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A234309
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a(n) = |{2 < k <= n/2: 2^{phi(k)} + 2^{phi(n-k)} - 1 is prime}|, where phi(.) is Euler's totient function.
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15
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0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 6, 5, 6, 5, 7, 7, 6, 7, 7, 8, 7, 7, 6, 6, 7, 9, 9, 6, 9, 12, 8, 6, 9, 9, 9, 8, 10, 8, 9, 6, 9, 8, 8, 10, 6, 8, 11, 8, 11, 8, 7, 10, 8, 7, 8, 7, 9, 9, 11, 11, 8, 8, 9, 10, 12, 7, 12, 10, 8, 5, 7, 9, 14, 9, 9, 9, 8, 7
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OFFSET
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1,8
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 5.
(ii) For any integer n > 1, 2^k +2^{phi(n-k)} - 1 is prime for some 0 < k < n, and 2^{sigma(j)} + 2^{phi(n-j)} - 1 is prime for some 0 < j < n, where sigma(j) is the sum of all positive divisors of j.
As phi(k) is even for any k > 2, part (i) of the conjecture implies that there are infinitely many primes of the form 4^a + 4^b - 1 with a and b positive integers (cf. A234310). Note that any Mersenne prime greater than 3 has the form 2^{2a+1} - 1 = 4^a + 4^a - 1.
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LINKS
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EXAMPLE
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a(6) = 1 since 2^{phi(3)} + 2^{phi(3)} - 1 = 2^2 + 2^2 - 1 = 7 is prime.
a(7) = 1 since 2^{phi(3)} + 2^{phi(4)} - 1 = 2^2 + 2^2 - 1 = 7 is prime.
a(8) = 2 since 2^{phi(3)} + 2^{phi(5)} - 1 = 2^2 + 2^4 - 1 = 19 and 2^{phi(4)} + 2^{phi(4)} - 1 = 2^2 + 2^2 - 1 = 7 are both prime.
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MATHEMATICA
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a[n_]:=Sum[If[PrimeQ[2^(EulerPhi[k])+2^(EulerPhi[n-k])-1], 1, 0], {k, 3, n/2}]
Table[a[n], {n, 1, 100}]
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CROSSREFS
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Cf. A000010, A000040, A000079, A000668, A233307, A233390, A233542, A233544, A233547, A233566, A233867, A233918, A234200, A234246, A234308, A234310, A234337, A234344, A234346, A234347, A234359, A234360, A234361
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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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