OFFSET
1,1
COMMENTS
Initial terms {2,3,5} and {3,5,7} give no prime, hence we start with a(1..3) = {5, 7, 11}.
Is the sequence infinite?
The Mathematica program is fast because it uses that fact that 7 must be one of the three numbers in the sum. Note that p = 5 (mod 6) for all terms but 7. - T. D. Noe, May 24 2013
LINKS
Zak Seidov and T. D. Noe, Table of n, a(n) for n = 1..500 (first 123 from Zak Seidov)
MATHEMATICA
t = {5, 11}; Do[u = Subsets[t, {2}]; w = Reverse[Sort[7 + Total /@ u]]; s = Select[w, PrimeQ, 1]; mx = s[[1]]; AppendTo[t, mx], {97}]; t = Insert[t, 7, 2] (* T. D. Noe, May 24 2013 *)
PROG
(PARI) N=50; v=vector(N); v[1]=5; v[2]=7; v[3]=11; n=3; while(n<N, m=0; for(i=1, n, for(j=i+1, n, for(k=j+1, n, t=v[i]+v[j]+v[k]; if(t>m && isprime(t), m=t)))); if(m>0, n++; v[n]=m)); v /* from Ralf Stephan */
CROSSREFS
KEYWORD
nonn
AUTHOR
Zak Seidov, May 23 2013
STATUS
approved