$f(x)(1+2x+2x^2+x^3)=frac{1}{(1-x)^3}$












-3












$begingroup$


We have $f(x) = sum_{i=0}^infty a_ix^i$. We know that $f(x)(1+2x+2x^2+x^3)=frac{1}{(1-x)^3}$.
We need to find $a_0, a_1, a_2$.
Also we need to find numbers $r, s, t$ from $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3.
I already get that $f(x) =sum_{i=0}^infty x^i times sum_{k=0}^infty x^{2k} times sum_{j=0}^infty x^{3j}$.
I am sure that this problem is connected to this.
Also there is a problem connected with two first ones: To write generating function for calculating number of integer solutions of the equation: $x_1 + 2x_2 + 3x_3 = n$.
Thank you and merry Christmas!






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    you can use Partal Fraction Decomposition
    $endgroup$
    – miracle173
    Dec 24 '18 at 8:17


















-3












$begingroup$


We have $f(x) = sum_{i=0}^infty a_ix^i$. We know that $f(x)(1+2x+2x^2+x^3)=frac{1}{(1-x)^3}$.
We need to find $a_0, a_1, a_2$.
Also we need to find numbers $r, s, t$ from $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3.
I already get that $f(x) =sum_{i=0}^infty x^i times sum_{k=0}^infty x^{2k} times sum_{j=0}^infty x^{3j}$.
I am sure that this problem is connected to this.
Also there is a problem connected with two first ones: To write generating function for calculating number of integer solutions of the equation: $x_1 + 2x_2 + 3x_3 = n$.
Thank you and merry Christmas!






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    you can use Partal Fraction Decomposition
    $endgroup$
    – miracle173
    Dec 24 '18 at 8:17
















-3












-3








-3


1



$begingroup$


We have $f(x) = sum_{i=0}^infty a_ix^i$. We know that $f(x)(1+2x+2x^2+x^3)=frac{1}{(1-x)^3}$.
We need to find $a_0, a_1, a_2$.
Also we need to find numbers $r, s, t$ from $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3.
I already get that $f(x) =sum_{i=0}^infty x^i times sum_{k=0}^infty x^{2k} times sum_{j=0}^infty x^{3j}$.
I am sure that this problem is connected to this.
Also there is a problem connected with two first ones: To write generating function for calculating number of integer solutions of the equation: $x_1 + 2x_2 + 3x_3 = n$.
Thank you and merry Christmas!






combinatorics







share|cite|improve this question











$endgroup$




We have $f(x) = sum_{i=0}^infty a_ix^i$. We know that $f(x)(1+2x+2x^2+x^3)=frac{1}{(1-x)^3}$.
We need to find $a_0, a_1, a_2$.
Also we need to find numbers $r, s, t$ from $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3.
I already get that $f(x) =sum_{i=0}^infty x^i times sum_{k=0}^infty x^{2k} times sum_{j=0}^infty x^{3j}$.
I am sure that this problem is connected to this.
Also there is a problem connected with two first ones: To write generating function for calculating number of integer solutions of the equation: $x_1 + 2x_2 + 3x_3 = n$.
Thank you and merry Christmas!






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 8:24







O.Joen

















asked Dec 24 '18 at 0:08









O.JoenO.Joen

13




13












  • $begingroup$
    you can use Partal Fraction Decomposition
    $endgroup$
    – miracle173
    Dec 24 '18 at 8:17




















  • $begingroup$
    you can use Partal Fraction Decomposition
    $endgroup$
    – miracle173
    Dec 24 '18 at 8:17


















$begingroup$
you can use Partal Fraction Decomposition
$endgroup$
– miracle173
Dec 24 '18 at 8:17






$begingroup$
you can use Partal Fraction Decomposition
$endgroup$
– miracle173
Dec 24 '18 at 8:17












2 Answers
2






active

oldest

votes


















0












$begingroup$

For a start.



