@Adam Ledger 

1. Any programmer or designer commits mistakes ==> bugs; e,g. one of the many possible cases was not considered. Once discovered, such bugs are fixed.

2. There are theoretical aspects. E.g. there is no general algorithms able to detect whether an expression is 0 or not. If such a zero expression is not detected, a series expansion could be wrong. In this case, a workaround (of the user) could be to rewrite the expression in a simpler form.
These "bugs" cannot be fixed, but maybe a more general (not "universal" though) method could be implemented.

You should post such problems and maybe some user will come with a confirmation/workaround/explanation.

@sand15 

Any name returned from the concatenation operator will be a global variable. See ?||

@Carl Love 

I am not sure whether ad hoc is the right term. If a user asks something about a function f : R --> R,  should the answer refer to the case f : X --> Y,  with X,Y Banach spaces, in order to be non "ad hoc"?

@Carl Love 

My point is that it is not very difficult to write a procedure with a local status for a variable.
 

@Carl Love 

Maybe something like this:

restart;
x:=10:
pl:=proc(f::uneval, r::uneval)
plot(f, r);
end:
pl(sin(x), x=-Pi..Pi);

@Carl Love 

I know, my answer was in the OP's context and it is correct.

@nm 

Because it is global.

@Carl Love 

Heinz'  is not that bad with a small change.

Heinz:= n-> min(select(d-> (d^2>n), Divisors(n))):

@Carl Love 

I wonder how useful is the type  `&+` due to the fact that the ordering of the terms in a sum could be difficult to predict.

@Al86 

This is not true. Take for instance p_0(t) = -t  or  p_0(t) = -t/2.
Not even when p_0 > 0.

@Al86 

It is not possible to maximize an expression depending on an unknown function (p_0 in this case).
It's just like asking for max { f(t): t in [0,1] } with no information about f().

Or, is the problem a variational one?

@Adam Ledger 

Your function F computes the set {p^t :  p is a prime, t is a power of 2, p^t | n} \ {2}.
This can be done much more efficiently using ifactors(n).

F:= n -> {seq(seq(p[1]^(2^j), j=0..ilog2(p[2])), p=ifactors(n)[2])} minus {2}:

@Markiyan Hirnyk 

I think that I have explained the problem crearly in my reply.
Note also that MultiSeries:-asympt(f,n,3) assuming x>0;  is even wronger.

@Markiyan Hirnyk 

Have you read my reply, or your "does not correspond to reality" was automatic?
a contains the asymptotic for real n --> oo. It is wrong. Please note that this result is used even when n::posint, so we cannot consider it as being correct in this case.