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%I #16 Oct 29 2022 12:53:16
%S 6,73,483,4074,4473,4623,7993,42813,69855,253782,799993,7999993,
%T 46000023,426000213,4600000023,6718967838,42600000213,46000000023,
%U 79999999993,426000000213
%N Numbers n such that 2*reversal(n)=sigma(n).
%C I. If p=8*10^n-7 is a prime then p is in the sequence because reversal(p)=4*10^n-3 & sigma(p)=8*10^n-6 so 2*reversal(p) =sigma(p). 73,7993,799993 & 7999993 are such terms.
%C II. If q=(2*10^n+1)/3 is a prime then (a): 69*q is in the sequence because 69*q=46*10^n+23; reversal (69*q)=32*10^n+64 & sigma(69*q)=96*q+96=64*10^n+128 so 2*reversal (69*q)=sigma(69*q). 483,4623 & 46000023 are such terms. (b): 639*q is in the sequence because 639*q=426*10^n+213; reversal (639*q)=312*10^n+624 & sigma(639*q)=936*q+936=624*10^n+1248 so 2*reversal(639*q)=sigma(639*q). 42813 & 426000213 are such terms.
%C a(21) > 10^12. - _Giovanni Resta_, Oct 28 2012
%e 253782 is in the sequence because reversal(253782)=287352; sigma(253782)=574704 & 2*287352=574704.
%t reversal[n_]:= FromDigits[Reverse[IntegerDigits[n]]]; Do[If[2* reversal[n]== DivisorSigma[1, n], Print[n]], {n, 1000000000}]
%t Select[Range[8*10^6],2*IntegerReverse[#]==DivisorSigma[1,#]&] (* The program generates the first 12 terms of the sequence. *) (* _Harvey P. Dale_, Oct 29 2022 *)
%Y Cf. A093170, A096507, A099190, A105322, A105323, A105325, A105326.
%K base,more,nonn
%O 1,1
%A _Farideh Firoozbakht_, Apr 16 2005
%E a(15)-a(19) from _Donovan Johnson_, Dec 21 2008
%E a(20) from _Giovanni Resta_, Oct 28 2012
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