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3-digit half-palindromes.
2

%I #15 Feb 23 2014 02:57:01

%S 22,51,87,91,102,121,145,169,187,190,212,220,225,245,247,248

%N 3-digit half-palindromes.

%C A positive integer m we call k-digit half-palindrome if there exist two bases 1<b<c such that m=[m_1 m_2...m_k]_b=[m_k m_(k-1)...m_1]_c, where m_i are digits in both of these bases with the condition m_1>0 and m_k>0 (see SeqFan Discussion list from Mar 03 2011, where we introduced "b,c-palindromes").

%C _Robert Israel_ showed (see SeqFan Discussion list from the same day) that every number of the form [n+1,n,n]_(2*n+1)is 3-digit half-palindrome with b=2*n+1 and c=2*n+2. Thus the sequence is infinite.

%C On the other hand, every number of the form [k*n+m+1,0,k*n+m]_(4*k*n+4*m+1), where k>=1,m>=0, is 3-digit half-palindrome with b=4*k*n+4*m+1 and c=4*k*n+4*m+3.

%e Let m=22. We have 22=[2 1 1]_3 and 22=[1 1 2]_4. Thus 22, by the definition, is a 3-digit half-palindrome.

%e Let m=91. We have 91=[3 3 1]_5 and 91 =[1 3 3]_8. Thus 91 is a 3-digit half palindrome.

%Y Cf. A002113, A006995, A059809

%K nonn,base

%O 1,1

%A _Vladimir Shevelev_, May 29 2011

%E Corrected by _R. J. Mathar_, Jul 02 2012