The Math Thread - Page 23

Imho, those numbers should exist for all bases >2.
Lets again take a quick look at the algorithm I also mentioned earlier, but adjust it to baseN:

We assume the last digit to be

p_0 = d,

and then continue:

r_0 = 0
d_i = (p_i*2)%N

r_(i+1) = (p_i*2)/N (/ used in the integer sense)
p_(i+1) = d_i+r_(i+1)

We can abort the algorithm when we get p_x == p_0 = d and r_x == r_0 = 0.

Now this isnt necessarily terminating for any d. But since p_i is in {0,...,N} and r_x in {0,1} there is only a finite number of combinations possible, so there must be cycles at some point, where
p_k == p_l && r_k == r_l && k < l
obviously, we can now state that for the lowest such k and l it holds true that for all m>k
p_m == p_(m+l-k) && r_m == r_(m+l-k)

This means we have a number of cycles. If there is a single m>k, where r_m == 0, we can take p_m as last digit of our number and generate a valid number in regards to the original question.

So all we are required to do is to proof, that in this cycle there is guaranteed to be a position where r_m != 1.
And this is rather trivial (I will make some leaps here, but it should be obvious, if not I can detail this) as the only cycle where r_m can stay permanently at 1 is the cycle of p_k = N-1 and r_k = 1. But as in this case it is necessary that p_(k-1) = N-1 and r_(k-1) = 1. Now this violates the rule of k being the lowest k to build such a cycle. And as we originally started with a different pair (d,0), it is clear that we can never enter this cycle -> there must be another cycle. -> There is a solution.

PS: There is the additional condition, that p_(m-1) != 0. But it also can be shown, that this is only a problem for N=2, as in all other cases simply taking m+1 produces a valid solution.

1/(1+e^x) + 1/(1+e^-x) = 1
multiply with (1+e^x)(1+e^-x)
-->
(1+e^-x) + (1+e^x) = (1+e^x)(1+e^-x)
2+ e^-x + e^x = 1 + e^x + e^-x + e^x * e^-x
as e^x*e^-x = e^0 = 1 -> both sides are equal, done