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Breaking of a periodic $pm1$ sequence into positive and negative parts.
Periodic sequencepartial sum of convergent seriesMaking all row sums and column sums non-negative by a sequence of movesProve that if $a_1 + a_2 + ldots$ converges then $a_1+2a_2+4a_4+8 a_8+ldots$ converges and $lim na_n=0$How can we have a circular sequence of $0$s and $1$s of length $d$ that are not periodic?Partial sums of periodic sequencesFind a sequence of positive integers given an averageIs this sequence always periodic?A question on the sums of finite subsequences of a sequence of positive realsProve that $x=0,122112122122…$ is irrational
$begingroup$
Let $A_n=(a_1,a_2,dots)$ be a periodic sequence of $a_i=pm1$, the period of length $2n$ containing equal numbers of positive and negative terms. The period is assumed to be irreducible, so that $2n$ is its least possible value.
Let a sequence of length $n $ be termed positive if all partial sums
$$
S_k=sum_i=1^ka_i,quad k=1..n
$$
are non-negative, and negative if all partial sums are non-positive.
The question: is it always possible to choose $i$ $(1le ile 2n)$, such that the sequence $(a_i,a_i+1,dots,a_i+n-1)$ is positive, and the sequence
$a_i+n,a_i+1,dots,a_i+2n-1$ is negative?
It seems to be true for small values of $n$, but I do not see how to prove this in general case. I could also overlook some counterexample.
sequences-and-series combinatorics
asked Mar 15 at 17:25
useruser
5,68111031
$endgroup$
add a comment |
$begingroup$
Let $A_n=(a_1,a_2,dots)$ be a periodic sequence of $a_i=pm1$, the period of length $2n$ containing equal numbers of positive and negative terms. The period is assumed to be irreducible, so that $2n$ is its least possible value.
Let a sequence of length $n $ be termed positive if all partial sums
$$
S_k=sum_i=1^ka_i,quad k=1..n
$$
are non-negative, and negative if all partial sums are non-positive.
The question: is it always possible to choose $i$ $(1le ile 2n)$, such that the sequence $(a_i,a_i+1,dots,a_i+n-1)$ is positive, and the sequence
$a_i+n,a_i+1,dots,a_i+2n-1$ is negative?
It seems to be true for small values of $n$, but I do not see how to prove this in general case. I could also overlook some counterexample.
sequences-and-series combinatorics
asked Mar 15 at 17:25
useruser
5,68111031
$endgroup$
add a comment |
$begingroup$
Let $A_n=(a_1,a_2,dots)$ be a periodic sequence of $a_i=pm1$, the period of length $2n$ containing equal numbers of positive and negative terms. The period is assumed to be irreducible, so that $2n$ is its least possible value.
Let a sequence of length $n $ be termed positive if all partial sums
$$
S_k=sum_i=1^ka_i,quad k=1..n
$$
are non-negative, and negative if all partial sums are non-positive.
The question: is it always possible to choose $i$ $(1le ile 2n)$, such that the sequence $(a_i,a_i+1,dots,a_i+n-1)$ is positive, and the sequence
$a_i+n,a_i+1,dots,a_i+2n-1$ is negative?
It seems to be true for small values of $n$, but I do not see how to prove this in general case. I could also overlook some counterexample.
sequences-and-series combinatorics
asked Mar 15 at 17:25
useruser
5,68111031
$endgroup$
Let $A_n=(a_1,a_2,dots)$ be a periodic sequence of $a_i=pm1$, the period of length $2n$ containing equal numbers of positive and negative terms. The period is assumed to be irreducible, so that $2n$ is its least possible value.
Let a sequence of length $n $ be termed positive if all partial sums
$$
S_k=sum_i=1^ka_i,quad k=1..n
$$
are non-negative, and negative if all partial sums are non-positive.
The question: is it always possible to choose $i$ $(1le ile 2n)$, such that the sequence $(a_i,a_i+1,dots,a_i+n-1)$ is positive, and the sequence
$a_i+n,a_i+1,dots,a_i+2n-1$ is negative?
It seems to be true for small values of $n$, but I do not see how to prove this in general case. I could also overlook some counterexample.
sequences-and-series combinatorics
sequences-and-series combinatorics
asked Mar 15 at 17:25
useruser
5,68111031
asked Mar 15 at 17:25
useruser
5,68111031
edited Mar 15 at 21:08
user
asked Mar 15 at 17:25
useruser
5,68111031
asked Mar 15 at 17:25
useruser
5,68111031
asked Mar 15 at 17:25
useruser
5,68111031
5,68111031
add a comment |
add a comment |
1 Answer
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$begingroup$
If I've understood your requirements, then I think the sequence with $n = 5$ and initial period $(+1,+1,-1,-1,+1,+1,+1,-1,-1,-1)$ is a counterexample. The partial sums for the subsequences of length $5$ starting at each index from $1$ to $10$ are listed in this table:
$$
beginarrayc
rmIndex&S_1&S_2&S_3&S_4&S_5&rmType\
hline
1&1&2&1&0&1&rmpositive\
2&1&0&-1&0&1\
3&-1&-2&-1&0&1\
4&-1&0&1&2&1\
5&1&2&3&2&1&rmpositive\
6&1&2&1&0&-1\
7&1&0&-1&-2&-1\
8&-1&-2&-3&-2&-1&rmnegative\
9&-1&-2&-1&0&-1&rmnegative\
10&-1&0&1&0&-1
endarray
$$
As you can see, in no case does a negative subsequence appear $5$ steps after a positive subsequence.
I don't know if $n = 5$ is minimal; this sequence was the first thing I tried.
answered Mar 15 at 23:00
FredHFredH
2,6041021
$endgroup$
$begingroup$
You are correct. Now I see that my actual problem is more complicated. Thanks!
