Closed formula of $(1+x_1)(1+x_1+x_2)^2dots(1+x_1+dots+x_n)^n$












1












$begingroup$


I am working on a larger problem which I have managed to reduce to a pretty neat generating function. The answer will be given by the coefficient of $x_1x_2dots x_n$ in



$F(x_1,dots,x_n)=(1+x_1)(1+x_1+x_2)^2dots (1+x_1+x_2+dots+x_n)^n$



Now, I have two questions about this function:




  1. I have been trying desperately to find a closed formula for this function, but without success. Am I missing something very obvious or is this as good as it gets?

  2. Is there a nice expression for the coefficient (call it $a_n$) of $x_1x_2,dots,x_n$? Here I might have made some progress. By writing each $(1+x_1+dots+x_i)^i$ as a multinomial I have been able to come up with this expression (hopefully correct...):
    begin{align}
    sum_{k_1+dots+k_n=n}left[binom{1}{k_1}binom{2}{k_2}dots binom{n}{k_n}(2-k_1)(3-(k_1+k_2))dots(n-(k_1+dots+k_{n-1})) right]
    end{align}
    but I don't seem to be able to reduce it to something nicer.


Any hints would be very much appreciated!






combinatorics generating-functions







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  • $begingroup$
    To save future readers some effort, I note that at present this sequence is not in OEIS.
    $endgroup$
    – Peter Taylor
    Dec 20 '18 at 13:02
















1












$begingroup$


I am working on a larger problem which I have managed to reduce to a pretty neat generating function. The answer will be given by the coefficient of $x_1x_2dots x_n$ in



$F(x_1,dots,x_n)=(1+x_1)(1+x_1+x_2)^2dots (1+x_1+x_2+dots+x_n)^n$



Now, I have two questions about this function:




  1. I have been trying desperately to find a closed formula for this function, but without success. Am I missing something very obvious or is this as good as it gets?

  2. Is there a nice expression for the coefficient (call it $a_n$) of $x_1x_2,dots,x_n$? Here I might have made some progress. By writing each $(1+x_1+dots+x_i)^i$ as a multinomial I have been able to come up with this expression (hopefully correct...):
    begin{align}
    sum_{k_1+dots+k_n=n}left[binom{1}{k_1}binom{2}{k_2}dots binom{n}{k_n}(2-k_1)(3-(k_1+k_2))dots(n-(k_1+dots+k_{n-1})) right]
    end{align}
    but I don't seem to be able to reduce it to something nicer.


Any hints would be very much appreciated!






combinatorics generating-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    To save future readers some effort, I note that at present this sequence is not in OEIS.
    $endgroup$
    – Peter Taylor
    Dec 20 '18 at 13:02














1












1








1


1



$begingroup$


I am working on a larger problem which I have managed to reduce to a pretty neat generating function. The answer will be given by the coefficient of $x_1x_2dots x_n$ in



$F(x_1,dots,x_n)=(1+x_1)(1+x_1+x_2)^2dots (1+x_1+x_2+dots+x_n)^n$



Now, I have two questions about this function:




  1. I have been trying desperately to find a closed formula for this function, but without success. Am I missing something very obvious or is this as good as it gets?

  2. Is there a nice expression for the coefficient (call it $a_n$) of $x_1x_2,dots,x_n$? Here I might have made some progress. By writing each $(1+x_1+dots+x_i)^i$ as a multinomial I have been able to come up with this expression (hopefully correct...):
    begin{align}
    sum_{k_1+dots+k_n=n}left[binom{1}{k_1}binom{2}{k_2}dots binom{n}{k_n}(2-k_1)(3-(k_1+k_2))dots(n-(k_1+dots+k_{n-1})) right]
    end{align}
    but I don't seem to be able to reduce it to something nicer.


Any hints would be very much appreciated!






combinatorics generating-functions







share|cite|improve this question









$endgroup$




I am working on a larger problem which I have managed to reduce to a pretty neat generating function. The answer will be given by the coefficient of $x_1x_2dots x_n$ in



$F(x_1,dots,x_n)=(1+x_1)(1+x_1+x_2)^2dots (1+x_1+x_2+dots+x_n)^n$



Now, I have two questions about this function:




  1. I have been trying desperately to find a closed formula for this function, but without success. Am I missing something very obvious or is this as good as it gets?

