I solved it!
The answer is: No, the sub-sequence 13579 cannot occur anywhere in this sequence.
Proof:
13579 are odd numbers.
Claim:
There can be no five subsequent odd numbers in the sequence.
Proof:
Suppose there were 5 subsequent odd numbers in the sequence. Consider the sequence modulo 2.
So then there would be a subsequence of the form
abcdef11111
The last digit of a number is odd iff the number itself is odd.
Thus it must hold that
1)a+b+c+d+e+f mod 2 = 1
2)b+c+d+e+f+1 mod 2 = 1
3)c+d+e+f+1+1 mod 2 = 1
4)d+e+f+1+1+1 mod 2 = 1
5)e+f+1+1+1+1 mod 2 = 1
From 1) and 2) follows
(a+b+c+d+e+f)+(b+c+d+e+f+1 )=1+1
=>
a+1=0
=>
a=1
From 2) and 3) follows
(b+c+d+e+f+1)+(c+d+e+f+1+1)=1+1
=>
b=1
And analogously from 3),4) and 4),5) it follows
c=1 and d=1
Plugging a=b=c=d=1 into equation 1) gives:
1+1+1+1+e+f=1
<=>
e+f=1
So either e=1 and f=0
or e=0 and f=1
Now let i be the FIRST index so that a_i a_(i+1) a_(i+2) a_(i+3) a_(i+4)
is a subsequence of 5 odd numbers.
Then 6 numbers abcdef with above properties occur in front of it, i.e
abcdef a_i a_(i+1) a_(i+2) a_(i+3) a_(i+4)
which is
1111ef a_i a_(i+1) a_(i+2) a_(i+3) a_(i+4)
modulo 2
Since i is the FIRST index so that a subsequence of 5 odd numbers start there
f cannot be odd itself, otherwise
f a_i a_(i+1) a_(i+2) a_(i+3) would already be a subsequence of 5 odd numbers starting at the lower index i-1.
And also e cannot be odd, since otherwise
abcde would already be a subsequence of 5 odd numbers starting at a lower index than i.
Therefore there cannot be any subsequence of 5 odd numbers in the sequence at all.
qed
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