A non-linear recurrence relation in two variablesCounting sequences - a recurrence relation.Solving a general two-term combinatorial recurrence relationDetermining a recurrence relationLinear Recurrence Relations in 2 Variables with Variable CoefficientsSolving a two dimensional non-homogenous linear recurrenceEliminating a variable from a two-variable linear recurrenceA two variable recurrence relation with conditionalsLinear two-dimensional recurrence relationCounting some binary trees with lots of extra stucturelinear recurrence relation for square of sequence given recursively
A non-linear recurrence relation in two variables
Counting sequences - a recurrence relation.Solving a general two-term combinatorial recurrence relationDetermining a recurrence relationLinear Recurrence Relations in 2 Variables with Variable CoefficientsSolving a two dimensional non-homogenous linear recurrenceEliminating a variable from a two-variable linear recurrenceA two variable recurrence relation with conditionalsLinear two-dimensional recurrence relationCounting some binary trees with lots of extra stucturelinear recurrence relation for square of sequence given recursively
$begingroup$
I'm interested in solving a particular non-linear recurrence in two variables:
$$lambda_j,k = (j+k) lambda_j,k-1 + beginpmatrix j+k \ 2 endpmatrix lambda_j-1, k-1.$$
Here $j geq -1$ and $k geq 0$, and we have initial conditions:
$lambda_-1,k = 0$ for all $k$;
$lambda_j,0 = 0$ for all $j>0$;
$lambda_0,0 = 1$.
This is a relation between the leading coefficients of certain polynomials occurring in a problem in modular invariant theory. $lambda_-1,k$ doesn't really make sense in context, but introducing it simplifies the relation and initial conditions nicely. Because of the context, I happen to know that for any prime $p$, $lambda_j,k neq 0$ mod $p$ unless $j+k geq p$. This suggests to me that there is a particularly simple solution, possibly a product of binomial coefficients.
Some starting points: Obviously $lambda_0,k = k!$ and it's easy to show that $lambda_j,k = 0$ if $j>k$ and $lambda_k,k = frac(2k)!2^k$ for $k geq 0$. I've tried solving it with generating functions in two variables, but that just produces a truly terrifying PDE of order 2 and degree 4.
co.combinatorics
asked 5 hours ago
Jon ElmerJon Elmer
233
$endgroup$
add a comment |
$begingroup$
I'm interested in solving a particular non-linear recurrence in two variables:
$$lambda_j,k = (j+k) lambda_j,k-1 + beginpmatrix j+k \ 2 endpmatrix lambda_j-1, k-1.$$
Here $j geq -1$ and $k geq 0$, and we have initial conditions:
$lambda_-1,k = 0$ for all $k$;
$lambda_j,0 = 0$ for all $j>0$;
$lambda_0,0 = 1$.
This is a relation between the leading coefficients of certain polynomials occurring in a problem in modular invariant theory. $lambda_-1,k$ doesn't really make sense in context, but introducing it simplifies the relation and initial conditions nicely. Because of the context, I happen to know that for any prime $p$, $lambda_j,k neq 0$ mod $p$ unless $j+k geq p$. This suggests to me that there is a particularly simple solution, possibly a product of binomial coefficients.
Some starting points: Obviously $lambda_0,k = k!$ and it's easy to show that $lambda_j,k = 0$ if $j>k$ and $lambda_k,k = frac(2k)!2^k$ for $k geq 0$. I've tried solving it with generating functions in two variables, but that just produces a truly terrifying PDE of order 2 and degree 4.
co.combinatorics
asked 5 hours ago
Jon ElmerJon Elmer
233
$endgroup$
$begingroup$
Great answer @Fedor Petrov, thanks a lot!
$endgroup$
– Jon Elmer
3 hours ago
add a comment |
$begingroup$
I'm interested in solving a particular non-linear recurrence in two variables:
$$lambda_j,k = (j+k) lambda_j,k-1 + beginpmatrix j+k \ 2 endpmatrix lambda_j-1, k-1.$$
Here $j geq -1$ and $k geq 0$, and we have initial conditions:
$lambda_-1,k = 0$ for all $k$;
$lambda_j,0 = 0$ for all $j>0$;
$lambda_0,0 = 1$.
This is a relation between the leading coefficients of certain polynomials occurring in a problem in modular invariant theory. $lambda_-1,k$ doesn't really make sense in context, but introducing it simplifies the relation and initial conditions nicely. Because of the context, I happen to know that for any prime $p$, $lambda_j,k neq 0$ mod $p$ unless $j+k geq p$. This suggests to me that there is a particularly simple solution, possibly a product of binomial coefficients.
Some starting points: Obviously $lambda_0,k = k!$ and it's easy to show that $lambda_j,k = 0$ if $j>k$ and $lambda_k,k = frac(2k)!2^k$ for $k geq 0$. I've tried solving it with generating functions in two variables, but that just produces a truly terrifying PDE of order 2 and degree 4.
co.combinatorics
asked 5 hours ago
Jon ElmerJon Elmer
233
$endgroup$
I'm interested in solving a particular non-linear recurrence in two variables:
$$lambda_j,k = (j+k) lambda_j,k-1 + beginpmatrix j+k \ 2 endpmatrix lambda_j-1, k-1.$$
Here $j geq -1$ and $k geq 0$, and we have initial conditions:
$lambda_-1,k = 0$ for all $k$;
$lambda_j,0 = 0$ for all $j>0$;
$lambda_0,0 = 1$.
