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A069282
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Numbers n such that phi(reversal(n)) = reversal(phi(n)). Ignore leading 0's.
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0
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1, 2, 3, 4, 5, 6, 7, 8, 9, 535, 767, 2110, 6188, 6991, 8299, 8816, 9928, 13580, 20502, 21100, 48991, 50805, 53035, 58085, 58585, 59395, 69991, 82428, 82770, 88188, 135800, 211000, 827700, 1358000, 2110000, 2753010, 3269623, 5808085, 5846485
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OFFSET
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1,2
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COMMENTS
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For an arithmetical function f, call the arguments n such that f(reverse(n)) = reverse(f(n)) the "palinpoints" of f. This sequence is the sequence of palinpoints of f(n) = phi(n).
If n is in the sequence and 10 divides n then for each natural number m, 10^m*n is in the sequence because phi(reversal(10^m*n))=phi(reversal(n))=reversal(phi(n)) =reversal(10^m*phi(n))=reversal(phi(10^m*n)). This sequence is infinite because the numbers 2110,13580,82770,8415570 are in the sequence and according to the above assertion all numbers of the form 2110*10^m, 13580*10^m, 82770*10^m & 8415570*10^m are in the sequence. If n is in the sequence and 10 doesn't divide n then it is obvious that the reversal of n is also in the sequence. If both numbers 5*10^(n-1)-1 & 7*10^n-9 are prime we can easily show that 7*10^n-9 is in the sequence. - Farideh Firoozbakht, Jan 18 2006
If both numbers 49*10^n-9 & 125*10^(n-2)-1 are prime then 49*10^n-9 is in the sequence (the proof is easy). 3 is the only known number n such that both numbers 49*10^n-9 & 125*10^(n-2)-1 are primes. - Farideh Firoozbakht, Jan 26 2006
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LINKS
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EXAMPLE
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Let f(n) = phi(n). Then f(6188) = 2304, f(8816) = 4032, so f(reverse(6188)) = reverse(f(6188)). Therefore 6188 belongs to the sequence.
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MATHEMATICA
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rev[n_] := FromDigits[Reverse[IntegerDigits[n]]]; f[n_] := EulerPhi[n]; Select[Range[10^6], f[rev[ # ]] == rev[f[ # ]] &]
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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