  2. Is there a nice expression for the coefficient (call it $a_n$) of $x_1x_2,dots,x_n$? Here I might have made some progress. By writing each $(1+x_1+dots+x_i)^i$ as a multinomial I have been able to come up with this expression (hopefully correct...):
    begin{align}
    sum_{k_1+dots+k_n=n}left[binom{1}{k_1}binom{2}{k_2}dots binom{n}{k_n}(2-k_1)(3-(k_1+k_2))dots(n-(k_1+dots+k_{n-1})) right]
    end{align}
    but I don't seem to be able to reduce it to something nicer.


Any hints would be very much appreciated!






combinatorics generating-functions




combinatorics generating-functions






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 20 '18 at 11:39









KurtKnödelKurtKnödel

474




474












  • $begingroup$
    To save future readers some effort, I note that at present this sequence is not in OEIS.
    $endgroup$
    – Peter Taylor
    Dec 20 '18 at 13:02


















  • $begingroup$
    To save future readers some effort, I note that at present this sequence is not in OEIS.
    $endgroup$
    – Peter Taylor
    Dec 20 '18 at 13:02
















$begingroup$
To save future readers some effort, I note that at present this sequence is not in OEIS.
$endgroup$
– Peter Taylor
Dec 20 '18 at 13:02




$begingroup$
To save future readers some effort, I note that at present this sequence is not in OEIS.
$endgroup$
– Peter Taylor
Dec 20 '18 at 13:02










2 Answers
2






active

oldest

votes


















1












$begingroup$

To obtain the monomial $x_1x_2cdots x_n$ from the product
$$
p = (1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_n)^n
$$
you have to pick the $x_n$ from one of the $(1+x_1+cdots+x_n)$-factors and there are $n$ factors to choose from. Now you are left with the coefficient of $x_1x_2cdots x_{n-1}$ in
$$
(1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_n)^{n-1}
$$
which is the same as the coefficient in
$$
(1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_{n-2})^{n-2}(1+x_1+cdots+x_{n-1})^{2n-2}.
$$
Now you have to pick the $x_{n-1}$ from one of the $(1+x_1+cdots+x_{n-1})$-factors and there are $2n-2$ to pick from.



Continuing this line of thought you get that the coefficient of $x_1x_2cdots x_n$ in $p$ is
$$
n(2n-2)(3n-5)(4n-9)...(nn-*).
$$
Note that by construction the constants $0,2,5,9,dots$ in this product are the sums
$$
0, 2, 2+3, 2+3+4, dots
$$
which are given by
$$
sum_{i=2}^k i = frac{k(k+1)}{2} - 1 = frac{(k-1)(k+2)}{2}.
$$
Hence, the desired coefficient is
$$
prod_{k=1}^n left(kn - frac{(k-1)(k+2)}{2}right).
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Aah, of course. I can't believe I couldn't figure this out myself. Thank you very much!
    $endgroup$
    – KurtKnödel
    Dec 20 '18 at 12:56



















1












$begingroup$

Write out all $n(n+1)/2$ factors in a row.

The $x_n$ comes from one of the final $n$ factors, the $x_{n-1}$ from one of the final $n+(n-1)$ factors but not the one used for $x_n$, and so on.

The product is
$$n(2n-2)(3n-5)(4n-9)(5n-14)...$$
But I can't help you with that.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, it all makes sense now:)
    $endgroup$
    – KurtKnödel
    Dec 20 '18 at 12:56











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

To obtain the monomial $x_1x_2cdots x_n$ from the product
$$
p = (1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_n)^n
$$
you have to pick the $x_n$ from one of the $(1+x_1+cdots+x_n)$-factors and there are $n$ factors to choose from. Now you are left with the coefficient of $x_1x_2cdots x_{n-1}$ in
$$
(1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_n)^{n-1}
$$
which is the same as the coefficient in
$$
(1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_{n-2})^{n-2}(1+x_1+cdots+x_{n-1})^{2n-2}.
$$
Now you have to pick the $x_{n-1}$ from one of the $(1+x_1+cdots+x_{n-1})$-factors and there are $2n-2$ to pick from.



