A360181 Numbers k such that the number of odd digits in k! is greater than or equal to the number of even digits.

0, 1, 11, 29, 36, 193, 281

Offset: 1

Author

Zhining Yang, Jan 28 2023

Keywords

• nonn
• base
• more
• less

Comments

If it exists, a(8) > 100000.

Examples

			11 is a term since 11! = 39916800, and the numbers of odd and even digits are both 4.
29 is a term since 29!=8841761993739701954543616000000, and the numbers of odd and even digits are 16 and 15 respectively.
		

Crossrefs

Cf. A034886, A352547, A360182.

Programs

Mathematica

Select[Range[0, 500],
 Count[IntegerDigits[#!], _?OddQ] >=
   Count[IntegerDigits[#!], _?EvenQ] &]

Python

from sympy import factorial as f
def ok(n):
    s=str(f(n))
    return(sum(1 for k in s if k in '02468')<=sum(1 for k in s if k in '13579'))
print([n for n in range(501) if ok(n)])

Python

from math import factorial
from itertools import count, islice
def A360181_gen(startvalue=0): # generator of terms >= startvalue
    f = factorial(m:=max(startvalue,0))
    for k in count(m):
        if len(s:=str(f)) <= sum(1 for d in s if d in {'1','3','5','7','9'})<<1:
            yield k
        f *= k+1
A360181_list = list(islice(A360181_gen(),7)) # Chai Wah Wu, May 10 2023