This is a relation between the leading coefficients of certain polynomials occurring in a problem in modular invariant theory. $lambda_-1,k$ doesn't really make sense in context, but introducing it simplifies the relation and initial conditions nicely. Because of the context, I happen to know that for any prime $p$, $lambda_j,k neq 0$ mod $p$ unless $j+k geq p$. This suggests to me that there is a particularly simple solution, possibly a product of binomial coefficients.
Some starting points: Obviously $lambda_0,k = k!$ and it's easy to show that $lambda_j,k = 0$ if $j>k$ and $lambda_k,k = frac(2k)!2^k$ for $k geq 0$. I've tried solving it with generating functions in two variables, but that just produces a truly terrifying PDE of order 2 and degree 4.
co.combinatorics
co.combinatorics
asked 5 hours ago
Jon ElmerJon Elmer
233
asked 5 hours ago
Jon ElmerJon Elmer
233
asked 5 hours ago
Jon ElmerJon Elmer
233
asked 5 hours ago
Jon ElmerJon Elmer
233
asked 5 hours ago
Jon ElmerJon Elmer
233
233
$begingroup$
Great answer @Fedor Petrov, thanks a lot!
$endgroup$
– Jon Elmer
3 hours ago
add a comment |
$begingroup$
Great answer @Fedor Petrov, thanks a lot!
$endgroup$
– Jon Elmer
3 hours ago
$begingroup$
Great answer @Fedor Petrov, thanks a lot!
$endgroup$
– Jon Elmer
3 hours ago
$begingroup$
Great answer @Fedor Petrov, thanks a lot!
$endgroup$
– Jon Elmer
3 hours ago
add a comment |
1 Answer
1
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$begingroup$
Denote $lambda_j,k=(j+k)! a_j,k$ (unless $j=-1,k=0$). Then we get $a_j,k=a_j,k-1+a_j-1,k-1/2$. Further denoting $a_j,k=2^-jb_j,k$ we get $b_j,k=b_j,k-1+b_j-1,k-1$ that looks like a Pascal triangle recurrence. So $b_j,k=binomkj$ (check the initial conditions also) and $lambda_j,k=2^-j(j+k)!binomkj$.
answered 4 hours ago
Fedor PetrovFedor Petrov
51.6k6121238
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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$begingroup$
Denote $lambda_j,k=(j+k)! a_j,k$ (unless $j=-1,k=0$). Then we get $a_j,k=a_j,k-1+a_j-1,k-1/2$. Further denoting $a_j,k=2^-jb_j,k$ we get $b_j,k=b_j,k-1+b_j-1,k-1$ that looks like a Pascal triangle recurrence. So $b_j,k=binomkj$ (check the initial conditions also) and $lambda_j,k=2^-j(j+k)!binomkj$.
answered 4 hours ago
Fedor PetrovFedor Petrov
51.6k6121238
$endgroup$
add a comment |
$begingroup$
Denote $lambda_j,k=(j+k)! a_j,k$ (unless $j=-1,k=0$). Then we get $a_j,k=a_j,k-1+a_j-1,k-1/2$. Further denoting $a_j,k=2^-jb_j,k$ we get $b_j,k=b_j,k-1+b_j-1,k-1$ that looks like a Pascal triangle recurrence. So $b_j,k=binomkj$ (check the initial conditions also) and $lambda_j,k=2^-j(j+k)!binomkj$.
answered 4 hours ago
Fedor PetrovFedor Petrov
51.6k6121238
$endgroup$
add a comment |
$begingroup$
Denote $lambda_j,k=(j+k)! a_j,k$ (unless $j=-1,k=0$). Then we get $a_j,k=a_j,k-1+a_j-1,k-1/2$. Further denoting $a_j,k=2^-jb_j,k$ we get $b_j,k=b_j,k-1+b_j-1,k-1$ that looks like a Pascal triangle recurrence. So $b_j,k=binomkj$ (check the initial conditions also) and $lambda_j,k=2^-j(j+k)!binomkj$.
answered 4 hours ago
Fedor PetrovFedor Petrov
51.6k6121238
$endgroup$
Denote $lambda_j,k=(j+k)! a_j,k$ (unless $j=-1,k=0$). Then we get $a_j,k=a_j,k-1+a_j-1,k-1/2$. Further denoting $a_j,k=2^-jb_j,k$ we get $b_j,k=b_j,k-1+b_j-1,k-1$ that looks like a Pascal triangle recurrence. So $b_j,k=binomkj$ (check the initial conditions also) and $lambda_j,k=2^-j(j+k)!binomkj$.
answered 4 hours ago
Fedor PetrovFedor Petrov
51.6k6121238
answered 4 hours ago
Fedor PetrovFedor Petrov
51.6k6121238
answered 4 hours ago
Fedor PetrovFedor Petrov
51.6k6121238
answered 4 hours ago
Fedor PetrovFedor Petrov
51.6k6121238
51.6k6121238
add a comment |
add a comment |
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$begingroup$
Great answer @Fedor Petrov, thanks a lot!
$endgroup$
– Jon Elmer
3 hours ago