login
A067723
Nontrivial palindromes k such that phi(k) is also a palindrome.
1
535, 767, 20502, 50805, 53035, 58085, 58585, 59395, 82428, 88188, 3269623, 5808085, 5846485, 8110118, 8666668, 8818188, 8872788, 8875788, 473040374, 515050515, 530303035, 535353535, 580303085, 580858085, 581585185, 585797585
OFFSET
1,1
COMMENTS
One-digit numbers (trivial palindromes) have been excluded from the sequence.
Conjecture: There are no terms with an even number of digits. - Klaus Brockhaus
There are no terms with an even number of digits. Proof: If m is in the sequence then phi(m) and m are palindromes since phi(m) is a palindrome so 10=phi(11) doesn't divide phi(m) hence 11 doesn't divide m but we know that 11 divides every palindromic number with an even number of digits so m can't have an even number of digits. - Farideh Firoozbakht, Feb 03 2006
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..757 (terms < 10^19)
EXAMPLE
phi(88188) = 29392, so 88188 is a term of the sequence.
MATHEMATICA
isp[n_] := Module[{a = IntegerDigits[n]}, a == Reverse[a]]; Select[Range[10^6], isp[ # ] && isp[EulerPhi[ # ]] &]
Select[Range[10, 59*10^7], AllTrue[{#, EulerPhi[#]}, PalindromeQ]&] (* Harvey P. Dale, Oct 24 2023 *)
PROG
(PARI) intreverse(n) = local(d, rev); rev=0; while(n>0, d=divrem(n, 10); n=d[1]; rev=10*rev+d[2]); rev; for(k=10, 6*10^8, if(k==intreverse(k), m=eulerphi(k); if(m==intreverse(m), print1(k, ", "))))
CROSSREFS
Cf. A000010.
Sequence in context: A165989 A183598 A252536 * A252275 A059949 A256088
KEYWORD
nonn,base
AUTHOR
Joseph L. Pe, Feb 05 2002
EXTENSIONS
Edited and extended by Klaus Brockhaus, Feb 11 2002
STATUS
approved