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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.

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1 Answer 1

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Denote $\sum b_l(x+r)^l=\sum c_l x^l$. The numbers $c_l$ are still integers, $c_0=\sum b_l r^l$ is divisible by $d$, and we have, denoting $dt+r=z$, $$ \frac1d\sum b_l(dt+r)^l=\frac1d\sum c_l (dt)^l=\frac{c_0}d+\sum_{l>0} c_l t\cdot (dt)^{l-1}=\\ \frac{c_0}d+\sum_{l>0} c_l t\cdot (z-r)^{l-1}= \frac{c_0}d+\sum_{l>0,0\leqslant j\leqslant l-1} c_l {l-1\choose j}(-r)^{l-1-j}z^jt, $$ that has desired form if you group all terms with $z^j$ together.

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  • $\begingroup$ Thanks for your solution. It's helpful to progress on my other problems. In given claim, $d\mid c_0=0$. $\endgroup$
    – Pruthviraj
    Jun 4, 2020 at 6:59

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