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%I #7 Aug 24 2021 07:00:19
%S 16,48,80,88,104,112,144,152,156,184,200,228,232,240,272,280,284,286,
%T 296,300,302,316,328,336,344,348,376,392,420,424,432,440,452,466,468,
%U 472,482,484,488,496,528,536,540,542,543,552,556,558,559,564,566,574,606
%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 4.
%e (64 base 2) = 10000 -> 1.
%o (Python)
%o from itertools import combinations
%o def ok(n):
%o b = bin(n)[2:]
%o for digs_to_remove in range(5):
%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 == 4)
%o return False
%o print(list(filter(ok, range(607)))) # _Michael S. Branicky_, Aug 24 2021
%Y Cf. A050423, A050425, A050426, A050427, A050429.
%K nonn,base
%O 1,1
%A _Clark Kimberling_
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