Finding the generating function to split $n$ into odd parts
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I have been recently working with generating functions in my discrete
mathematics course, and I was interested in one particular generating function.
I want to find the generating function for the number of ways one can split
$n$ into odd parts. I can see that the first coefficients of the sequence are
$$0,1,1,2,2,3,4,5,...$$
And so on. I've been trying to find a recursion, and while it seemed originally
that the number of ways looked to be the ceiling of
$frac{n}{2}.$ But it seems like the $6$th coefficient suggests otherwise. Is
there another possible recursion going on in this sequence? If so, how would I
go about finding the generating function for it?
sequences-and-series discrete-mathematics generating-functions
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add a comment |
$begingroup$
I have been recently working with generating functions in my discrete
mathematics course, and I was interested in one particular generating function.
I want to find the generating function for the number of ways one can split
$n$ into odd parts. I can see that the first coefficients of the sequence are
$$0,1,1,2,2,3,4,5,...$$
And so on. I've been trying to find a recursion, and while it seemed originally
that the number of ways looked to be the ceiling of
$frac{n}{2}.$ But it seems like the $6$th coefficient suggests otherwise. Is
there another possible recursion going on in this sequence? If so, how would I
go about finding the generating function for it?
sequences-and-series discrete-mathematics generating-functions
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3
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The number of partitions of $n$ into odd parts is the same as the number of partitions of $n$ into distinct parts; the generating function is $$prod_{kge 1}left(1+x^kright)=prod_{kge 1}frac1{1-x^{2k-1}}$$ (see for example here).
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– Brian M. Scott
Jun 13 '16 at 22:05
add a comment |
$begingroup$
I have been recently working with generating functions in my discrete
mathematics course, and I was interested in one particular generating function.
I want to find the generating function for the number of ways one can split
$n$ into odd parts. I can see that the first coefficients of the sequence are
$$0,1,1,2,2,3,4,5,...$$
And so on. I've been trying to find a recursion, and while it seemed originally
that the number of ways looked to be the ceiling of
$frac{n}{2}.$ But it seems like the $6$th coefficient suggests otherwise. Is
there another possible recursion going on in this sequence? If so, how would I
go about finding the generating function for it?
sequences-and-series discrete-mathematics generating-functions
$endgroup$
I have been recently working with generating functions in my discrete
mathematics course, and I was interested in one particular generating function.
I want to find the generating function for the number of ways one can split
$n$ into odd parts. I can see that the first coefficients of the sequence are
$$0,1,1,2,2,3,4,5,...$$
And so on. I've been trying to find a recursion, and while it seemed originally
that the number of ways looked to be the ceiling of
$frac{n}{2}.$ But it seems like the $6$th coefficient suggests otherwise. Is
there another possible recursion going on in this sequence? If so, how would I
go about finding the generating function for it?
sequences-and-series discrete-mathematics generating-functions
sequences-and-series discrete-mathematics generating-functions
asked Jun 13 '16 at 22:01
AlexaAlexa
20015
20015
3
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The number of partitions of $n$ into odd parts is the same as the number of partitions of $n$ into distinct parts; the generating function is $$prod_{kge 1}left(1+x^kright)=prod_{kge 1}frac1{1-x^{2k-1}}$$ (see for example here).
$endgroup$
– Brian M. Scott
Jun 13 '16 at 22:05
add a comment |
3
$begingroup$
The number of partitions of $n$ into odd parts is the same as the number of partitions of $n$ into distinct parts; the generating function is $$prod_{kge 1}left(1+x^kright)=prod_{kge 1}frac1{1-x^{2k-1}}$$ (see for example here).
$endgroup$
– Brian M. Scott
Jun 13 '16 at 22:05
3
3
$begingroup$
The number of partitions of $n$ into odd parts is the same as the number of partitions of $n$ into distinct parts; the generating function is $$prod_{kge 1}left(1+x^kright)=prod_{kge 1}frac1{1-x^{2k-1}}$$ (see for example here).
$endgroup$
– Brian M. Scott
Jun 13 '16 at 22:05
$begingroup$
The number of partitions of $n$ into odd parts is the same as the number of partitions of $n$ into distinct parts; the generating function is $$prod_{kge 1}left(1+x^kright)=prod_{kge 1}frac1{1-x^{2k-1}}$$ (see for example here).
$endgroup$
– Brian M. Scott
Jun 13 '16 at 22:05
add a comment |
1 Answer
1
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The generating function for this sequence is
$f(x)=prodlimits_{n=1}^{infty} cfrac{1}{1-x^{2n-1}}$
since
$f(x)=(1+x+x^{1+1}+cdots)(1+x^3+x^{3+3}+cdots)cdots (1+x^{2n-1}+x^{(2n-1)+(2n-1)})cdots = (1+x+x^2+cdots)(1+x^3+x^{2cdot3}+cdots)cdots (1+x^{2n-1}+x^{2(2n-1)})cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdotscfrac{1}{1-x^{2n-1}}dots=prodlimits_{n=1}^{infty} cfrac{1}{1-x^{2n-1}}$
Moreover, you can express this function as follows,
$prodlimits_{n=1}^{infty} (1+x^n)=(1+x)(1+x^2)(1+x^3)cdots=cfrac{1-x^2}{1-x}cfrac{1-x^4}{1-x^2}cfrac{1-x^6}{1-x^3}cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdots=f(x)$
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The explanation of the last expression of the function $f(x)$ needs more details, but I think it' s fine just like this.
