A134808 Cyclops numbers.

0, 101, 102, 103, 104, 105, 106, 107, 108, 109, 201, 202, 203, 204, 205, 206, 207, 208, 209, 301, 302, 303, 304, 305, 306, 307, 308, 309, 401, 402, 403, 404, 405, 406, 407, 408, 409, 501, 502, 503, 504, 505, 506, 507, 508, 509, 601, 602, 603, 604, 605, 606

Offset: 1

Author

Omar E. Pol, Nov 21 2007

Keywords

• base
• easy
• nonn

Comments

Numbers with middle digit 0, that have only one digit 0, and the total number of digits is odd; the digit 0 represents the eye of a cyclops.

Examples

			109 is a cyclops number because 109 has only one digit 0 and this 0 is the middle digit.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Brady Haran and Simon Pampena, Glitch Primes and Cyclops Numbers, Numberphile video (2015).

Crossrefs

Cf. A134809, A138131, A138148, A160717, A182809.

Programs

Mathematica

cyclopsQ[n_Integer, b_:10] := Module[{digitList = IntegerDigits[n, b], len, pos0s, flag}, len = Length[digitList]; pos0s = Select[Range[len], digitList[[#]] == 0 &]; flag = OddQ[len] && (Length[pos0s] == 1) && (pos0s == {(len + 1)/2}); Return[flag]]; Select[Range[0,999],cyclopsQ] (* Alonso del Arte, Dec 16 2010 *)
Reap[Do[id=IntegerDigits[n];If[Position[id,0]=={{(Length[id]+1)/2}},Sow[n]],{n,0,10^3}]][[2,1]] (* Zak Seidov, Dec 17 2010 *)
cycQ[n_]:=Module[{idn=IntegerDigits[n],len=IntegerLength[n]},OddQ[len] && DigitCount[ n,10,0]==1&&idn[[(len+1)/2]]==0]; Join[{0},Select[Range[ 0,700],cycQ]] (* Harvey P. Dale, Mar 07 2020 *)

PARI

a(n, {base=10}) = my (l=0); my (r=n-1); while (r >= (base-1)^(2*l), r -= (base-1)^(2*l); l++); return (base^(l+1) * ( (base^l-1)/(base-1) + if (base>2, fromdigits(digits(r \ ((base-1)^l), (base-1)), base)) ) + ( (base^l-1)/(base-1) + if (base>2, fromdigits(digits(r % ((base-1)^l), (base-1)), base)))) \\ Rémy Sigrist, Apr 29 2017

Python

from itertools import product
def cyclops(upto=float('inf'), upton=float('inf')): # generator
  yield 0
  c, n, half_digits, pow10 = 0, 1, 0, 10
  while 100**(half_digits+1) < upto and n < upton:
    half_digits += 1
    pow10 *= 10
    for left in product("123456789", repeat=half_digits):
      left_plus_eye = int("".join(left))*pow10
      for right in product("123456789", repeat=half_digits):
        c, n = left_plus_eye + int("".join(right)), n+1
        if c <= upto and n <= upton: yield c
print([c for c in cyclops(upto=606)])
print([c for c in cyclops(upton=52)]) # Michael S. Branicky, Jan 05 2021

Sage

def is_cyclops(n, base=10):
    dg = n.digits(base) if n > 0 else [0]
    return len(dg) % 2 == 1 and dg[len(dg)//2] == 0 and dg.count(0) == 1
is_A134808 = lambda n: is_cyclops(n)
# D. S. McNeil, Dec 17 2010