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A235728
a(n) = |{0 < k < n - 2: 2*m + 1, m*(m+1) - prime(m) and m*(m+1) + prime(m) are all prime with m = phi(k) + phi(n-k)/2}|, where phi(.) is Euler's totient function.
4
0, 0, 0, 0, 0, 2, 3, 2, 4, 4, 4, 4, 4, 5, 3, 6, 4, 7, 4, 3, 6, 5, 6, 4, 7, 4, 7, 3, 5, 5, 5, 6, 6, 6, 3, 6, 3, 4, 2, 2, 4, 3, 4, 5, 4, 3, 6, 4, 2, 4, 2, 4, 3, 3, 6, 4, 2, 6, 8, 6, 10, 4, 6, 7, 4, 6, 6, 8, 6, 6, 2, 9, 5, 9, 10, 12, 4, 10, 6, 10, 6, 9, 5, 11, 10, 7, 10, 10, 6, 9, 11, 7, 8, 8, 13, 6, 5, 5, 6, 9
OFFSET
1,6
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 5, and a(n) = 1 only for n = 191.
(ii) If n > 8 is not equal to 32, then there is a positive integer k < n - 2 such that 2*m + 1, m*(m+1) - prime(m) and m*(m+1) + prime(m) are all prime, where m = sigma(k) + phi(n-k)/2, and sigma(k) is the sum of all positive divisors of k.
(iii) If n > 444, then there is a positive integer k < n such that 2*m + 1, m^2 - prime(m) and m^2 + prime(m) are all prime, where m = sigma(k) + phi(n-k).
Clearly, part (i) of the conjecture implies that there are infinitely many odd primes p = 2*m + 1 with m*(m+1) - prime(m) = (p^2-1)/4 - prime((p-1)/2) and m*(m+1) + prime(m) = (p^2-1)/4 + prime((p-1)/2) both prime.
EXAMPLE
a(6) = 2 since phi(1) + phi(5)/2 = phi(3) + phi(3)/2 = 3 with 2*3 + 1 = 7, 3*4 - prime(3) = 7 and 3*4 + prime(3) = 17 all prime.
a(191) = 1 since phi(153) + phi(38)/2 = 105 with 2*105 + 1 = 211, 105*106 - prime(105) = 11130 - 571 = 10559 and 105*106 + prime(105) = 11130 + 571 = 11701 all prime.
MATHEMATICA
q[n_]:=PrimeQ[2n+1]&&PrimeQ[n(n+1)-Prime[n]]&&PrimeQ[n(n+1)+Prime[n]]
f[n_, k_]:=EulerPhi[k]+EulerPhi[n-k]/2
a[n_]:=Sum[If[q[f[n, k]], 1, 0], {k, 1, n-3}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 15 2014
STATUS
approved