A prime sequence can be partitioned into two sets of equal or consecutive sum Denote $P[n]$ as the prime sequence $\{p_1,p_2,\cdots,p_n\}$.
Conjecture:


*

*When $n=2k+1$ is odd, prime list $P[n]$ can be partitioned into two non-overlapping sublists, in which each sublist has equal sum $\operatorname{Total}[P[n]]/2$.

*When $n=2k$ is even, prime list $P[n]$ can be partitioned into two non-overlapping sublists, one sublist's sum is $(\operatorname{Total}[P[n]]-1)/2$, the other's is $(\operatorname{Total}[P[n]]+1)/2$.


For example:
$$
\begin{align*}
3-2 &= 1 \\
5-3-2 &= 0 \\
7 - 5 - 3 + 2 &= 1 \\
11 - 7 - 5 + 3 - 2 &= 0 \\
13 - 11 - 7 + 5 + 3 - 2 &= 1
\end{align*}
$$
and so on.
How could I write an efficient program to check it? Any clues to prove or disprove this conjecture?
BTW: I asked this question at mathgroup before, but I didn't describe it clearly.
 A: Scott Carnahan had an interesting idea; let's formalize it into an actual solution.  We will show that, given $n \ge 2$ a positive integer, $p_1, \cdots, p_n$ the first n primes, we have some $e_1, \cdots, e_n$ with $e_i = \pm 1$ such that $|e_1p_1 + e_2p_2 + \cdots + e_np_n| \le 1$.  (Note that we may further stipulate that $e_n = 1$.)  A simple parity argument from here suffices to prove the conjecture.
We will prove this by induction on $n$.  The cases $2 \le n \le 6$ are trivial to verify, and were provided already by a-boy.  We now fix $n \ge 7$.
We first need some asymptotics in the form of the Bertrand-Chebyshev theorem; we use the formulation that for $m > 1$ there is a prime between $m$ and $2m$.
Write $S_k = e_np_n + e_{n-1}p_{n-1} + \cdots + e_{n-k+1}p_{n-k+1}$, and let $M(k)$ be the minimum of $|S_k|$ over all tuples $(e_n, e_{n-1}, \cdots, e_{n-k+1})$.  We stipulated earlier that $e_n = 1$, so we have $M(1) = p_n$.  Two facts that will be useful to us in the future are that (1) $M(k+1) \le |M(k)-p_{n-k}|$ and (2) if $|a| \le |b|$, then $\min{\{|a+b|,|a-b|\}} \le |b|$.
We claim that $M(k) \le p_{n-k+1}$ for $k = 1, 2, \cdots, n-2$.  We prove this by induction on $k$.  The claim for $k = 1$ is trivial.  Now if $M(k) \le p_{n-k}$, then we are done, as $M(k+1) \le \min{\{|M(k)+p_{n-k}|,|M(k)-p_{n-k}|\}} \le p_{n-k}$ by fact (2).
Now suppose $p_{n-k} < M(k) \le p_{n-k+1}$.  Write $2m+1 = p_{n-k+1} \ge p_3 = 5$, so that $m > 1$.  In this case we know that $m < p_{n-k} < M(k) \le 2m$.  But then $M(k+1) \le M(k) - p_{n-k} \le 2m-(m+1) = (m-1) < p_{n-k}$ as desired.
The fact that $M(k) \le p_{n-k+1}$ is eminently useful.  
Indeed, we may use it to dispatch of the even case immediately.  Set $k = n-6$.  Then we have $M(n-6) \le 17$.  As all sums considered in $M(n-6)$ are sums of an even number of odd terms, we in fact have $M(n-6) \le 16$ and even.  Now we simply note that all odd numbers between -15 and 15 are realizable as sums and differences of the first 6 primes, which is left as an easy computational exercise.
In the odd case, we consider $k = n-5$.  Then $M(n-5) \le 13$.  For the same parity reasons as above, we have in fact $M(n-5) \le 12$.  And again, we note that all even numbers between -12 and 12 are realizable as sums and differences of the first 5 primes - another easy computational exercise.
The limits of $n-6$ and $n-5$ are the best possible for our small-case analysis.
If we were to establish an algorithm for this, we could just do the greedy algorithm on choosing $e_n$, then $e_{n-1}$, and so on, each time choosing $e_k$ so as to minimize $S_{k+1}$ (or randomly if $S_k = 0$).  Our claim that $M(k) \le p_{n-k}$ will continue to be satisfied by the greedy algorithm, as the proof of the claim does not involve changing prior $e_i$.  Thus our greedy-algorithm mimium modulus must satisfy the same inequality, and we continue until we are at $n-6$ or $n-5$, then finish as in our nonconstructive proof.
A: This is a special case of the subset-sum problem, which is NP-complete in general but probably tractable in your case. Wikipedia describes a pseudo-polynomial-time algorithm that may work for you. 
(Link: http://en.wikipedia.org/wiki/Subset_sum_problem#Pseudo-polynomial_time_dynamic_programming_solution)
A: Expanding on the comment above, consider Pn, the set of the first n primes,
and SSn, the set of subset sums of Pn. For n greater than 3, we see that
SSn is 6 numbers shy of being the interval [0, Sn], where Sn is the largest
subset sum, and these  numbers are 1,4,6 and their negatives subtracted from Sn.
Letting m=n+1, one sees this by noting SSm= SSn union the shifted set p_m + SSn.
(One also needs p_m less than S_n - 6, but that is easily established.)
Now the question can be answered by noting there is a subset sum equaling floor of
(Sn)/2, and setting the sign of those numbers in that sum to minus.
Added:  If a sum is realizable, one can use a tempered greedy algorithm
which subtracts the largest available prime from a running total as long
as the result is not 1, 4, or 6.
