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%I #23 Apr 08 2020 00:06:42
%S 1,2,3,4,5,6,7,9,99,999,9999,99999,999999,9999999,99999999,999999999,
%T 9999999999,99999999999,999999999999,9999999999999,99999999999999,
%U 999999999999999,9999999999999999,99999999999999999
%N Palindromes k such that k + 2 is also a palindrome.
%C Perhaps from 8th term onwards the only members are a(n) = 10^(n-7) - 1 for n > 7.
%C The above conjecture is true. Adding two to the least significant digit of a number can result in a carry of at most 1, which only happens if the digit of least significance is 8 or 9. If the least significant digit is 8, adding two results in that digit becoming 0, so the resulting number can't be palindromic. If only the k least and most significant digits are 9, the least significant digit will become 1 and all other adjacent digits 9 will turn into the digit 0 and produce a carry of 1. For the starting number to have been palindromic, the k most significant digits must also be 9's. Any digits that are not 9's between the 9's will not produce a carry on their own when increased by one through the previous carry, resulting in a nonpalindromic number with some 9's as most significant digits and a single 1 and 0's as least significant digits. - _Felix Fröhlich_, Jul 22 2014
%K base,nonn
%O 1,2
%A _Amarnath Murthy_, Apr 13 2003
%E Incorrect formula removed by _Felix Fröhlich_, Jul 24 2014
%E More terms from _Felix Fröhlich_, Jul 24 2014
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