A045887 Number of distinct even numbers visible as proper subsequences of n.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 2, 4, 2, 4

Offset: 0

Author

Felice Russo

Keywords

• base
• easy
• nonn

Examples

			a(10)=1 because we can form 0.
a(24)=2 because we can form 2, 4.
a(102)=4 because we can form 0, 2, 10, 12.
a(124)=5 because we can form the following even numbers: 2, 4, 12, 14, 24.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 0..10000
• Sean A. Irvine, Java program (github)

Crossrefs

Cf. A045888, A342845, A342846.

Programs

Python

from itertools import combinations
def a(n):
  s, eset = str(n), set()
  for i in range(len(s)):
    for j in range(i+1, len(s)+1):
      if s[j-1] in "02468":
        if len(s[i:j]) <= 2 and j-i < len(s):
          eset.add(int(s[i:j]))
        else:
          middle = s[i+1:j-1]
          for k in range(len(middle)+1):
            for c in combinations(middle, k):
              t = s[i] + "".join(c) + s[j-1]
              if len(t) < len(s):
                eset.add(int(t))
  return len(eset)
print([a(n) for n in range(105)]) # Michael S. Branicky, Mar 24 2021

Extensions

More terms from Fabian Rothelius, Feb 08 2001

a(102) and a(104) corrected by Reinhard Zumkeller, Jul 19 2011

a(102) and a(104) reverted to original values by Sean A. Irvine, Mar 23 2021