Finding the generating function to split $n$ into odd parts












0












$begingroup$


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




    $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jun 13 '16 at 22:01









AlexaAlexa

20015




20015








  • 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




    $begingroup$
    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










1 Answer
1






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oldest

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3












$begingroup$

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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  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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3












$begingroup$

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,



$prodlimits_{n=1}^{infty} (1+x^n)=(1+x)(1+x^2)(1+x^3)cdots=cfrac{1-x^2}{1-x}cfrac{1-x^4}{1-x^2}cfrac{1-x^6}{1-x^3}cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdots=f(x)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












$begingroup$

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,



$prodlimits_{n=1}^{infty} (1+x^n)=(1+x)(1+x^2)(1+x^3)cdots=cfrac{1-x^2}{1-x}cfrac{1-x^4}{1-x^2}cfrac{1-x^6}{1-x^3}cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdots=f(x)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

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,



$prodlimits_{n=1}^{infty} (1+x^n)=(1+x)(1+x^2)(1+x^3)cdots=cfrac{1-x^2}{1-x}cfrac{1-x^4}{1-x^2}cfrac{1-x^6}{1-x^3}cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdots=f(x)$






share|cite|improve this answer











$endgroup$



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,



$prodlimits_{n=1}^{infty} (1+x^n)=(1+x)(1+x^2)(1+x^3)cdots=cfrac{1-x^2}{1-x}cfrac{1-x^4}{1-x^2}cfrac{1-x^6}{1-x^3}cdots=cfrac{1}{1-x}cfrac{1}{1-x^3}cdots=f(x)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 2 '18 at 2:18

























answered Jun 13 '16 at 22:25









richarddedekindricharddedekind

711316




711316












  • $begingroup$
    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




$begingroup$
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


















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