OFFSET
1,2
COMMENTS
If k is in the sequence then m=3*(58/111*(10^(3*k)-1)-1) is a term of A072394.
Namely if k is a term of this sequence then for m=1/37*(58*10^(3*k)-169) we have sigma(m)=reversal(m)-m (see comment lines of A072394).
There is no further term up to 3000. Numbers corresponding to the larger terms are probable primes.
a(15) > 50000. - Robert Price, Oct 20 2014
MATHEMATICA
Do[If[PrimeQ[58/111*(10^(3 n) - 1) - 1], Print[n]], {n, 1874}]
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Farideh Firoozbakht, May 26 2010
EXTENSIONS
a(11)-a(14) from Robert Price, Oct 20 2014
STATUS
approved