Dear groupmembers,

The conjecture, which did not make a whole lot of sense anyway and was already showing itself unsupported empirically before the following result, turns out to be false. I had thought that perhaps there would be an argument modulo 19 or 19#.

The next record of 62 primes in succession occurs with 80671 (arrived c. 8AM today):

With the 19 numbers

a=2*103*157, b=3*173*233, c=5*41*193*239*271, d=7*113*167*257,

e=11*53*107*199*241*263, f=13*29*101, g=17*43*109*181*229, h=19*97*=

281, i=23*59, j=31, k=37*151, l=47*179*277, m=61*73*137, n=67*83*191*211*283, o=71*79*127*131*163*197*293, p=89, q=139, r=223*269 and s=227*251,

293#*(1/a+1/b+1/c+1/d+1/e+...+1/q+1/r+1/s+80652)=

164835030772135218479491263344471877765987572654819835433403742

893710632923823067093079399895258247669010672441740020208086391

is the largest of a sequence of 62 primes starting with 80671 addends of 1 =

and produced by multiplying each addend save one by the primes in sequence =

from 2 through 293.

A more reasonable conjecture is that requiring the largest prime in such a =

sequence to have no or at most one term in the sum as the primorial of the =

largest prime multiplied has a solution for 19 and for no larger prime. This I have not checked (and may not, since it isn't that interesting).

JGM

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