OFFSET
0,2
COMMENTS
Conjecture: a(n) > 0 for all n. This would follow from a form of Goldbach's (binary) conjecture. Checked up to 10^7; largest term in that range is a(9972987) = 4178506411.
Pomerance proves that x = n (mod phi(x)) has at least two solutions for each n, but this allows x < n and so does not prove the conjecture above.
a(n) > 0 for all n <= 10^9. The largest term in that range is a(990429171) = 1050844225771. - Donovan Johnson, Feb 18 2014
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
Carl Pomerance, On the congruences σ(n) ≡ a (mod n) and n ≡ a (mod φ(n)), Acta Arithmetica 26:3 (1974-1975), pp. 265-272.
MATHEMATICA
A234642[n_]:=NestWhile[# + 1 &, 1, Not[Mod[#, EulerPhi[#]] == n] &] (* JungHwan Min, Dec 23 2015 *)
A234642[n_]:=Catch[Do[If[Mod[k, EulerPhi[k]] == n, Throw[k]], {k, Infinity}]] (* JungHwan Min, Dec 23 2015 *)
xmp[n_]:=Module[{x=1}, While[Mod[x, EulerPhi[x]]!=n, x++]; x]; Array[xmp, 60, 0] (* Harvey P. Dale, Jan 04 2016 *)
PROG
(PARI) a(n)=my(k=n); while(k++%eulerphi(k)!=n, ); k
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Charles R Greathouse IV, Dec 28 2013
STATUS
approved