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EDIT: Thanks to RobPratt's insight, (https://oeis.org/A005251), I wrote a quick and dirty python program to generate the number of bin strings not containing a 3 length string;

sols=[0,1,1,1]

def alg(n):
    n=int(n)
    x=len(sols)
    while x <= n+3:
        sols.append(sols[x-1]+sols[x-2]+sols[x-4])
        x+=1
    return sols[n+3]


def generate(n):
    n=int(n)
    for i in range(0,n):
        print(alg(i))


while True:
    prompt = input("Input either 'alg n' or 'generate n' >>   ").split(" ")
    if prompt[0] == 'alg':
        print(alg(prompt[1]))
    else:
        generate(prompt[1])

Wish me luck for my exam!!!

ORIGINAL POST:

Hey Maths StackExchange!

I apologise if any of the things I say in this post are naive or stupid; Im a first year comp sci student with an upcoming discrete maths test and Im in no way a competent mathematician.

I was doing a problem in which I had to count the number of binary strings of length n which do not contain the factor '101'.

I played around with the problem a little bit, and I think I found a general formula.

There are obviously 2^n binary strings, so the number of strings which don't contain the factor is 2^n - {strings that do contain the factor}.

For a string of length n=6, for example, this would mean that the strings;

101XXX
X101XX
XX101X
XXX101

are defined as strings that contain the factor, where X ∈ {0,1}.

Im sure there is a more rigorous reason for this combinatorially, but by inspection I determined that there are always n-2 of these strings.

For each of these strings, I treated the X's as their own binary string, and given that there are n-3 X's in each string, the X's can form a possible 2^(n-3) binary strings.

Therefore, for an n length string, there are (n-2) formats, each containing 2^(n-3) possible strings, and therefore for an n length string there are; $$2^n - (n-2)2^{n-3}$$ strings not containing the factor '101'.

I also realise that this formula only relies on the length of the factor '101', so it could be generalised to; $$2^n - (n-(k-1))2^{n-k}$$ where k is the length of the factor.

The reason I am writing this is pretty much just to ask if my formula seems correct, because if it does, I intend to use it in my test.

Thanks!

noha
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    I think you're over counting. When you look at "all" strings of the form 101XXX and the string XXX101, you mistakenly think these sets are independent. They are not. Each contains 101101. – David G. Stork May 28 '23 at 03:23
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    You are overcounting the strings that contain 101. For example, consider 10101X. – RobPratt May 28 '23 at 03:23
  • Another possible source of overcounting: aren't there $2^{n-1}$ possible binary numbers of length $n$? After all, if you allow the first digit to be a $0$ then what you basically have is a binary number of length $n-1$, right? It's like talking about 3-digit integers while allowing the first digit to be a 0; you end up including 099, 098, 096 etc – H. sapiens rex May 28 '23 at 03:29
  • https://oeis.org/A005251 – RobPratt May 28 '23 at 03:33
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    Also, does "factor" mean what it normally means in this case? Because the question just sounds like determining the number of binary numbers of length $n$ that are coprime to 5 – H. sapiens rex May 28 '23 at 03:51
  • I expect that questions of this nature have been asked and answered on this website (several times) in the past. I suggest a search for those previous occurrences. – Gerry Myerson May 28 '23 at 03:54

0 Answers0