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A105323
Numbers of the form 41*(2*10^n+1) where (2*10^n+1)/3 is prime (n is in the sequence A096507).
3
861, 8241, 82000041, 8200000041, 82000000041, 8200000000041, 8200000000000000000041, 8200000000000000000000041, 8200000000000000000000000000000000000000041
OFFSET
1,1
COMMENTS
A105323=41*A093170=41*(2*10^A096507+1)=41*(2*10^(A056657+1)+1). If m is in the sequence then d(m)*reversal(m)=sigma(m) (see A104907). So this sequence is a subsequence of A104907.
EXAMPLE
861 is in the sequence because 861=41*(2*10^1+1); (2*10^1+1)/3=7 and 7 is prime.
MATHEMATICA
Do[If[PrimeQ[(2*10^n + 1)/3], Print[41*(2*10^n + 1)]], {n, 63}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Farideh Firoozbakht, Apr 16 2005
STATUS
approved