_Joseph L. Pe (JosephL.Pe(AT)hotmail.com), _, Jul 21 2002
_Joseph L. Pe (JosephL.Pe(AT)hotmail.com), _, Jul 21 2002
proposed
approved
editing
proposed
a(51) > 10^12. - Giovanni Resta, Oct 28 2012
Giovanni Resta, <a href="/A072393/b072393.txt">Table of n, a(n) for n = 1..50</a>
approved
editing
If m>1 and p=2*10^m+3 is prime then n=27*p is in the sequence because n-reversal(n)=27*(2*10^m+3)-reversal(27*(2*10^m+3))= (54*10^m+81)-(18*10^m+45)=36*10^m+36=18*(2*10^m+2)=phi(27)* phi(2*10^m+3)=phi(27*(2*10^m+3))=phi(n). Also if m>2 and p=(389*10^m+109)/3 is prime then 7*p is in the sequence (the proof is easy). Next term is greater than 2*10^8. - _Farideh Firoozbakht (mymontain(AT)yahoo.com), _, Jan 27 2006
More terms from _Farideh Firoozbakht (mymontain(AT)yahoo.com), _, Jan 27 2006
a(22)-a(29) from _Donovan Johnson (donovan.johnson(AT)yahoo.com), _, Dec 04 2011
proposed
approved
editing
proposed
91, 874, 3411, 9093, 40112, 44252, 54081, 67284, 80224, 90933, 91503, 4961782, 5400081, 5726691, 8750834, 9076921, 9155055, 54000081, 62023914, 90766921, 93079231, 430770922, 540000081, 636355044, 808618664, 907666921, 928709013, 4050394312, 4262971312
a(22)-a(29) from Donovan Johnson (donovan.johnson(AT)yahoo.com), Dec 04 2011
approved
editing
If m>1 and p=2*10^m+3 is prime then n=27*p is in the sequence because n-reversal(n)=27*(2*10^m+3)-reversal(27*(2*10^m+3))= (54*10^m+81)-(18*10^m+45)=36*10^m+36=18*(2*10^m+2)=phi(27)* phi(2*10^m+3)=phi(27*(2*10^m+3))=phi(n). Also if m>2 and p=(389*10^m+109)/3 is prime then 7*p is in the sequence (the proof is easy). Next term is greater than 2*10^8. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Jan 27 2006
base,nonn,new