Given $a_1,a_2,...,a_m$ positive integer. Denominator $d$ is smallest positive integer for $b_l$ integer coefficient.
$$\sum_{k=1}^m\binom{n}k a_k= \frac{1}d\sum_{l=1}^mb_ln^l$$
Now consider $n=dt+r$ where $d>r\ge 0$.
can it be shown that, above equation transform as
$$\frac{1}d\sum_{l=1}^mb_l(dt+r)^l=\sum_{u=0}^{m-1}(x_ut+y_u)(dt+r)^u$$
With $x_u$ and $y_u$ integers.
Example
Let $a_1=1,a_2=1,a_3=3$
then $\sum_{k=1}^3\binom{n}k a_k=(n^3-2n^2+3n)/2$ here $d=2$
Case$(1)$ $n=2t$,
$\frac{n^3-2n^2+3n}2=(t-1)n^2+n+t$
Case$(2)$ $n=2t+1$,
$\frac{n^3-2n^2+3n}2=(t-1)n^2+(t+2)n$
Thank you.