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%I #7 Feb 04 2014 04:18:35
%S 0,0,0,0,0,0,1,0,2,1,3,2,2,3,3,3,2,3,3,3,2,3,4,3,3,1,3,4,4,3,3,5,5,3,
%T 2,2,3,4,2,3,5,5,3,4,4,5,3,5,2,3,4,4,4,2,6,4,3,4,3,5,1,5,5,5,4,2,5,4,
%U 4,2,4,6,5,6,3,5,5,6,5,1,5,3,5,3,6,4,5,7,3,5,3,5,5,3,7,3,9,4,6,5
%N Number of ordered ways to write n = k + m with k > 0 and m > 0 such that both {3*k - 1, 3*k + 1} and {phi(m) - 1, phi(m) + 1} are twin prime pairs, where phi(.) is Euler's totient function.
%C Conjecture: (i) a(n) > 0 for all n > 8.
%C (ii) Any integer n > 6 can be written as k + m with k > 0 and m > 0 such that both {prime(k), prime(k) + 2} and {phi(m) - 1, phi(m) + 1} are twin prime pairs.
%C (iii) Each n = 12, 13, ... can be written as p + q (q > 0) with p, p + 6, phi(q) - 1 and phi(q) + 1 all prime.
%C (iv) If n > 2 is neither 10 nor 430, then n can be written as k + m with k > 0 and m > 0 such that both {3k - 1, 3*k + 1} and {6*m - 1, 6*m + 1} are twin prime pairs.
%C Note that each part of the above conjecture implies the twin prime conjecture.
%H Zhi-Wei Sun, <a href="/A237130/b237130.txt">Table of n, a(n) for n = 1..10000</a>
%e a(7) = 1 since 7 = 2 + 5 with 3*2 - 1 = 5, 3*2 + 1 =7, phi(5) - 1 = 3 and phi(5) + 1 = 5 all prime.
%e a(140) = 1 since 140 = 104 + 36 with 3*104 - 1 = 311, 3*104 + 1 = 313, phi(36) - 1 = 11 and phi(36) + 1 = 13 all prime.
%t PQ[n_]:=PrimeQ[EulerPhi[n]-1]&&PrimeQ[EulerPhi[n]+1]
%t a[n_]:=Sum[If[PrimeQ[3k-1]&&PrimeQ[3k+1]&&PQ[n-k],1,0],{k,1,n-1}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000010, A000040, A001359, A006512, A014574, A072281, A182662, A236531, A236566, A236831, A236968, A237127.
%K nonn
%O 1,9
%A _Zhi-Wei Sun_, Feb 04 2014
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