The problem with 19 turned out to be minimal. There is a unique (up to permutations) way to do with 19 for the first 56 primes what was done for 7 with the first 29. Further, right now I am conjecturing based upon a lot of empirical data that 19 sets the permanent record, but this is hedged. If true, I should be able to prove this (not by hand) over the weekend. If false, I might have a surprise on my computer by the time I get home or at least by Monday. The solution for 19 involves leaving 3 of the terms as primorials through the final multiplications, so there is a lot of room for the alternative problem where this is not acceptable.

JGM

--- On Thu, 2/7/13, James Merickel <moralforce120@...> wrote:

From: James Merickel <moralforce120@...>

Subject: [PrimeNumbers] Sequence of 30 primes (curio submission)

To: primenumbers@yahoogroups.com

Date: Thursday, February 7, 2013, 12:07 PM

Hi, folks. Just wanted to inform you all of a little thing I have computed:

16072758981106442684006718854529251552093=

abcdef+(1/a+1/b+1/c+1/d+1/e+1/f)*109#, where

a=7*13*37*83, b=17*19*89, c=11*79*101, d=23*29*67*103, e=47*61*107 and f=41*53*109.

Uniquely, it is the last of 30 primes beginning with 7=1+1+1+1+1+1+1 produced by sequentially multiplying all but one addend by the primes from 2 through 109.

The sequential build-up can be obtained from 485191936591420718030369 (just the smallest of 62 primes that work) by looking at the number's base-7 representation and taking the digits by order of increasing significance as the addend not to be multiplied by a prime (Digit equal to 0 means the 1st addend is skipped and the rest multiplied by the prime, etc., with the units digit corresponding to the prime 2).

Jim Merickel

P.S. Note that with the strict requirement that a term composed of one 1 and the rest of the addends 2 be prime would make the start of any larger analogous sequence start with at least 19. I doubt such a maximal sequence can be found for so large a start, but I will try to see if I am wrong.

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