

A135238


Numbers n such that phi(sigma(n)) = reversal(n).


2




OFFSET

1,2


COMMENTS

If both numbers 10^m3 & 5*10^(m1)1 are primes and n=3*(10^m3) then phi(sigma(n))=reversal(n), namely n is in the sequence (the proof is easy). Conjecture: n=2991 is the only such term of the sequence. there is no further term up to 35*10^7.
There are no other terms up to 10^10.  Donovan Johnson, Oct 24 2013
If p and 2*p1 are primes, where p = 3900000*100^t + 108900*10^t + 109, then 6*p3 is in the sequence. This happens at least for t=1 (2346534651), t=2 (234065340651), t=11, and t=76.  Giovanni Resta, Aug 09 2019


LINKS

Table of n, a(n) for n=1..9.


EXAMPLE

phi(sigma(880374)) = phi(1920960) = 473088 = reversal(880374), so 880374 is in the sequence.


MATHEMATICA

reversal[n_]:=FromDigits[Reverse[IntegerDigits[n]]]; Do[If[EulerPhi[DivisorSigma[1, n]]==reversal[n], Print[n]], {n, 350000000}]


PROG

(PARI) isok(n) = eulerphi(sigma(n)) == fromdigits(Vecrev(digits(n))); \\ Michel Marcus, Aug 09 2019


CROSSREFS

Cf. A071525, A000010, A000203.
Sequence in context: A054874 A174736 A324567 * A133376 A179056 A160814
Adjacent sequences: A135235 A135236 A135237 * A135239 A135240 A135241


KEYWORD

nonn,base,more


AUTHOR

Farideh Firoozbakht, Dec 26 2007


EXTENSIONS

a(7) from Donovan Johnson, Oct 24 2013
a(8)a(9) from Giovanni Resta, Aug 09 2019


STATUS

approved



