%I #24 Feb 22 2014 10:09:08
%S 21,291,2991,43954,211552,439954,43564354,43999954,464057476,
%T 43999999954,435600004354,439560043954,439999999954
%N Numbers n such that sigma(n) = reversal(n+2).
%C If p=10^m-3 is prime then 3*p is in the sequence.
%C Let f(m,r)=22*((10^m-1)/(10^(m+2)-1))*(10^((m+2)r)-1)-1. If m>1 and p=f(m,r) is prime, then 2*p is in the sequence. a(11) > 2*10^11. - _Giovanni Resta_, Dec 13 2013
%C According to comments and known terms of A230768, the next three terms are 435600004354, 439560043954, 439999999954 and a(14)>10^12.
%C Let g(m,r,s) = 22*(10^(m+r+4)*(10^(m+2)-1)*(10^(s*(m+r+4))-1)/(10^(m+r+4)-1)+10^(m+2)-1) - 1 = [21.9(m).78.0(r)](s).21.9(m).77, where m, r and s are nonnegative integers and "." means concatenation and x(y) means x repreated y times.
%C If p=g(m,r,s) is prime, then 2*p is in the sequence.
%C This generalizes the above f(m,r) = [21.9(m-2).78](r)-1 = g(m-2,0,r-1), for r>0. Also a(4)=2*g(1,0,0), a(6)=2*g(2,0,0), a(7)=2*g(0,0,1), a(8)=2*g(4,0,0), a(10)=2*g(7,0,0), a(11)=2*g(0,4,1), a(12)=2*g(1,2,1) and a(13)=2*g(8,0,0). - _Farideh Firoozbakht_, Feb 14 2014
%e sigma(21)=32=reversal(23)=reversal(21+2).
%t Do[If[c=FromDigits[Reverse[IntegerDigits[2+n]]]; c>n && DivisorSigma[1,n] == c, Print[n]], {n, 44000000}]
%Y Cf. A000203, A004086, A230009, A230768.
%K nonn,base,more
%O 1,1
%A _Farideh Firoozbakht_, Dec 06 2013
%E a(9)-a(10) from _Giovanni Resta_, Dec 13 2013
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