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A209320
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Number of ways to write 2n = p+q with p and q both prime, p+1 and q-1 both practical.
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9
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0, 0, 1, 2, 3, 2, 2, 2, 2, 3, 4, 5, 3, 2, 3, 3, 5, 7, 3, 3, 4, 4, 5, 8, 4, 3, 5, 2, 4, 8, 3, 4, 6, 2, 4, 7, 3, 4, 7, 2, 4, 9, 4, 4, 9, 5, 3, 9, 3, 5, 8, 3, 4, 10, 4, 6, 8, 5, 4, 14, 2, 4, 8, 2, 6, 10, 4, 4, 7, 4, 4, 10, 5, 4, 8, 3, 4, 9, 5, 5, 7, 3, 3, 13, 6, 5, 7, 4, 2, 11, 5, 5, 9, 4, 2, 9, 3, 6, 10, 7
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OFFSET
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1,4
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COMMENTS
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Conjecture: a(n)>0 for all n>2.
As p+q=(p+1)+(q-1), this unifies Goldbach's conjecture and its analog involving practical numbers.
The conjecture has been verified for n up to 10^7.
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LINKS
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EXAMPLE
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a(8) = 2 since 2*8 = 3+13 = 11+5 with 3, 5, 11, 13 all prime and 3+1, 13-1, 11+1, 5-1 all practical.
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MATHEMATICA
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f[n_]:=f[n]=FactorInteger[n]
Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2])
Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}]
pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0)
a[n_]:=a[n]=Sum[If[PrimeQ[2n-Prime[k]]==True&&pr[Prime[k]+1]==True&&pr[2n-Prime[k]-1]==True, 1, 0], {k, 1, PrimePi[2n-2]}]
Do[Print[n, " ", a[n]], {n, 1, 100}]
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CROSSREFS
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Cf. A005153, A002372, A045917, A208243, A208244, A208246, A208249, A209253, A209254, A209312, A219185, A219312, A219315.
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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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