Consider that
$$f(x)(1+2x+2x^2+x^3)(1-x)^3=1$$ must hold for any $x$ that is say
$$(1-x-x^2+x^4+x^5-x^6)f(x)=1$$ Replace $f(x)$ by a few terms (since you just want $a_0,a_1,a_2$) and identify the coefficients.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Your finding is correct. Indeed:
    $$f(x)(1+2x+2x^2+x^3)=frac{1}{(1-x)^3} Rightarrow \\
    f(x)(1+x)(1+x+x^2)=frac{1}{(1-x)^3} Rightarrow \
    f(x)=frac{1}{(1-x)(1-x^2)(1-x^3)}=sum_{i=0}^{infty} x^isum_{j=0}^{infty} x^{2j}sum_{k=0}^{infty} x^{3k} Rightarrow\
    i+2j+3k=n, i,j,k,nge 0.$$
    So, if you need to find a few terms:
    $$begin{align}n&=0Rightarrow (0,0,0) Rightarrow a_0=1;\
    n&=1 Rightarrow (1,0,0) Rightarrow a_1=1;\
    n&=2 Rightarrow (2,0,0), (0,1,0) Rightarrow a_2=2;\
    n&=3 Rightarrow (3,0,0), (1,1,0), (0,0,1) Rightarrow a_3=3.end{align}$$
    Wolfram answer.



    For more, you can see here.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you very much for your comment. Could I have one more minute of your attention for explaining how can i get to $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3?
      $endgroup$
      – O.Joen
      Dec 24 '18 at 10:03











    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    0












    $begingroup$

    For a start.



    Consider that
    $$f(x)(1+2x+2x^2+x^3)(1-x)^3=1$$ must hold for any $x$ that is say
    $$(1-x-x^2+x^4+x^5-x^6)f(x)=1$$ Replace $f(x)$ by a few terms (since you just want $a_0,a_1,a_2$) and identify the coefficients.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      For a start.



      Consider that
      $$f(x)(1+2x+2x^2+x^3)(1-x)^3=1$$ must hold for any $x$ that is say
      $$(1-x-x^2+x^4+x^5-x^6)f(x)=1$$ Replace $f(x)$ by a few terms (since you just want $a_0,a_1,a_2$) and identify the coefficients.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        For a start.



        Consider that
        $$f(x)(1+2x+2x^2+x^3)(1-x)^3=1$$ must hold for any $x$ that is say
        $$(1-x-x^2+x^4+x^5-x^6)f(x)=1$$ Replace $f(x)$ by a few terms (since you just want $a_0,a_1,a_2$) and identify the coefficients.






        share|cite|improve this answer









        $endgroup$



        For a start.



        Consider that
        $$f(x)(1+2x+2x^2+x^3)(1-x)^3=1$$ must hold for any $x$ that is say
        $$(1-x-x^2+x^4+x^5-x^6)f(x)=1$$ Replace $f(x)$ by a few terms (since you just want $a_0,a_1,a_2$) and identify the coefficients.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 7:44









        Claude LeiboviciClaude Leibovici

        122k1157134




        122k1157134























            0












            $begingroup$

            Your finding is correct. Indeed:
            $$f(x)(1+2x+2x^2+x^3)=frac{1}{(1-x)^3} Rightarrow \\
            f(x)(1+x)(1+x+x^2)=frac{1}{(1-x)^3} Rightarrow \
            f(x)=frac{1}{(1-x)(1-x^2)(1-x^3)}=sum_{i=0}^{infty} x^isum_{j=0}^{infty} x^{2j}sum_{k=0}^{infty} x^{3k} Rightarrow\
            i+2j+3k=n, i,j,k,nge 0.$$
            So, if you need to find a few terms:
            $$begin{align}n&=0Rightarrow (0,0,0) Rightarrow a_0=1;\
            n&=1 Rightarrow (1,0,0) Rightarrow a_1=1;\
            n&=2 Rightarrow (2,0,0), (0,1,0) Rightarrow a_2=2;\
            n&=3 Rightarrow (3,0,0), (1,1,0), (0,0,1) Rightarrow a_3=3.end{align}$$
            Wolfram answer.