Continuing this line of thought you get that the coefficient of $x_1x_2cdots x_n$ in $p$ is
$$
n(2n-2)(3n-5)(4n-9)...(nn-*).
$$
Note that by construction the constants $0,2,5,9,dots$ in this product are the sums
$$
0, 2, 2+3, 2+3+4, dots
$$
which are given by
$$
sum_{i=2}^k i = frac{k(k+1)}{2} - 1 = frac{(k-1)(k+2)}{2}.
$$
Hence, the desired coefficient is
$$
prod_{k=1}^n left(kn - frac{(k-1)(k+2)}{2}right).
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Aah, of course. I can't believe I couldn't figure this out myself. Thank you very much!
    $endgroup$
    – KurtKnödel
    Dec 20 '18 at 12:56
















1












$begingroup$

To obtain the monomial $x_1x_2cdots x_n$ from the product
$$
p = (1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_n)^n
$$
you have to pick the $x_n$ from one of the $(1+x_1+cdots+x_n)$-factors and there are $n$ factors to choose from. Now you are left with the coefficient of $x_1x_2cdots x_{n-1}$ in
$$
(1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_n)^{n-1}
$$
which is the same as the coefficient in
$$
(1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_{n-2})^{n-2}(1+x_1+cdots+x_{n-1})^{2n-2}.
$$
Now you have to pick the $x_{n-1}$ from one of the $(1+x_1+cdots+x_{n-1})$-factors and there are $2n-2$ to pick from.



Continuing this line of thought you get that the coefficient of $x_1x_2cdots x_n$ in $p$ is
$$
n(2n-2)(3n-5)(4n-9)...(nn-*).
$$
Note that by construction the constants $0,2,5,9,dots$ in this product are the sums
$$
0, 2, 2+3, 2+3+4, dots
$$
which are given by
$$
sum_{i=2}^k i = frac{k(k+1)}{2} - 1 = frac{(k-1)(k+2)}{2}.
$$
Hence, the desired coefficient is
$$
prod_{k=1}^n left(kn - frac{(k-1)(k+2)}{2}right).
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Aah, of course. I can't believe I couldn't figure this out myself. Thank you very much!
    $endgroup$
    – KurtKnödel
    Dec 20 '18 at 12:56














1












1








1





$begingroup$

To obtain the monomial $x_1x_2cdots x_n$ from the product
$$
p = (1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_n)^n
$$
you have to pick the $x_n$ from one of the $(1+x_1+cdots+x_n)$-factors and there are $n$ factors to choose from. Now you are left with the coefficient of $x_1x_2cdots x_{n-1}$ in
$$
(1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_n)^{n-1}
$$
which is the same as the coefficient in
$$
(1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_{n-2})^{n-2}(1+x_1+cdots+x_{n-1})^{2n-2}.
$$
Now you have to pick the $x_{n-1}$ from one of the $(1+x_1+cdots+x_{n-1})$-factors and there are $2n-2$ to pick from.



Continuing this line of thought you get that the coefficient of $x_1x_2cdots x_n$ in $p$ is
$$
n(2n-2)(3n-5)(4n-9)...(nn-*).
$$
Note that by construction the constants $0,2,5,9,dots$ in this product are the sums
$$
0, 2, 2+3, 2+3+4, dots
$$
which are given by
$$
sum_{i=2}^k i = frac{k(k+1)}{2} - 1 = frac{(k-1)(k+2)}{2}.
$$
Hence, the desired coefficient is
$$
prod_{k=1}^n left(kn - frac{(k-1)(k+2)}{2}right).
$$






share|cite|improve this answer











$endgroup$



To obtain the monomial $x_1x_2cdots x_n$ from the product
$$
p = (1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_n)^n
$$
you have to pick the $x_n$ from one of the $(1+x_1+cdots+x_n)$-factors and there are $n$ factors to choose from. Now you are left with the coefficient of $x_1x_2cdots x_{n-1}$ in
$$
(1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_n)^{n-1}
$$
which is the same as the coefficient in
$$
(1+x_1)(1+x_1+x_2)^2cdots (1+x_1+cdots+x_{n-2})^{n-2}(1+x_1+cdots+x_{n-1})^{2n-2}.
$$
Now you have to pick the $x_{n-1}$ from one of the $(1+x_1+cdots+x_{n-1})$-factors and there are $2n-2$ to pick from.



