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A236138
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a(n) = |{0 < k < n: p = phi(k) + phi(n-k)/3 - 1, prime(p-1) - (p-1) and prime(p-1) - 2*prime((p-1)/2) are all prime}|, where phi(.) is Euler's totient function.
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3
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 3, 0, 1, 3, 1, 4, 0, 5, 2, 2, 3, 4, 1, 4, 2, 5, 3, 1, 6, 4, 3, 0, 4, 5, 3, 3, 4, 5, 2, 4, 2, 2, 4, 3, 4, 1, 2, 2, 3, 5, 3, 0, 3, 2, 4, 1, 2, 2, 4, 0, 4, 1, 3, 3, 2, 0, 4, 1, 3, 2, 3, 1, 5, 3, 5, 1, 4, 2, 3, 5, 4, 4, 5, 4, 1, 2, 2, 3, 3, 7, 3, 2, 3
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OFFSET
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1,16
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 73.
(ii) For any integer n > 472, there is a positive integer k < n such that p = phi(k) + phi(n-k)/4 - 1, prime(p) - 2*prime((p-1)/2) and prime(p) - 2*prime((p+1)/2) are all prime.
Clearly, part (i) of the conjecture implies that there are infinitely many odd primes p with prime(p-1) - (p-1) and prime(p-1) - 2*prime((p-1)/2) both prime, and part (ii) implies that there are infinitely many odd primes p with prime(p) - 2*prime((p-1)/2) and prime(p) - 2*prime((p+1)/2) both prime.
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LINKS
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EXAMPLE
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a(20) = 1 since phi(11) + phi(9)/3 - 1 = 11, prime(10) - 10 = 29 - 10 = 19 and prime(10) - 2*prime(5) = 29 - 2*11 = 7 are all prime.
a(293) = 1 since phi(267) + phi(26)/3 - 1 = 176 + 12/3 - 1 = 179, prime(178) - 178 = 1061 - 178 = 883 and prime(178) - 2*prime(89) = 1061 - 2*461 = 139 are all prime.
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MATHEMATICA
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PQ[n_]:=n>0&&PrimeQ[n]
p[n_]:=n>2&&PrimeQ[n]&&PrimeQ[Prime[n-1]-(n-1)]&&PQ[Prime[n-1]-2*Prime[(n-1)/2]]
f[n_, k_]:=EulerPhi[k]+EulerPhi[n-k]/3-1
a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n-1}]
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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