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%I #24 Dec 17 2013 16:29:03
%S 62,206,6002,6439562,60000002,6004356002,29355232298,60000000002,
%T 292369967108,600439956002,643956439562,6303723993362
%N Numbers n such that sigma(n) = reversal(n+7).
%C If p=3*10^m+1 is prime then 2*p is in the sequence.
%C Also if p=322*10^m-219 is prime then 2*p is in the sequence. 206, 6002 and 6439562 are of this form. It is interesting that 6002 = 2*(3*10^3+1) = 2*(322*10^1-219) is of both two forms.
%C Let f(r,s,t) = 3*10^(t*(s+4)+2*r+1)+10^(r+1)*(22*(10^(s+2)-1)*(10^(t*(s+4))-1)/ (10^(s+4)-1))+1, if p=f(r,s,t) is prime and (i) r, s and t are nonnegative integers or (ii) s=t=0 and 2r is a positive integer or (iii) r=t=0 and s=-3, then 2*p is in the sequence. Note, very interestingly, this pattern is a generalization of the previous two forms mentioned above. In fact we have 3*10^(m+1)+1 = f(m/2,s,0) and 322*10^m-219 = f(0,m-3,1). - _Farideh Firoozbakht_, Dec 17 2013
%F a(1)=2*f(0,s,0), a(2)=2*f(0,-3,1), a(3)=2*f(1,s,0), a(4)=2*f(0,1,1), a(5)=2*f(3,s,0), a(6)=2*f(2,0,1), a(8)=2*f(9/2,s,0), a(10)=2*f(2,2,1), a(11)=2*f(0,1,2), where f(r,s,t) = 3*10^(t*(s+4)+2*r+1) +10^(r+1)*(22*(10^(s+2)-1)*(10^(t*(s+4))-1) / (10^(s+4)-1))+1.
%e sigma(206) = 312 = reversal(213) = reversal(206+7).
%t Do[If[c=FromDigits[Reverse[IntegerDigits[7+n]]]; c>n && DivisorSigma[1, n] == c, Print[n]], {n, 500000000}]
%Y Cf. A000203, A004086, A230009, A230021.
%K nonn,base,more
%O 1,1
%A _Farideh Firoozbakht_, Dec 12 2013
%E a(6)-a(12) from _Giovanni Resta_, Dec 13 2013
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