            For more, you can see here.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much for your comment. Could I have one more minute of your attention for explaining how can i get to $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3?
              $endgroup$
              – O.Joen
              Dec 24 '18 at 10:03
















            0












            $begingroup$

            Your finding is correct. Indeed:
            $$f(x)(1+2x+2x^2+x^3)=frac{1}{(1-x)^3} Rightarrow \\
            f(x)(1+x)(1+x+x^2)=frac{1}{(1-x)^3} Rightarrow \
            f(x)=frac{1}{(1-x)(1-x^2)(1-x^3)}=sum_{i=0}^{infty} x^isum_{j=0}^{infty} x^{2j}sum_{k=0}^{infty} x^{3k} Rightarrow\
            i+2j+3k=n, i,j,k,nge 0.$$
            So, if you need to find a few terms:
            $$begin{align}n&=0Rightarrow (0,0,0) Rightarrow a_0=1;\
            n&=1 Rightarrow (1,0,0) Rightarrow a_1=1;\
            n&=2 Rightarrow (2,0,0), (0,1,0) Rightarrow a_2=2;\
            n&=3 Rightarrow (3,0,0), (1,1,0), (0,0,1) Rightarrow a_3=3.end{align}$$
            Wolfram answer.



            For more, you can see here.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you very much for your comment. Could I have one more minute of your attention for explaining how can i get to $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3?
              $endgroup$
              – O.Joen
              Dec 24 '18 at 10:03














            0












            0








            0





            $begingroup$

            Your finding is correct. Indeed:
            $$f(x)(1+2x+2x^2+x^3)=frac{1}{(1-x)^3} Rightarrow \\
            f(x)(1+x)(1+x+x^2)=frac{1}{(1-x)^3} Rightarrow \
            f(x)=frac{1}{(1-x)(1-x^2)(1-x^3)}=sum_{i=0}^{infty} x^isum_{j=0}^{infty} x^{2j}sum_{k=0}^{infty} x^{3k} Rightarrow\
            i+2j+3k=n, i,j,k,nge 0.$$
            So, if you need to find a few terms:
            $$begin{align}n&=0Rightarrow (0,0,0) Rightarrow a_0=1;\
            n&=1 Rightarrow (1,0,0) Rightarrow a_1=1;\
            n&=2 Rightarrow (2,0,0), (0,1,0) Rightarrow a_2=2;\
            n&=3 Rightarrow (3,0,0), (1,1,0), (0,0,1) Rightarrow a_3=3.end{align}$$
            Wolfram answer.



            For more, you can see here.






            share|cite|improve this answer









            $endgroup$



            Your finding is correct. Indeed:
            $$f(x)(1+2x+2x^2+x^3)=frac{1}{(1-x)^3} Rightarrow \\
            f(x)(1+x)(1+x+x^2)=frac{1}{(1-x)^3} Rightarrow \
            f(x)=frac{1}{(1-x)(1-x^2)(1-x^3)}=sum_{i=0}^{infty} x^isum_{j=0}^{infty} x^{2j}sum_{k=0}^{infty} x^{3k} Rightarrow\
            i+2j+3k=n, i,j,k,nge 0.$$
            So, if you need to find a few terms:
            $$begin{align}n&=0Rightarrow (0,0,0) Rightarrow a_0=1;\
            n&=1 Rightarrow (1,0,0) Rightarrow a_1=1;\
            n&=2 Rightarrow (2,0,0), (0,1,0) Rightarrow a_2=2;\
            n&=3 Rightarrow (3,0,0), (1,1,0), (0,0,1) Rightarrow a_3=3.end{align}$$
            Wolfram answer.



            For more, you can see here.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 24 '18 at 9:47









            farruhotafarruhota

            20.4k2739




            20.4k2739












            • $begingroup$
              Thank you very much for your comment. Could I have one more minute of your attention for explaining how can i get to $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3?
              $endgroup$
              – O.Joen
              Dec 24 '18 at 10:03


















            • $begingroup$
              Thank you very much for your comment. Could I have one more minute of your attention for explaining how can i get to $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3?
              $endgroup$
              – O.Joen
              Dec 24 '18 at 10:03
















            $begingroup$
            Thank you very much for your comment. Could I have one more minute of your attention for explaining how can i get to $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3?
            $endgroup$
            – O.Joen
            Dec 24 '18 at 10:03




            $begingroup$
            Thank you very much for your comment. Could I have one more minute of your attention for explaining how can i get to $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3?
            $endgroup$
            – O.Joen
            Dec 24 '18 at 10:03


















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