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A172102
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Prime partial sums of Chen primes (starting with 1).
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0
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3, 11, 29, 59, 101, 239, 619, 809, 4253, 5323, 5923, 6551, 29131, 37277, 48341, 54413, 58711, 60937, 70537, 101063, 110533, 214993, 224603, 417203, 445069, 466537, 473867, 511391, 519089, 534629, 633449, 686269, 713771, 741913, 770767, 1000537
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OFFSET
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1,1
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COMMENTS
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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?
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LINKS
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FORMULA
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{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}.
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EXAMPLE
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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).
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MAPLE
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isA001358 := proc(n) return ( numtheory[bigomega](n) = 2 ); end proc:
isA109611 := proc(n) isprime(n) and ( isprime(n+2) or isA001358(n+2) ); end proc:
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:
A118482 := proc(n) option remember ; 1+add( A109611(j), j=1..n) ; end proc:
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:
for n from 1 to 10000000 do if isA172102(n) then printf("%d, \n", n) ; end if; end do ; (End)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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