Continuing this line of thought you get that the coefficient of $x_1x_2cdots x_n$ in $p$ is
$$
n(2n-2)(3n-5)(4n-9)...(nn-*).
$$
Note that by construction the constants $0,2,5,9,dots$ in this product are the sums
$$
0, 2, 2+3, 2+3+4, dots
$$
which are given by
$$
sum_{i=2}^k i = frac{k(k+1)}{2} - 1 = frac{(k-1)(k+2)}{2}.
$$
Hence, the desired coefficient is
$$
prod_{k=1}^n left(kn - frac{(k-1)(k+2)}{2}right).
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 12:18

























answered Dec 20 '18 at 12:11









ChristophChristoph

12.5k1642




12.5k1642












  • $begingroup$
    Aah, of course. I can't believe I couldn't figure this out myself. Thank you very much!
    $endgroup$
    – KurtKnödel
    Dec 20 '18 at 12:56


















  • $begingroup$
    Aah, of course. I can't believe I couldn't figure this out myself. Thank you very much!
    $endgroup$
    – KurtKnödel
    Dec 20 '18 at 12:56
















$begingroup$
Aah, of course. I can't believe I couldn't figure this out myself. Thank you very much!
$endgroup$
– KurtKnödel
Dec 20 '18 at 12:56




$begingroup$
Aah, of course. I can't believe I couldn't figure this out myself. Thank you very much!
$endgroup$
– KurtKnödel
Dec 20 '18 at 12:56











1












$begingroup$

Write out all $n(n+1)/2$ factors in a row.

The $x_n$ comes from one of the final $n$ factors, the $x_{n-1}$ from one of the final $n+(n-1)$ factors but not the one used for $x_n$, and so on.

The product is
$$n(2n-2)(3n-5)(4n-9)(5n-14)...$$
But I can't help you with that.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, it all makes sense now:)
    $endgroup$
    – KurtKnödel
    Dec 20 '18 at 12:56
















1












$begingroup$

Write out all $n(n+1)/2$ factors in a row.

The $x_n$ comes from one of the final $n$ factors, the $x_{n-1}$ from one of the final $n+(n-1)$ factors but not the one used for $x_n$, and so on.

The product is
$$n(2n-2)(3n-5)(4n-9)(5n-14)...$$
But I can't help you with that.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, it all makes sense now:)
    $endgroup$
    – KurtKnödel
    Dec 20 '18 at 12:56














1












1








1





$begingroup$

Write out all $n(n+1)/2$ factors in a row.

The $x_n$ comes from one of the final $n$ factors, the $x_{n-1}$ from one of the final $n+(n-1)$ factors but not the one used for $x_n$, and so on.

The product is
$$n(2n-2)(3n-5)(4n-9)(5n-14)...$$
But I can't help you with that.






share|cite|improve this answer









$endgroup$



Write out all $n(n+1)/2$ factors in a row.

The $x_n$ comes from one of the final $n$ factors, the $x_{n-1}$ from one of the final $n+(n-1)$ factors but not the one used for $x_n$, and so on.

The product is
$$n(2n-2)(3n-5)(4n-9)(5n-14)...$$
But I can't help you with that.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 12:04









Empy2Empy2

33.6k12462




33.6k12462












  • $begingroup$
    Thank you, it all makes sense now:)
    $endgroup$
    – KurtKnödel
    Dec 20 '18 at 12:56


















  • $begingroup$
    Thank you, it all makes sense now:)
    $endgroup$
    – KurtKnödel
    Dec 20 '18 at 12:56
















$begingroup$
Thank you, it all makes sense now:)
$endgroup$
– KurtKnödel
Dec 20 '18 at 12:56




$begingroup$
Thank you, it all makes sense now:)
$endgroup$
– KurtKnödel
Dec 20 '18 at 12:56


















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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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