This question is about the beaviour of 4-genus of knots with respect to connected sum.


Let us indicate with $T(k)$ a Torus knot of type $(2,k)$, $k$ is an odd integer.
Fix an orientation for every $T(k)$ so that $T(-k)$ represents the same knot with reversed orientation.


$T(k)\sharp T(-k)$ is a slice knot. More generally if $K^\* $ denotes the mirror of $K$ then $K\sharp K^\*$ is slice (infact ribbon).


My question is:


* Is it true that a knot of the form $T(a\_1)\sharp\dots\sharp T(a\_n)$ is slice if and only 
if n is even, say $n=2k$, and we can arrange the coefficients so that for every $i\leq n$
we have $a\_{k+i}=-a\_i$?


Note that for any pair of knots $K\_1$ and $K\_2$, if both $K\_1$ and $K\_1\sharp K\_2$ are slice then $K\_2$ is also slice. Therefore one only needs to show that there exists
$i$ and $j$ such that $a\_i=-a\_j$.


Of course we can generalize this problem:


* Which connected sums of torus knots are slice?


Here are some links for definitions:


Torus knot: <http://en.wikipedia.org/wiki/Torus_knot>


Slice knot: <http://en.wikipedia.org/wiki/Slice_knot>


Slice genus: <http://en.wikipedia.org/wiki/Slice_genus>



When is a connected sum of torus knots a slice knot? 
I believe the answer to your last two questions is *yes*, and it follows from Litherland's (1979) computation of the Tristram-Levine invariants of torus knots. See Kearton's survey here: 


<http://www.maths.ed.ac.uk/~aar/slides/durham.pdf>


i.e. a connect sum of torus knots is slice if and only if the prime summands appear in balancing mirror-reflected pairs. 