$endgroup$
– user
Mar 15 at 23:19
add a comment |
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1 Answer
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$begingroup$
If I've understood your requirements, then I think the sequence with $n = 5$ and initial period $(+1,+1,-1,-1,+1,+1,+1,-1,-1,-1)$ is a counterexample. The partial sums for the subsequences of length $5$ starting at each index from $1$ to $10$ are listed in this table:
$$
beginarrayc
rmIndex&S_1&S_2&S_3&S_4&S_5&rmType\
hline
1&1&2&1&0&1&rmpositive\
2&1&0&-1&0&1\
3&-1&-2&-1&0&1\
4&-1&0&1&2&1\
5&1&2&3&2&1&rmpositive\
6&1&2&1&0&-1\
7&1&0&-1&-2&-1\
8&-1&-2&-3&-2&-1&rmnegative\
9&-1&-2&-1&0&-1&rmnegative\
10&-1&0&1&0&-1
endarray
$$
As you can see, in no case does a negative subsequence appear $5$ steps after a positive subsequence.
I don't know if $n = 5$ is minimal; this sequence was the first thing I tried.
answered Mar 15 at 23:00
FredHFredH
2,6041021
$endgroup$
$begingroup$
You are correct. Now I see that my actual problem is more complicated. Thanks!
$endgroup$
– user
Mar 15 at 23:19
add a comment |
$begingroup$
If I've understood your requirements, then I think the sequence with $n = 5$ and initial period $(+1,+1,-1,-1,+1,+1,+1,-1,-1,-1)$ is a counterexample. The partial sums for the subsequences of length $5$ starting at each index from $1$ to $10$ are listed in this table:
$$
beginarrayc
rmIndex&S_1&S_2&S_3&S_4&S_5&rmType\
hline
1&1&2&1&0&1&rmpositive\
2&1&0&-1&0&1\
3&-1&-2&-1&0&1\
4&-1&0&1&2&1\
5&1&2&3&2&1&rmpositive\
6&1&2&1&0&-1\
7&1&0&-1&-2&-1\
8&-1&-2&-3&-2&-1&rmnegative\
9&-1&-2&-1&0&-1&rmnegative\
10&-1&0&1&0&-1
endarray
$$
As you can see, in no case does a negative subsequence appear $5$ steps after a positive subsequence.
I don't know if $n = 5$ is minimal; this sequence was the first thing I tried.
answered Mar 15 at 23:00
FredHFredH
2,6041021
$endgroup$
$begingroup$
You are correct. Now I see that my actual problem is more complicated. Thanks!
$endgroup$
– user
Mar 15 at 23:19
add a comment |
$begingroup$
If I've understood your requirements, then I think the sequence with $n = 5$ and initial period $(+1,+1,-1,-1,+1,+1,+1,-1,-1,-1)$ is a counterexample. The partial sums for the subsequences of length $5$ starting at each index from $1$ to $10$ are listed in this table:
$$
beginarrayc
rmIndex&S_1&S_2&S_3&S_4&S_5&rmType\
hline
1&1&2&1&0&1&rmpositive\
2&1&0&-1&0&1\
3&-1&-2&-1&0&1\
4&-1&0&1&2&1\
5&1&2&3&2&1&rmpositive\
6&1&2&1&0&-1\
7&1&0&-1&-2&-1\
8&-1&-2&-3&-2&-1&rmnegative\
9&-1&-2&-1&0&-1&rmnegative\
10&-1&0&1&0&-1
endarray
$$
As you can see, in no case does a negative subsequence appear $5$ steps after a positive subsequence.
I don't know if $n = 5$ is minimal; this sequence was the first thing I tried.
answered Mar 15 at 23:00
FredHFredH
2,6041021
$endgroup$
If I've understood your requirements, then I think the sequence with $n = 5$ and initial period $(+1,+1,-1,-1,+1,+1,+1,-1,-1,-1)$ is a counterexample. The partial sums for the subsequences of length $5$ starting at each index from $1$ to $10$ are listed in this table:
$$
beginarrayc
rmIndex&S_1&S_2&S_3&S_4&S_5&rmType\
hline
1&1&2&1&0&1&rmpositive\
2&1&0&-1&0&1\
3&-1&-2&-1&0&1\
4&-1&0&1&2&1\
5&1&2&3&2&1&rmpositive\
6&1&2&1&0&-1\
7&1&0&-1&-2&-1\
8&-1&-2&-3&-2&-1&rmnegative\
9&-1&-2&-1&0&-1&rmnegative\
10&-1&0&1&0&-1
endarray
$$
As you can see, in no case does a negative subsequence appear $5$ steps after a positive subsequence.
I don't know if $n = 5$ is minimal; this sequence was the first thing I tried.
answered Mar 15 at 23:00
FredHFredH
2,6041021
answered Mar 15 at 23:00
FredHFredH
2,6041021
answered Mar 15 at 23:00
FredHFredH
2,6041021
answered Mar 15 at 23:00
FredHFredH
2,6041021
2,6041021
$begingroup$
You are correct. Now I see that my actual problem is more complicated. Thanks!
$endgroup$
– user
Mar 15 at 23:19
add a comment |
$begingroup$
You are correct. Now I see that my actual problem is more complicated. Thanks!
$endgroup$
– user
Mar 15 at 23:19
$begingroup$
You are correct. Now I see that my actual problem is more complicated. Thanks!
$endgroup$
– user
Mar 15 at 23:19
$begingroup$
You are correct. Now I see that my actual problem is more complicated. Thanks!
$endgroup$
– user
Mar 15 at 23:19
add a comment |
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