How many 15 digit binary sequences are there with no exactly two adjacent 0's ?

I want to count  the number of 15 digit serieses that have no exactly 2 adjacent zeros.
example for an allowed series( to be counted):
exampled for an allowed ones(not to be counted):
their number should be 1597
Only need to know how to calculate this?
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ste5anSenior DeveloperCommented:
If you disallow exactly 2 consecutive 0's
10747 (see comments)

if you disallow 2 or more consecutive 0's
aboo_sAuthor Commented:
ste5an I didn't really get the connection, if you could explain!
ozo :
F(n+2) = number of binary sequences of length n that have no consecutive 0's.
F(15+2)=f(17)=0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597=1597 if you disregard the first 0.
now why is this true, can this be explained! you see I do have the answer already but the way we got it is what I need.
Can you please explain? A bonus would be explaining also how they are related to Fibonacci?
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for n>=2
binary sequences of length n that have no consecutive 0's
= binary sequences of length n starting with 1 that have no consecutive 0's +  binary sequences of length n starting with 0 that have no consecutive 0's
= binary sequences of length n starting with 1 that have no consecutive 0's +  binary sequences of length n starting with 01 that have no consecutive 0's
= binary sequences of length n-1 that have no consecutive 0's + binary sequences of length n-2 that have no consecutive 0's
Which is the Fibonacci recurrence
aboo_sAuthor Commented:
consider n=4 with 3 zeros:
all cases is: 2^4=16
not allowed:
which leaves 14 cases.
0 1 1 2 3 5 8 13
F(7)=13 ??
Besides even if it is true ,why?

if I want to know number of 4 digits sequences that have exactly 2 zeros but consecutive I would do the following:
All 4 digit sequences that have 2 zeros is : 2 of 4 = 6
number of sequences that have consecutive zeros is three, 0011, 1001,0011 --->  6-3=3
so answer is 3.
How could such a calculation (straight forward) be done to either prove the Fibonacci theory or such give the answer to the original question.
I really need to understand this.
If you disallow 3 consecutive zeros, you would get the tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) with a(0)=1, a(1)=2, a(2)=4.
for the same reason that no 2 consecutive zeros gives you F(n+2) where F(n) = F(n-1) + F(n-2) with F(0+2)=1, F(1+2)=2

If you want n digit sequences that have exactly 2 zeros but consecutive, that would be the same as n-1 digit sequences that have exactly 1 zero, which would be C(n-1,1) = n-1

If you disallow exactly 3 consecutive zero's, but allow 4 or more consecutive zero's, that would be  a(n) = 2*a(n-1) - a(n-4) + a(n-5)

But if you are asking about different sequences, and there seem to be at least four different kinds of sequences that you have talked about so far,
then perhaps it would be better to ask them as separate questions.
"If you have follow-up or related questions, post a new question for each of them."

Or if there's a general class of sequences that you are actually interested in, of which the ones you mentioned are specific instances, then please define the general class.
aboo_sAuthor Commented:
Sorry for the confusion I was just trying to get things straight (unsuccessfully it seems), anyhow:
If I want  to count these series of numbers
000000000000101             ----> here a jump over 100
so on
I get  10747 of these as you said (I first thought I would get 1597) anyway my question was and remains how?
Now if you happen to know the answer then I would very much appreciate it if you shared your knowledge with me.
On the other hand if you do not know the answer to my query then I would be thankful just the same for what I learned from you(which is this weird connection to the Fibo series). So if there is nothing else you can help me with on this issue I will gladly close the question granting you full points and go on in my research some other place.

And again sorry for any confusion I may have caused.
The number of binary sequences of length n, with no subsequence of exactly 2 0's (but allowing subsequences of more than 2 0's) is a(n+3), where  a(n) = Sum{a(k): k=0,1,2,...,n-4,n-2,n-1}; (skipping a(n-3)); with a(0)=0, a(1)=0, a(2)=1, a(3)=1

Specifically, the sequences of length N would be:
1{sequences of length N-1} -> a(n-1) where n=N+3
01{sequences of length N-2} -> a(n-2)
0001{sequences of length N-4} -> a(n-4)
00001{sequences of length N-5} -> a(n-5)
000...0001{sequences of length 2} -> a(5)
000...00001{sequences of length 1} -> a(4)
000...000001{sequences of length 0} -> a(3)
000...000000{sequences of length 0} -> a(3)=a(2)

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aboo_sAuthor Commented:
wonderful site ozo
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