login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

a(n) = phi(n - phi(n) + 3).
2

%I #24 Sep 08 2022 08:45:56

%S 2,2,2,4,2,6,2,6,2,6,2,10,2,10,4,10,2,8,2,8,4,8,2,18,4,16,4,18,2,20,2,

%T 18,8,12,6,18,2,22,6,18,2,20,2,18,8,18,2,24,4,20,10,30,2,24,6,24,8,20,

%U 2,46,2,24,8,24,8,42,2,24,12,42,2,32,2,40,18,42

%N a(n) = phi(n - phi(n) + 3).

%C This is the Euler Phi function of 3 more than the Cototient of n.

%C If n is noncomposite, a(n) = 2. Proof: For n = 1, phi(1 - phi(1) + 3) = phi(1-1+3) = phi(3) = 2. For n = p, phi(p - phi(p) + 3) = phi(p - (p-1) + 3) = phi(4) = 2.

%C If n is the product of twin primes, a(n) is the arithmetic mean of the prime factors. Equivalently, when n is the product of twin primes, a(n) +- 1 represents the largest and the smallest prime factors of n respectively.

%H Antti Karttunen, <a href="/A186749/b186749.txt">Table of n, a(n) for n = 1..65537</a>

%F a(n) = phi(n - phi(n) + 3) = A000010(n - A000010(n) + 3) = A000010(A051953(n) + 3).

%e a(15) = 4, Since phi(15 - phi(15) + 3) = 4. Note that 15 is the product of twin primes and that a(15) = 4 is the arithmetic mean of the prime factors of 15: (3+5)/2 = 4.

%p with(numtheory); seq( phi(k - phi(k) + 3), k=1..70);

%t Table[EulerPhi[n - EulerPhi[n] + 3], {n, 100}]

%o (Magma) [EulerPhi(n-EulerPhi(n)+3): n in [1..100]]; // _Vincenzo Librandi_, Dec 08 2015

%o (PARI) A186749(n) = eulerphi(n - eulerphi(n) + 3); \\ _Antti Karttunen_, Mar 04 2018

%o (GAP) List([1..70],n->Phi(n-Phi(n)+3)); # _Muniru A Asiru_, Mar 04 2018

%Y Cf. A000010, A051953.

%K nonn

%O 1,1

%A _Wesley Ivan Hurt_, Aug 29 2013