminiscent of the Liouville numbers.
> Most values of f for a(n) growing quickly enough will be transcendantal, I
> presume.
>
>
>> According to Popular Computing, my old friend Hermann P. Robinson
>> computed that f(a) = .18190589020080121567...
>>
>> Problem 22 asks for more digits, and I'm asking for someone
>> to enter this sequence (the sequence of decimal digits,
>> with keyword cons, as usual).
>
>
> and keyword base, of course.
>
>
>
>> If a = primes, f(a)=oo, so we don't consider that one!
>>
>>
> But a variant with a function of the exponent growing quicker than 2 would
> converge.
> One can write the c(m) function as
>
> m/10^(2*(1 + Floor[Log[10, m]]))
>
> so
>
> m/10^(3*(1 + Floor[Log[10, m]]))
>
> should work.
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/