%I #17 Apr 07 2020 21:01:42
%S 9,153,165,195,2289,2409,2457,2661,2709,2829,3171,3219,3339,3591,
%T 34785,35793,36273,36465,37833,38313,38505,39321,39513,39993,41925,
%U 42405,42597,43413,43605,44085,45453,45645,46125,47133,50115,50595,50787
%N Positive integers n that are palindromic in base 2 and whose binary representation has the same number of 0's as 1's.
%C Every term of this sequence corresponds to a different term of sequence A031443 (Numbers that in base 2 have the same number of 0's as 1's). (See formula.) - _Leroy Quet_, Sep 05 2008
%H Chai Wah Wu, <a href="/A143905/b143905.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A031443(n)*2^A070939(A031443(n)) + A030101(A031443(n)). - _Leroy Quet_, Sep 05 2008
%F Intersection of A031443 and A006995. - _R. J. Mathar_, Sep 05 2008
%e 165 in binary is 10100101. This binary representation is a palindrome. And it has both four 0's and four 1's. So 165 is in the sequence.
%t Select[Range[100000], Reverse[IntegerDigits[ #, 2]] == IntegerDigits[ #, 2] && DigitCount[ #, 2, 0] == DigitCount[ #, 2, 1] &] (* _Stefan Steinerberger_, Sep 05 2008 *)
%o (PARI) isok(n) = {my(b = binary(n)); (Vecrev(b) == b) && (hammingweight(n) == #b/2);} \\ _Michel Marcus_, Aug 01 2017
%Y Cf. A006995, A031443, A143906.
%K base,nonn
%O 1,1
%A _Leroy Quet_, Sep 04 2008
%E More terms from _Stefan Steinerberger_ and _R. J. Mathar_, Sep 05 2008