

A115926


Numbers n such that phi(n)=reversal(n)n.


1



37, 397, 1853, 15503, 48776, 198683, 200882, 1061361, 3542805, 3564217, 3868867, 3962197, 4438616, 19844683, 198444683, 202195682, 309520655, 431092646, 439419646, 500729929, 535973599, 3566790217, 3963436297, 4149753226, 17296101143, 39560402197
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OFFSET

1,1


COMMENTS

All primes of the form 4*10^n3 are in the sequence because if 4*10^n3 is prime then phi(4*10^n3)=(4*10^n4) =(8*10^n7)(4*10^n3)=reversal(4*10^n3)(4*10^n3).
Also if n>1 and p=(94*10^n+113)/9 is prime then 19*p is in the sequence (the proof is easy). Next term is greater than 125*10^6.
If p=(1/303)*(232*10^(4n)+71) is prime then 7*p is in the sequence (the proof is easy). The first four such terms happen for n=2, 101, 104 & 444 and numbers of digits of these terms of the sequence are 9, 405, 417 & 1777 respectively.  Farideh Firoozbakht, Jan 02 2008
a(32) > 10^12.  Giovanni Resta, Oct 28 2012


LINKS

Giovanni Resta, Table of n, a(n) for n = 1..31


EXAMPLE

If n=37, phi(37)= 36 = 7337


MATHEMATICA

Do[If[EulerPhi[n]==FromDigits[Reverse[IntegerDigits[n]]]n, Print[n]], {n, 125000000}]
Do[If[EulerPhi[n]==FromDigits[Reverse[IntegerDigits[n]]]n, Print[n]], {n, 600000000}]  Jessica M. Cornwall (jmc510(AT)psu.edu), Apr 05 2006


CROSSREFS

Cf. A072393.
Sequence in context: A283629 A264626 A201789 * A083818 A090023 A254682
Adjacent sequences: A115923 A115924 A115925 * A115927 A115928 A115929


KEYWORD

more,nonn,base


AUTHOR

Farideh Firoozbakht, Jan 31 2006


EXTENSIONS

More terms from Jessica M. Cornwall (jmc510(AT)psu.edu), Apr 05 2006
a(22)a(31) from Giovanni Resta, Oct 28 2012


STATUS

approved



