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Numbers n such that phi(n) + sigma(n) = n + reversal(n).
8

%I #10 Feb 11 2014 19:05:28

%S 1,2,3,5,7,11,101,131,151,181,191,313,353,373,383,727,757,787,797,919,

%T 929,10301,10501,10601,11311,11411,12421,12721,12821,13331,13831,

%U 13931,14341,14741,15451,15551,16061,16361,16561,16661,17471,17971,18181

%N Numbers n such that phi(n) + sigma(n) = n + reversal(n).

%C Note that all terms so far are palindromes.

%C It is obvious that if n is a term of the sequence greater than 1 then n is prime iff n is a palindrome. Do there exist composite terms in the sequence? - _Farideh Firoozbakht_, Jan 28 2006 Answer: Yes, see next comment.

%C Giovanni Resta writes (Sep 06 2006): The smallest composite number such that n+rev(n)=phi(n)+sigma(n) is n = 3197267223 = 3 * 79 * 677 * 19927 with rev(n) = 3227627913, phi(n) = 2101316256, sigma(n) = 4323578880 and 3197267223+3227627913 = 6424895136 = 2101316256+4323578880.

%H Vincenzo Librandi, <a href="/A069217/b069217.txt">Table of n, a(n) for n = 1..700</a>

%F If p is prime and rev(p)=p then p+rev(p)=2p=phi(p)+sigma(p) so all palindromic primes are in the sequence. - _Farideh Firoozbakht_, Sep 12 2006

%e phi(101) + sigma(101) = 202 = 101 + 101 = 101 + reversal(101).

%t Select[Range[5*10^4], EulerPhi[ # ] + DivisorSigma[1, # ] == # + FromDigits[Reverse[IntegerDigits[ # ]]] &]

%Y Contains composite terms, so is strictly different from A002385.

%K base,nonn

%O 1,2

%A _Joseph L. Pe_, Apr 11 2002