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– richarddedekind
Jun 13 '16 at 22:29
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
The generating function for this sequence is
$f(x)=prodlimits_{n=1}^{infty} cfrac{1}{1-x^{2n-1}}$
since
$f(x)=(1+x+x^{1+1}+cdots)(1+x^3+x^{3+3}+cdots)cdots (1+x^{2n-1}+x^{(2n-1)+(2n-1)})cdots = (1+x+x^2+cdots)(1+x^3+x^{2cdot3}+cdots)cdots (1+x^{2n-1}+x^{2(2n-1)})cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdotscfrac{1}{1-x^{2n-1}}dots=prodlimits_{n=1}^{infty} cfrac{1}{1-x^{2n-1}}$
Moreover, you can express this function as follows,
$prodlimits_{n=1}^{infty} (1+x^n)=(1+x)(1+x^2)(1+x^3)cdots=cfrac{1-x^2}{1-x}cfrac{1-x^4}{1-x^2}cfrac{1-x^6}{1-x^3}cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdots=f(x)$
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The explanation of the last expression of the function $f(x)$ needs more details, but I think it' s fine just like this.
$endgroup$
– richarddedekind
Jun 13 '16 at 22:29
add a comment |
$begingroup$
The generating function for this sequence is
$f(x)=prodlimits_{n=1}^{infty} cfrac{1}{1-x^{2n-1}}$
since
$f(x)=(1+x+x^{1+1}+cdots)(1+x^3+x^{3+3}+cdots)cdots (1+x^{2n-1}+x^{(2n-1)+(2n-1)})cdots = (1+x+x^2+cdots)(1+x^3+x^{2cdot3}+cdots)cdots (1+x^{2n-1}+x^{2(2n-1)})cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdotscfrac{1}{1-x^{2n-1}}dots=prodlimits_{n=1}^{infty} cfrac{1}{1-x^{2n-1}}$
Moreover, you can express this function as follows,
$prodlimits_{n=1}^{infty} (1+x^n)=(1+x)(1+x^2)(1+x^3)cdots=cfrac{1-x^2}{1-x}cfrac{1-x^4}{1-x^2}cfrac{1-x^6}{1-x^3}cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdots=f(x)$
$endgroup$
$begingroup$
The explanation of the last expression of the function $f(x)$ needs more details, but I think it' s fine just like this.
$endgroup$
– richarddedekind
Jun 13 '16 at 22:29
add a comment |
$begingroup$
The generating function for this sequence is
$f(x)=prodlimits_{n=1}^{infty} cfrac{1}{1-x^{2n-1}}$
since
$f(x)=(1+x+x^{1+1}+cdots)(1+x^3+x^{3+3}+cdots)cdots (1+x^{2n-1}+x^{(2n-1)+(2n-1)})cdots = (1+x+x^2+cdots)(1+x^3+x^{2cdot3}+cdots)cdots (1+x^{2n-1}+x^{2(2n-1)})cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdotscfrac{1}{1-x^{2n-1}}dots=prodlimits_{n=1}^{infty} cfrac{1}{1-x^{2n-1}}$
Moreover, you can express this function as follows,
$prodlimits_{n=1}^{infty} (1+x^n)=(1+x)(1+x^2)(1+x^3)cdots=cfrac{1-x^2}{1-x}cfrac{1-x^4}{1-x^2}cfrac{1-x^6}{1-x^3}cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdots=f(x)$
$endgroup$
The generating function for this sequence is
$f(x)=prodlimits_{n=1}^{infty} cfrac{1}{1-x^{2n-1}}$
since
$f(x)=(1+x+x^{1+1}+cdots)(1+x^3+x^{3+3}+cdots)cdots (1+x^{2n-1}+x^{(2n-1)+(2n-1)})cdots = (1+x+x^2+cdots)(1+x^3+x^{2cdot3}+cdots)cdots (1+x^{2n-1}+x^{2(2n-1)})cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdotscfrac{1}{1-x^{2n-1}}dots=prodlimits_{n=1}^{infty} cfrac{1}{1-x^{2n-1}}$
Moreover, you can express this function as follows,
$prodlimits_{n=1}^{infty} (1+x^n)=(1+x)(1+x^2)(1+x^3)cdots=cfrac{1-x^2}{1-x}cfrac{1-x^4}{1-x^2}cfrac{1-x^6}{1-x^3}cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdots=f(x)$
edited Dec 2 '18 at 2:18
answered Jun 13 '16 at 22:25
richarddedekindricharddedekind
711316
711316
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The explanation of the last expression of the function $f(x)$ needs more details, but I think it' s fine just like this.
$endgroup$
– richarddedekind
Jun 13 '16 at 22:29
add a comment |
$begingroup$
The explanation of the last expression of the function $f(x)$ needs more details, but I think it' s fine just like this.
$endgroup$
– richarddedekind
Jun 13 '16 at 22:29
$begingroup$
The explanation of the last expression of the function $f(x)$ needs more details, but I think it' s fine just like this.
$endgroup$
– richarddedekind
Jun 13 '16 at 22:29
$begingroup$
The explanation of the last expression of the function $f(x)$ needs more details, but I think it' s fine just like this.
$endgroup$
– richarddedekind
Jun 13 '16 at 22:29
add a comment |
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$begingroup$
The number of partitions of $n$ into odd parts is the same as the number of partitions of $n$ into distinct parts; the generating function is $$prod_{kge 1}left(1+x^kright)=prod_{kge 1}frac1{1-x^{2k-1}}$$ (see for example here).
$endgroup$
– Brian M. Scott
Jun 13 '16 at 22:05