# Are large numbers the sum of two or more large primes? [Hoping for reasonable constants]

Is it true that for all $n>N$ that n is the sum of two or more distinct primes that are either large or (for parity reasons) 2?

I feel like I've seen a result allowing this with $p\gg n^e$ for reasonably small e (0.4?). In my case I could be much more lax: even $p>23$ would suffice. But I'd like an effective N, actually very small if possible ($N=10^7$ would be ideal).

I stress that the number of summands need not be small; I'm not trying anything like Goldbach. Seventeen primes or even two hundred would be fine.

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@Jack Huizenga: Yes, they need to be distinct. Let me edit that in. –  Charles Aug 22 '11 at 6:17
Ramaré showed in his thesis that at most seven primes suffice. The original result, of such a number of summands, is Schnirelmann's constant. numdam.org/item?id=ASNSP_1995_4_22_4_645_0 He improved Riesel and Vaughan (19 primes). Distinctness, or size considerations, becomes irrelevant fairly quickly, though I do not know an explicit way. The tactic of šnirel’man can also be used, for $p>23$ rather than all primes. The principal technique, is to show sums of two primes have positive density, and extrapolate by summation of these. –  Junkie Aug 22 '11 at 8:22
What would be possible parity reason? –  Fedor Petrov Aug 22 '11 at 14:09
@Fedor Petrov: one cannot write an odd integer as the sum of (say) 10 primes that are all large, as the question preferred. At least one of the primes must be equal to 2, by parity. –  Junkie Aug 22 '11 at 23:44

By computer enumeration, every number in [96, 250] can be written as the sum of distinct primes from {2, 29, 31, 37, ...}. Strong induction shows that every integer $n>95$ can be written as a sum of distinct primes from the set using Bertrand's postulate (since $2\cdot96<250$).
@Gerhard Paseman: I happen not to need those for my result. I know that six primes suffice for some N, and I suspect that they can all be taken larger than $\sqrt[4]n$ (or 2). –  Charles Aug 22 '11 at 21:37