A050427 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.

8, 24, 40, 44, 52, 56, 72, 76, 78, 92, 100, 114, 116, 120, 136, 140, 142, 143, 148, 150, 151, 158, 164, 168, 172, 174, 188, 196, 210, 212, 216, 220, 226, 233, 234, 236, 241, 242, 244, 248, 264, 268, 270, 271, 276, 278, 279, 282, 283, 287, 303, 308, 310, 318

Offset

1,1

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Example

(72 base 2) = 1001000 -> 1001.

Maple

filter:= proc(n) local L, Lp, m, i, c;
  L:= convert(n, base, 2);
  m:= nops(L);
  for i from 1 to m do
    Lp:= subsop(i=NULL, L);
    if Lp[-1] = 1 and andmap(t -> Lp[t]=Lp[-t], [$1..(m-1)/2]) then return false fi
  od;
  for c in combinat:-choose(m, 2) do
    Lp:= subsop(seq(i=NULL, i=c), L);
    if Lp[-1] = 1 and andmap(t -> Lp[t]=Lp[-t], [$1..(m-2)/2]) then return false fi;
  od;
  for c in combinat:-choose(m, 3) do
    Lp:= subsop(seq(i=NULL, i=c), L);
    if Lp[-1] = 1 and andmap(t -> Lp[t]=Lp[-t], [$1..(m-2)/2]) then return true fi;
  od;
false
end proc;
select(filter, [$8..1000]); # Robert Israel, Mar 18 2026

Prog

(Python)
from itertools import combinations
def ok(n):
    b = bin(n)[2:]
    for digs_to_remove in range(4):
        for skip in combinations(range(len(b)), digs_to_remove):
            newb = "".join(b[i] for i in range(len(b)) if i not in skip)
            if len(newb) > 0 and newb[0] == '1' and newb == newb[::-1]:
                return (digs_to_remove == 3)
    return False
print(list(filter(ok, range(320)))) # Michael S. Branicky, Aug 24 2021

Crossrefs

Cf. A050425, A050426, A050428, A050429.

Sequence in context: A134223 A304475 A316300 * A329549 A031046 A173080

Adjacent sequences: A050424 A050425 A050426 * A050428 A050429 A050430

Keyword

nonn,base

Author

Clark Kimberling

Status

approved