Conjecture: Suppose p[i] is the i-th prime, p[1]=2, g[i]=p[i + 1] - p[i], h[1]=g[1], h[n] = g[1]-g[2]+g[3]-g[4]+...+ (-1)^(n-1) g[n] = h[n-1]+(-1)^(n-1) g[n], then {h[n]} changes sign infinitely often,however lazily, and h[n] never just equal to 0, for n=1,2,3,...

http://att.newsmth.net/att.php?s.749.91400.300.nb

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Some motivation would be helpful... but if I understand your obscure notation then $h_n=-2+2\cdot 3-2\cdot 5 + \cdots \pm 2p_n \mp p_{n+1}$ is always odd, hence nonzero. – Eben Freeman Aug 4 2010 at 5:21
@Eben, yes, and the numbers $2-3+5-\dots\pm p_{n+1}$ are tabulated at research.att.com/~njas/sequences/A008347 although I'm not sure anything there gets us any farther on the question of sign changes. – Gerry Myerson Aug 4 2010 at 5:40
Yes,$h_n=-2+2(3-5 + \cdots \pm p_n) \mp p_{n+1}$ is always odd – a-boy Aug 4 2010 at 7:22
The terms of -h[2n] are in sequence oeis.org/classic/A121573, prime-gap race; difference of the cumulative sums of gaps above and below prime(2n). – tdnoe Aug 5 2010 at 16:07