A050421 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.

12, 20, 24, 26, 28, 38, 39, 40, 44, 48, 50, 57, 58, 60, 68, 70, 71, 72, 74, 75, 78, 79, 80, 84, 86, 87, 92, 96, 98, 100, 105, 106, 108, 110, 113, 117, 118, 121, 122, 124, 132, 134, 135, 138, 139, 140, 141, 144, 152, 154, 155, 158, 159, 160, 164, 166, 167, 172

Offset

1,1

Links

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

Example

(20 base 2) = 101000 -> 0000.

Maple

filter:= proc(n) local L, R, k, m;
  L:= convert(n, base, 2);
  R:= ListTools:-Reverse(L);
  m:= nops(L);
  if L = R then return false fi;
  if ormap(i -> subsop(i=NULL, L)=subsop(m+1-i=NULL, R), [$1..m]) then return false fi;
  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)])
end proc:
select(filter, [$4..200]); # Robert Israel, Oct 28 2020

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 == newb[::-1]:
                return (digs_to_remove == 2)
    return False
print(list(filter(ok, range(173)))) # Michael S. Branicky, Aug 24 2021

Crossrefs

Cf. A050426, A050420, A050422, A050423, A050424.

Sequence in context: A038528 A052020 A075078 * A307517 A206552 A332832

Adjacent sequences: A050418 A050419 A050420 * A050422 A050423 A050424

Keyword

nonn,base

Author

Clark Kimberling

Status

approved