A234200 - OEIS

%I #12 Dec 21 2013 09:46:54

%S 0,0,0,0,1,2,1,3,3,2,3,2,2,3,2,2,2,1,3,2,2,2,2,2,4,2,2,4,4,3,2,4,4,3,

%T 2,3,7,2,4,4,3,7,3,6,5,3,6,5,4,3,4,3,7,4,6,3,3,4,6,7,3,7,4,6,8,2,4,6,

%U 7,8,5,2,2,10,6,3,7,7,3,7,6,2,7,4,2,6,7,9,8,4,1,3,2,4,5,8,10,4,10,7

%N a(n) = |{0 < k < n/2: k*phi(n-k) - 1 and k*phi(n-k) + 1 are both prime}|, where phi(.) is Euler's totient function.

%C Conjecture: (i) a(n) > 0 for all n > 4.

%C (ii) If n > 3 is different from 9 and 29, then k*sigma(n-k) - 1 and k*sigma(n-k) + 1 are both prime for some 0 < k < n.

%C Obviously, either of the two parts implies the twin prime conjecture. We have verified part (i) for n up to 10^8.

%H Zhi-Wei Sun, <a href="/A234200/b234200.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;29f53124.1312">New representation problems involving Euler's totient function</a>, a message to Number Theory List, Dec. 18, 2013.

%e a(5) = 1 since 2*phi(3) - 1 = 3 and 2*phi(3) + 1 = 5 are both prime.

%e a(7) = 1 since 3*phi(4) - 1 = 5 and 3*phi(4) + 1 = 7 are both prime.

%e a(18) = 1 since 5*phi(13) - 1 = 59 and 5*phi(13) + 1 = 61 are both prime.

%e a(91) = 1 since 13*phi(78) - 1 = 311 and 13*phi(78) + 1 = 313 are both prime.

%e a(101) = 1 since 6*phi(95) - 1 = 431 and 6*phi(95) + 1 = 433 are both prime.

%t TQ[n_]:=PrimeQ[n-1]&&PrimeQ[n+1]

%t a[n_]:=Sum[If[TQ[k*EulerPhi[n-k]],1,0],{k,1,(n-1)/2}]

%t Table[a[n],{n,1,100}]

%Y Cf. A000010, A000040, A001359, A006512, A014574, A233542, A233544, A233547, A233566, A233567, A233867, A233918.

%K nonn

%O 1,6

%A _Zhi-Wei Sun_, Dec 21 2013