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%I #11 Aug 24 2021 07:52:31
%S 12,20,24,26,28,38,39,40,44,48,50,57,58,60,68,70,71,72,74,75,78,79,80,
%T 84,86,87,92,96,98,100,105,106,108,110,113,117,118,121,122,124,132,
%U 134,135,138,139,140,141,144,152,154,155,158,159,160,164,166,167,172
%N Numbers for which in base 2 the least number of digits that can be removed to leave a palindrome (possibly beginning with 0) is 2.
%H Robert Israel, <a href="/A050421/b050421.txt">Table of n, a(n) for n = 1..10000</a>
%e (20 base 2) = 101000 -> 0000.
%p filter:= proc(n) local L,R,k,m;
%p L:= convert(n,base,2);
%p R:= ListTools:-Reverse(L);
%p m:= nops(L);
%p if L = R then return false fi;
%p if ormap(i -> subsop(i=NULL,L)=subsop(m+1-i=NULL,R), [$1..m]) then return false fi;
%p ormap(p -> subsop(p[1]=NULL,subsop(p[2]=NULL,L))=subsop(m-p[1]=NULL,subsop(m+1-p[2]=NULL,R)), [seq(seq([i,j],j=i+1..m),i=1..m-1)])
%p end proc:
%p select(filter, [$4..200]); # _Robert Israel_, Oct 28 2020
%o (Python)
%o from itertools import combinations
%o def ok(n):
%o b = bin(n)[2:]
%o for digs_to_remove in range(4):
%o for skip in combinations(range(len(b)), digs_to_remove):
%o newb = "".join(b[i] for i in range(len(b)) if i not in skip)
%o if len(newb) > 0 and newb == newb[::-1]:
%o return (digs_to_remove == 2)
%o return False
%o print(list(filter(ok, range(173)))) # _Michael S. Branicky_, Aug 24 2021
%Y Cf. A050426, A050420, A050422, A050423, A050424.
%K nonn,base
%O 1,1
%A _Clark Kimberling_
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