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A115749
Numbers n such that sigma(n)=8*reversal(n).
0
861, 951, 2070, 8241, 900051, 8864151, 9000051, 82000041, 8200000041, 82000000041
OFFSET
1,1
COMMENTS
If p=3*10^n+17 is prime then 3*p is in the sequence because sigma(3*p)=4*(3*10^n+18)=12*10^n+72=8*(15*10^(n-1)+9)=8* reversal(9*10^n+51)=8*reversal(3*p). Also if p=(2*10^n+1)/3 is prime then 123*p is in the sequence (the proof is easy). Next term is greater than 13*10^7.
a(11) > 10^12. - Giovanni Resta, Oct 28 2012
EXAMPLE
82000041 is in the sequence because sigma(82000041)
=112000224=8*14000028=8*reversal(82000041).
MATHEMATICA
Do[If[DivisorSigma[1, n]==8*FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 130000000}]
CROSSREFS
KEYWORD
base,more,nonn
AUTHOR
Farideh Firoozbakht, Feb 12 2006
EXTENSIONS
a(9)-a(10) from Donovan Johnson, Dec 21 2008
STATUS
approved