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 A235917 a(n) = |{0 < k < n - 2: p = prime(k) + phi(n-k)/2, p^2 - 1 - prime(p) and (p^2 - 1)/2 - prime(p) are all prime}|, where phi(.) is Euler's totient function. 2
 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 3, 1, 4, 2, 4, 3, 1, 2, 2, 3, 3, 1, 2, 2, 1, 1, 2, 4, 1, 5, 2, 2, 3, 2, 6, 2, 1, 3, 3, 2, 4, 5, 4, 2, 5, 3, 4, 2, 3, 4, 4, 3, 3, 2, 1, 4, 3, 2, 3, 4, 5, 7, 3, 5, 1, 6, 1, 7, 3, 6, 5, 3, 5, 2, 3, 4, 5, 3, 8, 6, 4, 2, 6, 4, 8, 3, 7, 5, 6, 6, 4, 3, 5, 6, 4, 3 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,12 COMMENTS Conjecture: (i) a(n) > 0 for all n > 14. (ii) For any integer n > 12, there is a positive integer k < n such that p = prime(k) + phi(n-k), (p^2 - 1)/2 - prime(p) and (p^2 - 1)/4 - prime(p) are all prime. Clearly, part (i) of the conjecture implies that there are infinitely many primes p with p^2 - 1 - prime(p) and (p^2 - 1)/2 - prime(p) both prime. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 EXAMPLE a(10) = 1 since prime(5) + phi(5)/2 = 11 + 2 = 13, 13^2 - 1 - prime(13) = 168 - 41 = 127 and (13^2 - 1)/2 - prime(13) = 84 - 41 = 43 are all prime. a(71) = 1 since prime(19) + phi(52)/2 = 67 + 12 = 79, 79^2 - 1 - prime(79) = 6240 - 401 = 5839 and (79^2 - 1)/2 - prime(79) = 3120 - 401 = 2719 are all prime. MATHEMATICA PQ[n_]:=n>0&&PrimeQ[n] p[n_]:=PrimeQ[n]&&PQ[n^2-1-Prime[n]]&&PQ[(n^2-1)/2-Prime[n]] f[n_, k_]:=Prime[k]+EulerPhi[n-k]/2 a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n-3}] Table[a[n], {n, 1, 100}] CROSSREFS Cf. A000010, A000040, A234694, A235912. Sequence in context: A089990 A071427 A248813 * A093949 A324518 A108807 Adjacent sequences: A235914 A235915 A235916 * A235918 A235919 A235920 KEYWORD nonn AUTHOR Zhi-Wei Sun, Jan 17 2014 STATUS approved

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Last modified May 28 06:10 EDT 2023. Contains 362992 sequences. (Running on oeis4.)