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A230194
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Number of ways to write n = x + y + z (x, y, z > 0) such that all the 11 integers 6*x-1, 6*x+1, 6*x-5, 6*x+5, 6*y-1, 6*y-5, 6*y+5, 6*(x+y)+5, 6*z-1, 6*z-5 and 6*z+5 are prime.
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1
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0, 0, 0, 0, 0, 1, 1, 3, 3, 3, 4, 3, 6, 5, 3, 3, 3, 5, 4, 4, 4, 2, 9, 10, 9, 7, 5, 12, 8, 2, 8, 6, 6, 7, 9, 4, 3, 10, 11, 2, 4, 6, 10, 9, 11, 9, 4, 10, 17, 9, 1, 4, 7, 6, 6, 6, 2, 5, 14, 13, 7, 5, 14, 6, 3, 5, 4, 12, 11, 14, 5, 2, 16, 11, 5, 9, 6, 8, 11, 23, 15, 3, 23, 18, 17, 9, 8, 20, 5, 10, 14, 3, 14, 15, 16, 9, 8, 24, 10, 7
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OFFSET
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1,8
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COMMENTS
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Conjecture: a(n) > 0 for all n > 5.
Let S be the set of those primes p with p-4 and p+6 also prime. Since each element of S has the form 6*k-1 with k > 0, the conjecture implies that 6*n-3 with n > 5 can be expressed as a sum of three primes in the set S. If n = x + y + z, then 6*n = (6*(x+y)+5) + (6*z-5). So a(n) > 0 implies Goldbach's conjecture for the even number 6*n.
Let T be the set of those primes p with p+4 and p-6 also prime. Clearly each element of T has the form 6*k+1 with k > 0. We conjecture that 6*n+3 with n > 5 can be expressed as a sum of three primes in the set T.
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LINKS
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EXAMPLE
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a(30) = 2 since 30 = 2 + 14 + 14 = 18 + 4 + 8, and 6*2-1 = 11, 6*2+1 = 13, 6*2-5 = 7, 6*2+5 = 17, 6*14-1 = 83, 6*14-5 = 79, 6*14+5 = 89, 6*(2+14)+5 = 101, 6*18-1 = 107, 6*18+1 = 109, 6*18-5 = 103, 6*18+5 = 113, 6*4-1 = 23, 6*4-5 = 19, 6*4+5 = 29, 6*(18+4)+5 = 137, 6*8-1 = 47, 6*8-5 = 43 and 6*8+5 = 53 are all prime.
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MATHEMATICA
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SQ[n_]:=PrimeQ[6n-1]&&PrimeQ[6n-5]&&PrimeQ[6n+5]
a[n_]:=Sum[If[SQ[i]&&PrimeQ[6i+1]&&SQ[j]&&PrimeQ[6(i+j)+5]&&SQ[n-i-j], 1, 0], {i, 1, n-2}, {j, 1, n-1-i}]
Table[a[n], {n, 1, 100}]
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CROSSREFS
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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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