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A233547
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a(n) = |{0 < k < n/2: phi(k)*phi(n-k) - 1 and phi(k)*phi(n-k) + 1 are both prime}|, where phi(.) is Euler's totient function.
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15
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0, 0, 0, 0, 0, 1, 2, 1, 3, 4, 3, 2, 3, 2, 3, 1, 1, 2, 1, 5, 2, 3, 1, 2, 1, 1, 3, 4, 5, 4, 3, 2, 3, 2, 5, 2, 5, 5, 3, 5, 3, 1, 5, 3, 7, 6, 3, 2, 4, 7, 5, 1, 4, 6, 6, 5, 2, 4, 6, 9, 9, 6, 8, 5, 8, 8, 6, 6, 9, 4, 8, 6, 8, 5, 7, 9, 7, 9, 5, 7, 3, 9, 5, 6, 7, 7, 10, 5, 12, 7, 5, 7, 5, 7, 5, 7, 8, 4, 7, 13
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OFFSET
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1,7
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 5.
(ii) For any n > 3, sigma(k)*phi(n-k) - 1 and sigma(k)*phi(n-k) + 1 are both prime for some 0 < k < n, where sigma(k) is the sum of all (positive) divisors of k.
(iii) For any n > 5 not equal to 35, there is a positive integer k < n such that phi(k)*phi(n-k) - 1 is a Sophie Germain prime.
Note that part (i) implies the twin prime conjecture. We have verified it for n up to 10^7.
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LINKS
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EXAMPLE
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a(6) = 1 since phi(1)*phi(5) = 1*4 = 4 with 4 - 1 and 4 + 1 twin primes.
a(8) = 1 since phi(1)*phi(7) = 1*6 = 6 with 6 - 1 and 6 + 1 twin primes.
a(16) = 1 since phi(2)*phi(14) = 1*6 = 6 with 6 - 1 and 6 + 1 twin primes.
a(17) = 1 since phi(3)*phi(14) = 2*6 = 12 with 12 - 1 and 12 + 1 twin primes.
a(19) = 1 since phi(1)*phi(18) = 1*6 = 6 with 6 - 1 and 6 + 1 twin primes.
a(23) = 1 since phi(2)*phi(21) = 1*12 = 12 with 12 - 1 and 12 + 1 twin primes.
a(25) = 1 since phi(11)*phi(14) = 10*6 = 60 with 60 - 1 and 60 + 1 twin primes.
a(26) = 1 since phi(7)*phi(19) = 6*18 = 108 with 108 - 1 and 108 + 1 twin primes.
a(42) = 1 since phi(14)*phi(28) = 6*12 = 72 with 72 - 1 and 72 +1 twin primes.
a(52) = 1 since phi(14)*phi(38) = 6*18 = 108 with 108 - 1 and 108 + 1 twin primes.
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MATHEMATICA
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TQ[n_]:=PrimeQ[n-1]&&PrimeQ[n+1]
a[n_]:=Sum[If[TQ[EulerPhi[k]*EulerPhi[n-k]], 1, 0], {k, 1, (n-1)/2}]
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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