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A069215
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Numbers n such that phi(n) = reversal(n).
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13
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1, 21, 63, 270, 291, 2991, 6102, 46676013, 69460293, 2346534651, 6313047393, 23400000651, 80050617822, 234065340651, 234659934651, 2340000000651, 2530227348360, 2934000006591
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OFFSET
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1,2
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COMMENTS
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If 10^n-3 is prime (n is in the sequence A089765) and m=3*(10^n-3) then m is in this sequence, for example 299999999999999991 is a term of this sequence because 299999999999999991=3*(10^17-3) and 17 is in the sequence A089675. So 3*(10^A089675-3) is a subsequence of this sequence, A101700 is this subsequence. - Farideh Firoozbakht, Dec 26 2004
If p=(79*10^(4n+1)-83)/101 is prime then 3p is in the sequence. The proof is easy. 21, 2346534651 & 3*(79*10^2697-83)/101 are the first three such terms. - Farideh Firoozbakht, Apr 22 2008, Aug 16 2008
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LINKS
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EXAMPLE
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phi(291) = 192.
phi(6102) = 2016 = reversal(6102), so 6102 belongs to the sequence.
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MATHEMATICA
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Do[If[EulerPhi[n] == FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 1, 10^5}]
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PROG
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(PARI) for( n=1, 1e9, A004086(n)==eulerphi(n) & print1(n", "))
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CROSSREFS
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KEYWORD
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nonn,base,hard,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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