Elliptic curves over $\mathbb{Q}$ with no rational torsion and $\mu$-invariant equal to 1 at $p=3$ How to find out examples over elliptic curves over $\mathbb{Q}$  with no rational torsion and $\mu$-invariant equal to 1 at $p=3$ $?$
 A: EDIT: There was a problem with my original answer.  Details below the bottom line.

If you have any elliptic curve $E/{\mathbb Q}$ with a point of order 3, then we have an exact sequence of Galois modules:
$$
0 \to {\mathbb Z}/3{\mathbb Z} \to E[3] \to \mu_3 \to 0.
$$
Here the ${\mathbb Z}/3{\mathbb Z}$ term arises from the point of order 3 defined over ${\mathbb Q}$. The $\mu_3$ term is then forced upon us since the determinant of this Galois representation is the mod 3 cyclotomic character.
Now here is the trick.  There is an isogenous curve $E'/{\mathbb Q}$ such that 
$$
0 \to \mu_3 \to E'[3] \to {\mathbb Z}/3{\mathbb Z} \to 0
$$ 
(i.e. the two outer terms have switched!),
and moreover this sequence is not split.  
I'll say in a moment how to find this curve.  But let me first point out that this new curve should have $\mu$-invariant equal to 1 -- this follows from Greenberg's conjecture on $\mu$-invariants.  The $\mu$-invariant should be the smallest $n$ such that $E[p^n]$ has a cyclic sub which is odd and ramified.  In this case, the sub of $\mu_3$ (which is odd and ramified) gives that $\mu(E')$ is at least 1. Since the sequence is not split, there is no cyclic sub of size 9 (and so conjecturally the curve has $\mu$-invariant 1). (EDIT: Nope.  This isn't true. See below. I should be assuming that 3-isogeny class for $E$ has only two curves in it.)
Now, how to find this curve?  Just take $E$ and mod out by that ${\mathbb Z}/3{\mathbb Z}$ sub of $E[3]$.  The resulting curve then has $\mu_3$ as a sub and ${\mathbb Z}/3{\mathbb Z}$ as quotient.  If this extension is non-split, we are done.  If not, mod out by ${\mathbb Z}/3{\mathbb Z}$ again.  And keep repeating until the extension is non-split.  (If this continued indefinitely, then the Tate module of $E$ at 3 would be reducible which it is not.) 

EDIT: OK. The problem is that just because the sequence 
$$
0 \to \mu_3 \to E'[3] \to {\mathbb Z}/3{\mathbb Z} \to 0
$$ 
is not split does not mean that $E'[9]$ doesn't contain a cyclic sub of size 9.  In fact, if the process described above of modding out by ${\mathbf Z}/3{\mathbf Z}$ takes two steps, then it will contain such a sub and have $\mu$-invariant strictly greater than 1.
In fact, this problem comes up in the example of conductor 19 in Jeff H's answer.  In this case, there are 3 isogenous curves of conductor 19 -- 19a1, 19a2, and 19a3 as in Cremona's tables.  I'll call them $E_1$, $E_2$ and $E_3$ respectively.  For $E_1$ we have $E_1[3] \cong \mu_3 \times {\mathbb Z}/3{\mathbb Z}$.  Modding out $E_1$ by the $\mu_3$ gives $E_3$, and for $E_3$ we have a non-split extension.
$$
0 \to {\mathbb Z}/3{\mathbb Z} \to E_3[3] \to \mu_3 \to 0.
$$
Modding out $E_1$ by the ${\mathbb Z}/3{\mathbb Z}$ gives $E_2$, and for $E_2$ we have a non-split extension.
$$
0 \to \mu_3 \to E_2[3] \to {\mathbb Z}/3{\mathbb Z} \to 0.
$$
So if we had started my solution above with $E = E_3$ (which is a curve with a point of order 3), we would have to first mod out by ${\mathbb Z}/3{\mathbb Z}$ and we arrive at $E_1$ which again has a point of order 3.  So we again mod out by ${\mathbb Z}/3{\mathbb Z}$ and we arrive at $E_2$ which has no point of order 3.  However, $E_2$ has a cyclic isogeny of degree 9 to $E_3$ whose kernel is odd and ramified.  Thus $\mu(E_2)$ is at least 2 (and exactly 2 by Greenberg's conjecture).  And every other curve in this isogeny class has a point of order 3.
OK.  How to fix this?  Well the problem is that third curve in the isogeny class.  So instead, we need a curve with a point of order 3 whose 3-isogeny class has size 2.  This way we are guaranteed from the start that there is no cyclic sub of size 9.  After a quick peek at Cremona's tables, I see that 44a2 works.  It has $\mu$-invariant equal to 1 and no 3-torsion.
A: You can find many examples like this using Rob Pollack's tables of elliptic curve Iwasawa invariants: http://math.bu.edu/people/rpollack/Data/curves1-5000 
Just search the table for a curve with $\mu=1$ at $p=3$ (this is relatively rare), and then check whether it has rational torsion at LMFDB. 
Here's what appears to be the example with smallest conductor:
http://www.lmfdb.org/EllipticCurve/Q/19.a1

Edit: This particular example is incorrect, but this method is still probably the easiest way to find such examples.

