整系数多项式除以2^k求和为整数

hbghlyj

$p\inZ[x]$, 求证$f(p)\inZ$. $$f(p)=\sum_{k=0}^\infty p(k) / 2^k$$

1. a=Table[Random[Integer,10],{5}];
2. p[x_]:=Sum[a[[k]] x^(k-1),{k,1,Length[a]}];
3. q[x_]:=p[x+1]-p[x];
4. F=Sum[p[x]/2^x,{x,0,Infinity}]
5. G=(2 p[x] /. x->0) + Sum[q[x]/2^x,{x,0,Infinity}]

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tommywong

$\displaystyle F(n)
=\frac{p(n)}{1/2-1}+\frac{1}{(1/2-1)^2}\sum_{k=1}^{deg(p)} \frac{(-1)^k (1/2)^{k-1}}{(1/2-1)^{k-1}}\Delta^k(p(n))$
$\displaystyle =-2p(n)-4\sum_{k=1}^{deg(p)}\Delta^k(p(n))$

$\displaystyle f(p)=\sum_{k=0}^\infty p(k) / 2^k
=p(0)-\dfrac{1}{2}F(0)=2\sum_{k=0}^{deg(p)}\Delta^k(p(0))$

hbghlyj

Summation by parts
$$ \sum_{k=0}^n f(k) [g(k+1)-g(k)] = [f(n+1) g(n+1) - f(0) g(0)] - \sum_{k=0}^n g(k+1) [f(k+1)-f(k)] $$
for $g(k)=2^{-k}$ and $f(k)=p(k)$ gives in the limit $n \to \infty$ the formula
$$ -\sum_{k=0}^{\infty} p(k)/2^{k+1}= -p(0) - \sum_{k=0}^{\infty} (p(k+1)-p(k))/2^{k+1} \; . $$
Multiply by $-2$ to get
$$  \sum_{k=0}^{\infty} p(k)/2^k = 2 p(0) + \sum_{k=0}^{\infty} q(k)/2^k \; , $$
where $q(k)=p(k+1)-p(k)$ is a polynomial of degree $n-1$.
By induction on the degree of the polynomial using that for degree polynomials $p$ the result is $1$ proves the result.

hbghlyj

The sequence $a_n = f( (1+x)^n)$ is A207047.
The sequence $b_n = f(x^n)$ is A002050. It counts the number of simplices in a Barycentric subdivision of a $n$ simplex.