Are these numbers known?

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Are the numbers appearing as coefficients in the following sequence of polynomials known? Is there a known recurrence relation to compute them?



beginalign*
(x+y) &= x+y \
(x+y)(x+2y) &= x^2+3yx+2y^2 \
(x+y)(x+2y)(x+3y) &= x^3+6yx^2+11y^2x+6y^3 \
(x+y)(x+2y)(x+3y)(x+4y) &= x^4+10yx^3+35y^2x^2+50y^3x+24y^4 \
&textetc.
endalign*






algebra-precalculus







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  • 3




    Looks like en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind, see also oeis.org/A094638.
    – Martin R
    1 hour ago















up vote
5
down vote

favorite












Are the numbers appearing as coefficients in the following sequence of polynomials known? Is there a known recurrence relation to compute them?



beginalign*
(x+y) &= x+y \
(x+y)(x+2y) &= x^2+3yx+2y^2 \
(x+y)(x+2y)(x+3y) &= x^3+6yx^2+11y^2x+6y^3 \
(x+y)(x+2y)(x+3y)(x+4y) &= x^4+10yx^3+35y^2x^2+50y^3x+24y^4 \
&textetc.
endalign*






algebra-precalculus







share|cite|improve this question

















  • 3




    Looks like en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind, see also oeis.org/A094638.
    – Martin R
    1 hour ago













up vote
5
down vote

favorite









up vote
5
down vote

favorite











Are the numbers appearing as coefficients in the following sequence of polynomials known? Is there a known recurrence relation to compute them?



beginalign*
(x+y) &= x+y \
(x+y)(x+2y) &= x^2+3yx+2y^2 \
(x+y)(x+2y)(x+3y) &= x^3+6yx^2+11y^2x+6y^3 \
(x+y)(x+2y)(x+3y)(x+4y) &= x^4+10yx^3+35y^2x^2+50y^3x+24y^4 \
&textetc.
endalign*






algebra-precalculus







share|cite|improve this question













Are the numbers appearing as coefficients in the following sequence of polynomials known? Is there a known recurrence relation to compute them?



beginalign*
(x+y) &= x+y \
(x+y)(x+2y) &= x^2+3yx+2y^2 \
(x+y)(x+2y)(x+3y) &= x^3+6yx^2+11y^2x+6y^3 \
(x+y)(x+2y)(x+3y)(x+4y) &= x^4+10yx^3+35y^2x^2+50y^3x+24y^4 \
&textetc.
endalign*






algebra-precalculus




algebra-precalculus






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asked 1 hour ago









Zen

3114




3114







  • 3




    Looks like en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind, see also oeis.org/A094638.
    – Martin R
    1 hour ago













  • 3




    Looks like en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind, see also oeis.org/A094638.
    – Martin R
    1 hour ago








3




3




Looks like en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind, see also oeis.org/A094638.
– Martin R
1 hour ago





Looks like en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind, see also oeis.org/A094638.
– Martin R
1 hour ago











1 Answer
1






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That are the Unsigned Stirling numbers of the first kind $ n brack k $, which are defined as the coefficients in
the expansion
$$ tag*
x(x+1)(x+2) cdots (x+n-1) = sum_k=0^n n brack k x^k
$$



They satisfy the recurrence relation
$$
n+1 brack k = n n brack k + n brack k-1
$$
for $k > 0$, with the initial conditions
$$
 0 brack 0 = 1 , , quad  n brack 0 =  0 brack n = 0
$$
for $n > 0$.



$(*)$ with $n+1$ instead of $n$ and $frac xy$ instead of $x$
leads to
$$
(x+y)(x+2y) cdots (x+ny) = sum_k=0^n n+1 brack k+1 x^k y^n-k
$$






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    1 Answer
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    1 Answer
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    up vote
    6
    down vote













    That are the Unsigned Stirling numbers of the first kind $ n brack k $, which are defined as the coefficients in
    the expansion
    $$ tag*
    x(x+1)(x+2) cdots (x+n-1) = sum_k=0^n n brack k x^k
    $$



    They satisfy the recurrence relation
    $$
    n+1 brack k = n n brack k + n brack k-1
    $$
    for $k > 0$, with the initial conditions
    $$
     0 brack 0 = 1 , , quad  n brack 0 =  0 brack n = 0
    $$
    for $n > 0$.



    $(*)$ with $n+1$ instead of $n$ and $frac xy$ instead of $x$
    leads to
    $$
    (x+y)(x+2y) cdots (x+ny) = sum_k=0^n n+1 brack k+1 x^k y^n-k
    $$






    share|cite|improve this answer


























      up vote
      6
      down vote













      That are the Unsigned Stirling numbers of the first kind $ n brack k $, which are defined as the coefficients in
      the expansion
      $$ tag*
      x(x+1)(x+2) cdots (x+n-1) = sum_k=0^n n brack k x^k
      $$



      They satisfy the recurrence relation
      $$
      n+1 brack k = n n brack k + n brack k-1
      $$
      for $k > 0$, with the initial conditions
      $$
       0 brack 0 = 1 , , quad  n brack 0 =  0 brack n = 0
      $$
      for $n > 0$.



      $(*)$ with $n+1$ instead of $n$ and $frac xy$ instead of $x$
      leads to
      $$
      (x+y)(x+2y) cdots (x+ny) = sum_k=0^n n+1 brack k+1 x^k y^n-k
      $$






      share|cite|improve this answer
























        up vote
        6
        down vote










        up vote
        6
        down vote









        That are the Unsigned Stirling numbers of the first kind $ n brack k $, which are defined as the coefficients in
        the expansion
        $$ tag*
        x(x+1)(x+2) cdots (x+n-1) = sum_k=0^n n brack k x^k
        $$



        They satisfy the recurrence relation
        $$
        n+1 brack k = n n brack k + n brack k-1
        $$
        for $k > 0$, with the initial conditions
        $$
         0 brack 0 = 1 , , quad  n brack 0 =  0 brack n = 0
        $$
        for $n > 0$.



        $(*)$ with $n+1$ instead of $n$ and $frac xy$ instead of $x$
        leads to
        $$
        (x+y)(x+2y) cdots (x+ny) = sum_k=0^n n+1 brack k+1 x^k y^n-k
        $$






        share|cite|improve this answer














        That are the Unsigned Stirling numbers of the first kind $ n brack k $, which are defined as the coefficients in
        the expansion
        $$ tag*
        x(x+1)(x+2) cdots (x+n-1) = sum_k=0^n n brack k x^k
        $$



        They satisfy the recurrence relation
        $$
        n+1 brack k = n n brack k + n brack k-1
        $$
        for $k > 0$, with the initial conditions
        $$
         0 brack 0 = 1 , , quad  n brack 0 =  0 brack n = 0
        $$
        for $n > 0$.



        $(*)$ with $n+1$ instead of $n$ and $frac xy$ instead of $x$
        leads to
        $$
        (x+y)(x+2y) cdots (x+ny) = sum_k=0^n n+1 brack k+1 x^k y^n-k
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 42 mins ago

























        answered 1 hour ago









        Martin R

        25.2k32844




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