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 A234200 a(n) = |{0 < k < n/2: k*phi(n-k) - 1 and k*phi(n-k) + 1 are both prime}|, where phi(.) is Euler's totient function. 8
 0, 0, 0, 0, 1, 2, 1, 3, 3, 2, 3, 2, 2, 3, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 4, 2, 2, 4, 4, 3, 2, 4, 4, 3, 2, 3, 7, 2, 4, 4, 3, 7, 3, 6, 5, 3, 6, 5, 4, 3, 4, 3, 7, 4, 6, 3, 3, 4, 6, 7, 3, 7, 4, 6, 8, 2, 4, 6, 7, 8, 5, 2, 2, 10, 6, 3, 7, 7, 3, 7, 6, 2, 7, 4, 2, 6, 7, 9, 8, 4, 1, 3, 2, 4, 5, 8, 10, 4, 10, 7 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,6 COMMENTS Conjecture: (i) a(n) > 0 for all n > 4. (ii) If n > 3 is different from 9 and 29, then k*sigma(n-k) - 1 and k*sigma(n-k) + 1 are both prime for some 0 < k < n. Obviously, either of the two parts implies the twin prime conjecture. We have verified part (i) for n up to 10^8. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 Zhi-Wei Sun, New representation problems involving Euler's totient function, a message to Number Theory List, Dec. 18, 2013. EXAMPLE a(5) = 1 since 2*phi(3) - 1 = 3 and 2*phi(3) + 1 = 5 are both prime. a(7) = 1 since 3*phi(4) - 1 = 5 and 3*phi(4) + 1 = 7 are both prime. a(18) = 1 since 5*phi(13) - 1 = 59 and 5*phi(13) + 1 = 61 are both prime. a(91) = 1 since 13*phi(78) - 1 = 311 and 13*phi(78) + 1 = 313 are both prime. a(101) = 1 since 6*phi(95) - 1 = 431 and 6*phi(95) + 1 = 433 are both prime. MATHEMATICA TQ[n_]:=PrimeQ[n-1]&&PrimeQ[n+1] a[n_]:=Sum[If[TQ[k*EulerPhi[n-k]], 1, 0], {k, 1, (n-1)/2}] Table[a[n], {n, 1, 100}] CROSSREFS Cf. A000010, A000040, A001359, A006512, A014574, A233542, A233544, A233547, A233566, A233567, A233867, A233918. Sequence in context: A046819 A159945 A089216 * A102746 A287618 A123143 Adjacent sequences:  A234197 A234198 A234199 * A234201 A234202 A234203 KEYWORD nonn AUTHOR Zhi-Wei Sun, Dec 21 2013 STATUS approved

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Last modified October 22 08:00 EDT 2019. Contains 328315 sequences. (Running on oeis4.)