%I #21 Jan 08 2013 11:48:12
%S 2,11,17,47,71,157,157,167,203569,203569,2803083484951
%N a(n) = smallest final prime in a chain of 2n+1 consecutive primes such that sum of the last 1, 3, 5, ..., 2n+1 terms in the sequence is also a prime.
%C Conjecture: a(n) exists for all n. (This may follow from the Green-Tao theorem, Dirichlet's theorem and Dickson's conjecture.)
%e Consider the chain of following consecutive prime numbers 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157
%e Take the sum of an odd number of primes out of this sequence starting at the end:
%e S(1) = 157
%e S(3) = 157 + 151 + 149 = 457
%e S(5) = 157 + 151 + 149 + 139 + 137 = 733
%e S(7) = 157 + 151 + 149 + 139 + 137 + 131 +127 = 991
%e S(9) = 157 + 151 + 149 + 139 + 137 + 131 +127 + 113 + 109 = 1213
%e S(11) = 157 + 151 + 149 + 139 + 137 + 131 +127 + 113 + 109 + 107 + 103 = 1423
%e All of these are prime numbers.
%e Currently a(10) is the last known term, a chain of 21 primes found after searching up to 4*10^13. The 21 consecutive primes are 2803083484321, 2803083484343, 2803083484349, 2803083484363, 2803083484391, 2803083484429, 2803083484499, 2803083484507, 2803083484633, 2803083484637, 2803083484639, 2803083484673, 2803083484697, 2803083484703, 2803083484763, 2803083484777, 2803083484781, 2803083484819, 2803083484921, 2803083484937, 2803083484951, where the sums S(21), S(19), S(17), S(15) . . . . to S(1): 58864753177133, 53258586208469, 47652419239757, 42046252270937, 36440085301931, 30833918332661, 25227751363349, 19621584393949, 14015417424409, 8409250454809, 2803083484951 respectively are also primes.
%t (* This program is not convenient for n > 9 *) run[m_, n_] := Prime /@ Range[m + 2n, m, -1]; ok[ru_List] := (test = True; For[k = 1, k <= Length[ru], k = k+2, s = Total[ru[[1 ;; k]]]; If[! PrimeQ[s], test = False; Break[]]]; test); a[n_] := a[n] = Catch[For[m = 1, m <= 10^5, m++, r = run[m, n]; If[ok[r ], Throw[r[[1]]]]]]; Table[Print[a[n]]; a[n], {n, 0, 9}] (* _Jean-François Alcover_, Jan 08 2013 *)
%Y Cf. A182121.
%K nonn
%O 0,1
%A _Vikram Pandya_, Jan 04 2013