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%I #3 Mar 30 2012 18:40:50
%S 3,11,29,59,101,239,619,809,4253,5323,5923,6551,29131,37277,48341,
%T 54413,58711,60937,70537,101063,110533,214993,224603,417203,445069,
%U 466537,473867,511391,519089,534629,633449,686269,713771,741913,770767,1000537
%N Prime partial sums of Chen primes (starting with 1).
%C 43 is the first prime which is not a Chen prime, hence this sequence begins the same as prime sums of the first n primes (see A013916). The subset consisting of Chen prime partial sums of Chen primes begins a(1) = 3 = A109611(2), a(2) = 11 = A109611(5), a(3) = 29 = A109611(10), a(4) = 59 = A109611(10), a(5) = 101 = A109611(21), a(6) = 239 = A109611(40), a(7) = 809 = A109611(95). Which are the next Chen prime partial sums of Chen primes?
%F {p: p prime and for some k, p = SUM [i=1..k] {q such that q + 2 is either a prime or a semiprime} = {p: p in A000040 and p in A118482}.
%e a(7) = 1+2+3+5+7+11+13+17+19+23+29+31+37+41+47+53+59+67+71+83 = 619 is prime, which is the sum of the first 19 Chen primes (starting with 1).
%p Contribution from _R. J. Mathar_, Feb 07 2010: (Start)
%p isA001358 := proc(n) return ( numtheory[bigomega](n) = 2 ); end proc:
%p isA109611 := proc(n) isprime(n) and ( isprime(n+2) or isA001358(n+2) ); end proc:
%p A109611 := proc(n) option remember; local a; if n = 1 then 2; else a := nextprime( procname(n-1) ) ; while not isA109611(a) do a := nextprime(a) ; end do ; return a; end if; end proc:
%p A118482 := proc(n) option remember ; 1+add( A109611(j),j=1..n) ; end proc:
%p isA172102 := proc(n) if isprime(n) then for j from 1 do if A118482(j) > n then return false; elif A118482(j) = n then return true; end if; end do ; else false ; end if; end proc:
%p for n from 1 to 10000000 do if isA172102(n) then printf("%d,\n",n) ; end if; end do ; (End)
%Y Cf. A000040, A001358, A109611, A118482.
%K easy,nonn
%O 1,1
%A _Jonathan Vos Post_, Jan 25 2010
%E Extended by _R. J. Mathar_, Feb 07 2010
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