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A218754
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Number of ways to write n=p+q(3+(-1)^n)/2 with q<=n/2 and p, q, p^2+3pq+q^2 all prime.
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13
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0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 2, 2, 1, 3, 1, 3, 1, 1, 2, 1, 0, 3, 3, 2, 3, 3, 0, 3, 0, 3, 2, 1, 1, 4, 1, 2, 2, 1, 2, 0, 2, 2, 2, 3, 0, 4, 1, 1, 2, 0, 1, 2, 3, 5, 0, 2, 1, 3, 4, 1, 1, 2, 2, 6, 2, 2, 4, 1, 2, 3, 2, 3, 3, 3, 2, 4, 1, 2, 5, 0, 3, 4, 2, 3, 4, 3, 1, 4, 3
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OFFSET
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1,17
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COMMENTS
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Conjecture: a(n)>0 for all n>=1188.
This conjecture is stronger than both Goldbach's conjecture and Lemoine's conjecture.
Zhi-Wei Sun also made the following conjecture: Given any positive odd integer d, there is a prime p(d) such that for any prime p>p(d) there is a prime q<p such that p^2+dpq+q^2 is prime. For example, we may take p(1)=5, p(3)=2, p(5)=61, p(7)=3, p(9)=13, p(11)=7, p(13)=3, p(15)=163, p(17)=13, p(19)=5, p(21)=p(23)=2, p(25)=89, p(27)=3, p(29)=53.
Conjecture verified for d up to 100 and p up to 10^7. - Mauro Fiorentini, Sep 23 2023
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LINKS
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EXAMPLE
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For n=72 we have a(72)=1 since the only primes p and q with p+q=72, q<=36 and p^2+3pq+q^2 prime are p=67 and q=5.
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MATHEMATICA
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a[n_]:=a[n]=Sum[If[PrimeQ[q]==True&&PrimeQ[n-q(3-(-1)^n)/2]&&PrimeQ[q^2+3q(n-q(3-(-1)^n)/2)+(n-q(3-(-1)^n)/2)^2]==True, 1, 0], {q, 1, n/2}]
Do[Print[n, " ", a[n]], {n, 1, 20000}]
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CROSSREFS
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Cf. A000034 = 1,2,1,2,... = (3-(-1)^n)/2. (Note: Offset shifted w.r.t. use in the definition of this sequence.) - M. F. Hasler, Nov 05 2012
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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