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%I #15 Jun 26 2015 04:47:51
%S 5602,42346,184650339,356930335,453038125082
%N Numbers n such that sigma(n+1) = reverse(sigma(n)).
%C a(6) > 10^13. - _Giovanni Resta_, Jun 26 2015
%e sigma(5602 + 1) = 6048 = reverse(8406) = reverse(sigma(5602)), so 5602 is a term of the sequence.
%t Do[ If[FromDigits[Reverse[IntegerDigits[DivisorSigma[1, n]]]] == DivisorSigma[1, n + 1], Print[n]], {n, 1, 10^7}]
%o (PARI) isok(n) = digits(sigma(n+1)) == Vecrev(digits(sigma(n))); \\ _Michel Marcus_, Jun 26 2015
%Y Cf. A028980 (sigma(n) = reverse(sigma(n))).
%K base,more,nonn
%O 1,1
%A _Joseph L. Pe_, Sep 19 2002
%E a(3)-a(4) from _Donovan Johnson_, Feb 01 2009
%E a(5) from _Giovanni Resta_, Jun 26 2015
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