A132793 -id:A132793 - cp's OEIS frontend

A132794 Numbers n such that sigma(phi(n)) -phi(n) -1 = phi(sigma(n) -n -1).

8, 16, 64, 256, 16384, 262144, 1048576, 4294967296

Offset: 1

Paolo P. Lava and Giorgio Balzarotti, Aug 31 2007

Used sigma(n)-n-1, namely the sum of proper divisors minus 1.
a(8) > 10^8. - Michel Marcus, Nov 01 2014
Every 2^(A000043+1) is a term. Proof sketch: Let ch=A048050 and n=2^k, then ch(phi(2^k))=phi(ch(2^k)), ch(2^(k-1))=phi(2^k-2), 2^(k-1)-2=phi(2^(k-1)-1), since phi(prime)=prime-1 the condition is satisfied by every k=A000043+1 or n=2^(A000043+1). See link. - Jon Maiga, Dec 14 2018
Conjecture: a(n)=2^(A000043(n)+1), if true the next terms are: 4294967296, 4611686018427387904, 1237940039285380274899124224... - Jon Maiga, Dec 14 2018
a(9) > 6.5*10^11. - Giovanni Resta, Dec 01 2019

Jon Maiga, Euler Phi, Chowla and Mersenne primes, 2018.

Cf. A000010, A000043, A000203, A001229, A018784, A033632, A048050, A132793.

Filtered([4..1000000],n->Sigma(Phi(n))-Phi(n)-1=Phi(Sigma(n)-n-1)); # Muniru A Asiru, Dec 16 2018

[n: n in [2..30000] | DivisorSigma(1,n) ne n+1 and DivisorSigma(1, EulerPhi(n)) - EulerPhi(n) - 1 eq EulerPhi(DivisorSigma(1,n) - n -1) ]; // G. C. Greubel, Dec 13 2018

with(numtheory); P:=proc(n) local a,i; for i from 1 to n do
a:=phi(sigma(i)-i-1); if a>0 then
if sigma(phi(i))-phi(i)-1=a then print(i);
fi; fi; od; end: P(10^7);

ch[n_]:=DivisorSigma[1,n]-n-1
test[n_]:=ch[n]!=0 && ch[EulerPhi[n]] == EulerPhi[ch[n]]
Flatten[Position[Range[300000], Integer_ ? test]] (* Jon Maiga, Dec 14 2018 *)

isok(n) = ((s=(sigma(n)-n-1)) != 0) && (sigma(eulerphi(n))-eulerphi(n)-1 == eulerphi(s)); \\ Michel Marcus, Nov 01 2014

a(1) corrected and a(6)-a(7) from Michel Marcus, Nov 01 2014

a(8) from Giovanni Resta, Dec 01 2019

cp's OEIS Frontend

A132794 Numbers n such that sigma(phi(n)) -phi(n) -1 = phi(sigma(n) -n -1).

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