[seqfan] Re: A computational challenge from 1973

[Olivier Gerard]

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.