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May 30, 2017 at 16:41 history edited HenrikRüping CC BY-SA 3.0
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May 30, 2017 at 16:40 vote accept HenrikRüping
May 30, 2017 at 16:36 comment added HenrikRüping @ Fedor: I guess there is an index shift somewhere. In your reformulation it should be "less than $2^k$", since otherwise $p(t)=1$ would be an obvious counterexample. Furthermore it follows that $p = f/(t-1)^k$ must be divisible by $t+1$ (Otherwise plug in $t=-1$).
May 30, 2017 at 15:18 answer added Fedor Petrov timeline score: 1
May 30, 2017 at 15:03 comment added Fedor Petrov Equivalent reformulation: if a polynomial $f(t)$ with integer coefficients is divisible by $(t-1)^{k}$, then the sum of absolute coefficients is not less than $2^{k+1}$.
May 30, 2017 at 14:34 history edited HenrikRüping CC BY-SA 3.0
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May 30, 2017 at 14:21 comment added ThiKu The first sentence means "... of {\em integer} valued sequences ...", right?
May 30, 2017 at 13:58 history asked HenrikRüping CC BY-SA 3.0