|
Search: id:A067723
|
|
|
| A067723 |
|
Nontrivial palindromes n such that phi(n) is also a palindrome. |
|
+0
2
|
|
| 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
(list; graph; listen)
|
|
|
OFFSET
|
1,1
|
|
|
COMMENT
|
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 palindrome since phi(m) is 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 be with an even number of digits. - Farideh Firoozbakht, Feb 03 2006
|
|
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[ # ]] &]
|
|
PROGRAM
|
(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: A067803 A098258 A077085 this_sequence A059949 A077076 A033916
Adjacent sequences: A067720 A067721 A067722 this_sequence A067724 A067725 A067726
|
|
KEYWORD
|
base,easy,nonn
|
|
AUTHOR
|
Joseph L. Pe (joseph_l_pe(AT)hotmail.com), Feb 05 2002
|
|
EXTENSIONS
|
Edited and extended by Klaus Brockhaus (klaus-brockhaus(AT)t-online.de) Feb 11 2002
|
|
|
Search completed in 0.002 seconds
|
