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 A236074 a(n) = |{0 < k < n: p = phi(k) + phi(n-k)/6 + 1, prime(2*p) - 2*prime(p) and prime(p) - 2*prime((p-1)/2) are all prime}|, where phi(.) is Euler's totient function. 4
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 2, 1, 1, 2, 0, 1, 0, 0, 0, 0, 2, 2, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 4, 0, 0, 1, 2, 0, 1, 2, 1, 0, 3, 4, 0, 0, 0, 2, 1, 3, 1, 2, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,29 COMMENTS Conjecture: (i) a(n) > 0 for all n > 116. (ii) For any integer n > 196, there is a positive integer k < n such that p = phi(k) + phi(n-k)/6 + 1, prime(2*p) - 2*prime(p) and prime(p-1) - 2*prime((p-1)/2) are all prime. Clearly, part (i) (or part (ii)) implies that there are infinitely many odd primes p with prime(2*p) - 2*prime(p) and prime(p) - 2*prime((p-1)/2) (or prime(p-1) - 2*prime((p-1)/2), resp.) both prime. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 EXAMPLE a(30) = 1 since phi(4) + phi(26)/6 + 1 = 5, prime(2*5) - 2*prime(5) = 29 - 2*11 = 7 and prime(5) - 2*prime((5-1)/2) = 11 - 2*3 = 5 are all prime. a(204) = 1 since phi(159) + phi(45)/6 + 1 = 109, prime(2*109) - 2*prime(109) = 1361 - 2*599 = 163, and prime(109) - 2*prime((109-1)/2) = 599 - 2*251 = 97 are all prime. MATHEMATICA PQ[n_]:=PQ[n]=n>0&&PrimeQ[n] p[n_]:=PQ[n]&&PQ[Prime[2n]-2Prime[n]]&&PQ[Prime[n]-2*Prime[(n-1)/2]] f[n_, k_]:=EulerPhi[k]+EulerPhi[n-k]/6+1 a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n-1}] Table[a[n], {n, 1, 100}] CROSSREFS Cf. A000010, A000040, A234694, A235924, A236075. Sequence in context: A051778 A057554 A060575 * A099916 A099917 A137412 Adjacent sequences: A236071 A236072 A236073 * A236075 A236076 A236077 KEYWORD nonn AUTHOR Zhi-Wei Sun, Jan 19 2014 STATUS approved

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Last modified December 8 09:53 EST 2023. Contains 367663 sequences. (Running on oeis4.)