|
%I #15 Jul 14 2022 12:08:40
%S 5,24,130,792,5080,34584,247360,1817112,13918720,108848664,869866240,
%T 7169995032,59085276160,505735077144,4311229112320,37428004374552,
%U 335520388710400,2861870689152024,27669446179225600,223578655251963672,2398913308953149440,17708639883984065304
%N Number of n-digit terms in A333369.
%C Also the number of n-digit simbers, where a simber is a positive integer in which any odd digit, if present, occurs an odd number of times, and any even digit, if present, occurs an even number of times.
%H Michael S. Branicky, <a href="/A353735/b353735.txt">Table of n, a(n) for n = 1..1002</a>
%H Project Euler, <a href="https://projecteuler.net/problem=520">Problem 520: Simbers</a>.
%F From _Bernard Schott_, Jul 11 2022: (Start)
%F Conjecture 1: lim_{n->oo} a(2n+1)/a(2n-1) = 100.
%F Conjecture 2: lim_{n->oo} a(2n+2)/a(2n) = 81. (End)
%e There are five 1-digit terms in A333369: 1, 3, 5, 7, 9. Thus, a(1) = 5.
%o (Python)
%o def isA333369(n):
%o digits = list(map(int, str(n)))
%o return all(digits.count(d)%2 == d%2 for d in set(digits))
%o def a(n): return sum(1 for i in range(10**(n-1), 10**n) if isA333369(i))
%o print([a(n) for n in range(1, 7)]) # _Michael S. Branicky_, May 06 2022
%Y Cf. A333369, A353007, A353736.
%K nonn,base
%O 1,1
%A _Michael S. Branicky_, May 06 2022
|
