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A234359
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a(n) = |{2 < k < n-2: 5^{phi(k)} + 5^{phi(n-k)/2} - 1 is prime}|, where phi(.) is Euler's totient function.
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10
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0, 0, 0, 0, 0, 1, 2, 1, 2, 4, 2, 4, 4, 3, 4, 3, 6, 5, 4, 6, 7, 8, 6, 7, 11, 7, 10, 9, 9, 7, 10, 11, 8, 7, 11, 10, 9, 6, 11, 15, 4, 14, 5, 14, 11, 13, 9, 13, 6, 12, 10, 12, 11, 10, 10, 13, 9, 7, 11, 7, 11, 4, 11, 9, 10, 6, 11, 8, 4, 10, 12, 13, 9, 7, 9, 6, 12, 10
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OFFSET
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1,7
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COMMENTS
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Conjecture: For any integer a > 1, there is a positive integer N(a) such that if n > N(a) then a^{phi(k)} + a^{phi(n-k)/2} - 1 is prime for some 2 < k < n-2. Moreover, we may take N(2) = N(3) = ... = N(6) = N(8) = 5 and N(7) = 17.
Clearly, this conjecture implies that for each a = 2, 3, ... there are infinitely many primes of the form a^{2*k} + a^m - 1, where k and m are positive integers.
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LINKS
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EXAMPLE
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a(6) = 1 since 5^{phi(3)} + 5^{phi(3)/2} - 1 = 29 is prime.
a(11) = 2 since 5^{phi(4)} + 5^{phi(7)/2} - 1 = 149 and 5^{phi(7)} + 5^{phi(4)/2} - 1 = 15629 are both prime.
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MATHEMATICA
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f[n_, k_]:=5^(EulerPhi[k])+5^(EulerPhi[n-k]/2)-1
a[n_]:=Sum[If[PrimeQ[f[n, k]], 1, 0], {k, 3, n-3}]
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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