Let us look at the abelian group $V$ of integer valued sequences modulo sequence which are zero almost everywhere.


Let me fix one constant $k\in \mathbb{N}$, which I will omit from the notation.
One specific sequence is given by $a\_n = \binom{n+k}{n}$. Let $\Sigma^m a$ be the shift of $a$ by $m$, i.e. $(\Sigma^ma)\_n =a\_{n-m}$. Now I am interested in zero combinations of those shifts. 


*Question:* Given integers $\lambda\_m$ (such that almost all,but not all integers are zero) and such that


\[\sum\_m \lambda\_m(\Sigma^ma)=0 \in V,\]


then $\sum\_m |\lambda\_m|\ge 2^{k+1}$.


A example where the (conjectured) minimum is obtained is the following. Let $T:V\rightarrow V$ be the difference operator, i.e. it is defined via $(Ta)\_n=a\_{n+1}-a\_n$. Since $a$ is a polynomial of degree $k$, we have $T^{k+1}a =0$ and if we write the left hand side out, we get exactly $2^{k+1}$ as that sum.



Zero combinations of certain sequences
Consider the Laurent polynomial $f(t)=\sum \lambda\_m t^m$. It is easy to prove that your conditions are equivalent to a system of $k+1$ equations $f^{(i)}(1)=0$ for $i=0,\dots,k$. In other words, $f(t)=(t-1)^{k+1}h(t)$ for some Laurent polynomial $h$ with integer coefficients. If $h(-1)\ne 0$, we get $|f(-1)|\geqslant 2^{k+1}$, this implies your claim. But if $h(-1)\ne 0$, it may fail, for example, consider $k=2$ and $f(t)=(t-1)(t^2-1)(t^3-1)$.


