%I #36 May 13 2023 13:33:52
%S 0,1,11,29,36,193,281
%N Numbers k such that the number of odd digits in k! is greater than or equal to the number of even digits.
%C If it exists, a(8) > 100000.
%e 11 is a term since 11! = 39916800, and the numbers of odd and even digits are both 4.
%e 29 is a term since 29!=8841761993739701954543616000000, and the numbers of odd and even digits are 16 and 15 respectively.
%t Select[Range[0, 500],
%t Count[IntegerDigits[#!], _?OddQ] >=
%t Count[IntegerDigits[#!], _?EvenQ] &]
%o (Python)
%o from sympy import factorial as f
%o def ok(n):
%o s=str(f(n))
%o return(sum(1 for k in s if k in '02468')<=sum(1 for k in s if k in '13579'))
%o print([n for n in range(501) if ok(n)])
%o (Python)
%o from math import factorial
%o from itertools import count, islice
%o def A360181_gen(startvalue=0): # generator of terms >= startvalue
%o f = factorial(m:=max(startvalue,0))
%o for k in count(m):
%o if len(s:=str(f)) <= sum(1 for d in s if d in {'1','3','5','7','9'})<<1:
%o yield k
%o f *= k+1
%o A360181_list = list(islice(A360181_gen(),7)) # _Chai Wah Wu_, May 10 2023
%Y Cf. A034886, A352547, A360182.
%K nonn,base,more,less
%O 1,3
%A _Zhining Yang_, Jan 28 2023