%I
%S 8,24,40,44,52,56,72,76,78,92,100,114,116,120,136,140,142,143,148,150,
%T 151,158,164,168,172,174,188,196,210,212,216,220,226,233,234,236,241,
%U 242,244,248,264,268,270,271,276,278,279,282,283,287,303,308,310,318
%N Numbers for which in base 2 the least number of digits that can be removed to leave a base 2 palindromic number (beginning with 1) is 3.
%e (72 base 2) = 1001000 > 1001.
%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[0] == '1' and newb == newb[::1]:
%o return (digs_to_remove == 3)
%o return False
%o print(list(filter(ok, range(320)))) # _Michael S. Branicky_, Aug 24 2021
%Y Cf. A050425, A050426, A050428, A050429.
%K nonn,base
%O 1,1
%A _Clark Kimberling_
