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I don't know if this is true or not but I want this to be true and so far I don't have any counterexample.

Let $i$ be odd. Do there exist coefficients $a_k \in \{0,1\}$ such that

$$\sum_{k=1}^{i-1} (-1)^k \binom{i}{k} a_k = 1\text{?}$$

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Given that $i=2j+1$ is odd, $a_k$ and $a_{i-k}$ cancel out if both are equal. So you could equivalently ask for coefficients $b_k\in \{-1,0,1\}$ such that $$\sum_{k=1}^jb_k\binom{i}{k}=1.$$

I will ignore the cases of $k=0$ and $k>j$

If there is a prime $p$ such that $p|\binom {i}{k}$ with few exceptions, then it may be possible to rule out a sum of the form you seek being any value congruent to $1 \bmod p.$

This eliminates the sum being $1 \bmod p$ for a prime power $i=p^e$, for any $i=3p$ and for all $i=5p$ except, perhaps, $i=35$ and $i=55$

If $i=p^e$ is a prime or prime-power then $p|\binom{i}{k}$ so the sum can't never be anything other than a multiple of $p.$

If $i=3p$ then $p|\binom{3p}{k}$ with the exception that $\binom{3p}{p} \equiv 3 \bmod p$ so the sum can only be $0,3$ or $p-3$ $\bmod p.$

For $i=5p$ we have $\binom{5p}{p} \equiv 5 \bmod p$ and $\binom{5p}{2p} \equiv 10 \bmod p$ so the sum can only be $0,\pm5,\pm 10,\pm 15$ $\bmod p$ For $p=7$ we do have $15 \equiv 1 \bmod p.$ And $p=11$ is not obviously ruled out.

However we do see that the only chance for $\sum_{k=1}^{27}b_k\binom{55}k=1$ is $b_{11}=0$,$b_{22}=-1$ and the other terms (which are all multiples of $11$) adding to $\binom{55}{22}+1.$ Looking $\bmod 5$ seems to force $b_5=0$ and $b_{25}=1.$

In general, for $i=qp$ with $q<p$ and both prime, $\binom{qp}{rp} \equiv \binom{q}{r} \bmod p$ This will eliminate all but a finite number of $p$ for any given $q$. And in the cases not immediately ruled, out the possibilities to search over are restricted.

Similar considerations rule out $i=7p$ for $p >7$ prime, with the possible exceptions of $p=11, 13, 17, 19, 29, 31, 41, 43$, the primes dividing $N-1,N$ or $N+1$ for $N=7,14,21,35,42,49,56,63$ , these being the numbers $N$ which can be formed from some or all of $7,21,35=\binom{7}{1},\binom{7}{2},\binom{7}{3}$ with addition and subtraction.

I don't immediately see an obstacle to $i=105.$

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For $i = 35$, take $a_k = 1$ for $k = 11, 14, 15, 22, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34$, and $0$ otherwise.

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  • $\begingroup$ Nice example. Given Gerhard's comments and some fiddling I did with small values, maybe the question should be about finding (characterizing?) $i$ for which this does work. $\endgroup$ – Brian Hopkins Jul 14 at 23:23
  • $\begingroup$ If my computations are correct, $35$ is the least. $\endgroup$ – Robert Israel Jul 15 at 1:05
  • $\begingroup$ That's a nice counterexample. Can we argue that once we have used some coefficients in a combination to sum to 1, then there does not exist any combination of the remaining terms that could sum to 0. $\endgroup$ – neha mainali Jul 16 at 4:06
  • $\begingroup$ I believe 35 is the least for which it holds. $\endgroup$ – neha mainali Jul 16 at 4:07

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