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 A236968 Number of ordered ways to write n = k + m with k > 0 and m > 0 such that 6*k - 1, 6*k + 1 and k + phi(m) are all prime, where phi(.) is Euler's totient function. 3
 0, 1, 2, 2, 1, 2, 2, 3, 3, 1, 4, 4, 3, 5, 3, 1, 1, 4, 5, 6, 3, 1, 4, 4, 3, 2, 2, 3, 3, 5, 3, 6, 5, 1, 6, 1, 4, 6, 4, 1, 6, 7, 8, 6, 2, 2, 5, 8, 4, 4, 3, 3, 7, 8, 3, 5, 3, 4, 6, 7, 8, 9, 5, 2, 3, 2, 4, 7, 5, 2, 2, 6, 6, 8, 5, 1, 6, 2, 6, 7, 3, 3, 8, 8, 6, 5, 2, 5, 6, 9, 9, 5, 4, 1, 7, 2, 3, 9, 6, 3 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Conjecture: (i) a(n) > 0 for all n > 1. Also, any n > 12 can be written as k + m (k > 0 and m > 2) with 6*k - 1, 6*k + 1 and k + phi(m)/2 all prime. (ii)  Each integer n > 34 can be written as p + q (q > 0) with p and p + phi(q) both prime. Also, any integer n > 14 can be written as p + q (q > 2) with p, p + 6 and p + phi(q)/2 all prime. Clearly, part (i) of the conjecture implies that any integer n > 1 can be written as p + m - phi(m), where p is a prime and m is a positive integer. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 EXAMPLE a(17) = 1 since 17 = 7 + 10 with 6*7 - 1 = 41, 6*7 + 1 = 43 and 7 + phi(10) = 7 + 4 = 11 all prime. a(486) = 1 since 486 = 325 + 161 with 6*325 - 1 = 1949, 6*325 + 1 = 1951 and 325 + phi(161) = 325 + 132 = 457 all prime. MATHEMATICA p[n_, k_]:=PrimeQ[6k-1]&&PrimeQ[6k+1]&&PrimeQ[k+EulerPhi[n-k]] a[n_]:=Sum[If[p[n, k], 1, 0], {k, 1, n-1}] Table[a[n], {n, 1, 100}] CROSSREFS Cf. A000010, A000040, A001359, A002822, A006512, A182662, A199920, A236531, A236831. Sequence in context: A057536 A245574 A245573 * A265744 A245588 A014420 Adjacent sequences:  A236965 A236966 A236967 * A236969 A236970 A236971 KEYWORD nonn AUTHOR Zhi-Wei Sun, Feb 02 2014 STATUS approved

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Last modified October 15 08:45 EDT 2018. Contains 316210 sequences. (Running on oeis4.)