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Recursion series with two variable indices $u_n,k = u_n-1,k + u_n-2,k-1$
“Upper summation” binomial identity: different version from “Concrete Mathematics”Uniqueness in transfinite recursion.Closed-form solution of non-autonomous vector recurrence relationFunction that maps numbers to diagonal co-ordinatesGeneral Expression for an Element of a Triangular ArrayFinding Value of Pascal's Triangle Given Single IndexToward Explicit Formula from Recursion : Is Generating function “the only” answer?Binomial Approximation of Gaussian DistributionGenerate a row of a modified Pascal's triangle.How to find the closed form of the recursive equation, if 2 of the 3 roots of the characteristic equation are imaginary and only 1 root is real?
$begingroup$
I found a recursive formula that looks very similar to Pascal's triangle. Where in Pascal's triangle you have the recurring formula
$$u_n,k = u_n-1,k + u_n-1,k-1,$$
which gives the Binomial function $binomnk$ with the initial condition $u_1,0=u_1,1 = 1$. In my case the second term is one value lower in the index
$$u_n,k = u_n-1,k + u_n-2,k-1 , $$
with the initial condition $u_1,0 = u_2,0 = 1 $ and $ u_2,1 = 2$. 
It looks like the image above (where we miss the top 1). Where n is the height and k is the width. How can one calculate a closed form for this? I've done this bedore with single variable recursion formulas by using the characteristic polynomial, but I'm not sure how to start now.
functions binomial-coefficients recursion
asked Mar 21 at 23:31
user1792605user1792605
606
$endgroup$
add a comment |
$begingroup$
I found a recursive formula that looks very similar to Pascal's triangle. Where in Pascal's triangle you have the recurring formula
$$u_n,k = u_n-1,k + u_n-1,k-1,$$
which gives the Binomial function $binomnk$ with the initial condition $u_1,0=u_1,1 = 1$. In my case the second term is one value lower in the index
$$u_n,k = u_n-1,k + u_n-2,k-1 , $$
with the initial condition $u_1,0 = u_2,0 = 1 $ and $ u_2,1 = 2$.
It looks like the image above (where we miss the top 1). Where n is the height and k is the width. How can one calculate a closed form for this? I've done this bedore with single variable recursion formulas by using the characteristic polynomial, but I'm not sure how to start now.
functions binomial-coefficients recursion
asked Mar 21 at 23:31
user1792605user1792605
606
$endgroup$
add a comment |
$begingroup$
I found a recursive formula that looks very similar to Pascal's triangle. Where in Pascal's triangle you have the recurring formula
$$u_n,k = u_n-1,k + u_n-1,k-1,$$
which gives the Binomial function $binomnk$ with the initial condition $u_1,0=u_1,1 = 1$. In my case the second term is one value lower in the index
$$u_n,k = u_n-1,k + u_n-2,k-1 , $$
with the initial condition $u_1,0 = u_2,0 = 1 $ and $ u_2,1 = 2$.
It looks like the image above (where we miss the top 1). Where n is the height and k is the width. How can one calculate a closed form for this? I've done this bedore with single variable recursion formulas by using the characteristic polynomial, but I'm not sure how to start now.
functions binomial-coefficients recursion
asked Mar 21 at 23:31
user1792605user1792605
606
$endgroup$
I found a recursive formula that looks very similar to Pascal's triangle. Where in Pascal's triangle you have the recurring formula
$$u_n,k = u_n-1,k + u_n-1,k-1,$$
which gives the Binomial function $binomnk$ with the initial condition $u_1,0=u_1,1 = 1$. In my case the second term is one value lower in the index
$$u_n,k = u_n-1,k + u_n-2,k-1 , $$
with the initial condition $u_1,0 = u_2,0 = 1 $ and $ u_2,1 = 2$.
It looks like the image above (where we miss the top 1). Where n is the height and k is the width. How can one calculate a closed form for this? I've done this bedore with single variable recursion formulas by using the characteristic polynomial, but I'm not sure how to start now.
functions binomial-coefficients recursion
functions binomial-coefficients recursion
asked Mar 21 at 23:31
user1792605user1792605
606
asked Mar 21 at 23:31
user1792605user1792605
606
edited Mar 22 at 1:25
user1792605
asked Mar 21 at 23:31
user1792605user1792605
606
asked Mar 21 at 23:31
user1792605user1792605
606
asked Mar 21 at 23:31
user1792605user1792605
606
606
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Hint The entries in the $k$th antidiagonal are binomial coefficients $n choose k$.
answered Mar 22 at 1:49
TravisTravis
64k769151
$endgroup$
1
$begingroup$
Nice thanks. I got it it's $$ binomn-k+1k-1 + binomn-k+1k-2 $$
$endgroup$
– user1792605
Mar 22 at 2:44
$begingroup$
Nice work! NB using the first recurrence you mention (the one relating different binomial coefficients) you can simplify this to $$n - k + 2choosek - 1 .$$
$endgroup$
– Travis
Mar 22 at 5:39
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint The entries in the $k$th antidiagonal are binomial coefficients $n choose k$.
answered Mar 22 at 1:49
TravisTravis
64k769151
$endgroup$
1
$begingroup$
Nice thanks. I got it it's $$ binomn-k+1k-1 + binomn-k+1k-2 $$
$endgroup$
– user1792605
Mar 22 at 2:44
$begingroup$
Nice work! NB using the first recurrence you mention (the one relating different binomial coefficients) you can simplify this to $$n - k + 2choosek - 1 .$$
$endgroup$
– Travis
Mar 22 at 5:39
add a comment |
$begingroup$
Hint The entries in the $k$th antidiagonal are binomial coefficients $n choose k$.
answered Mar 22 at 1:49
TravisTravis
64k769151
$endgroup$
1
$begingroup$
Nice thanks. I got it it's $$ binomn-k+1k-1 + binomn-k+1k-2 $$
$endgroup$
– user1792605
Mar 22 at 2:44
$begingroup$
Nice work! NB using the first recurrence you mention (the one relating different binomial coefficients) you can simplify this to $$n - k + 2choosek - 1 .$$
$endgroup$
– Travis
Mar 22 at 5:39
add a comment |
$begingroup$
Hint The entries in the $k$th antidiagonal are binomial coefficients $n choose k$.
answered Mar 22 at 1:49
TravisTravis
64k769151
$endgroup$
Hint The entries in the $k$th antidiagonal are binomial coefficients $n choose k$.
answered Mar 22 at 1:49
TravisTravis
64k769151
answered Mar 22 at 1:49
TravisTravis
64k769151
answered Mar 22 at 1:49
TravisTravis
64k769151
answered Mar 22 at 1:49
TravisTravis
64k769151
64k769151
1
$begingroup$
Nice thanks. I got it it's $$ binomn-k+1k-1 + binomn-k+1k-2 $$
$endgroup$
– user1792605
Mar 22 at 2:44
$begingroup$
Nice work! NB using the first recurrence you mention (the one relating different binomial coefficients) you can simplify this to $$n - k + 2choosek - 1 .$$
$endgroup$
– Travis
Mar 22 at 5:39
add a comment |
1
$begingroup$
Nice thanks. I got it it's $$ binomn-k+1k-1 + binomn-k+1k-2 $$
$endgroup$
– user1792605
Mar 22 at 2:44
$begingroup$
Nice work! NB using the first recurrence you mention (the one relating different binomial coefficients) you can simplify this to $$n - k + 2choosek - 1 .$$
$endgroup$
– Travis
Mar 22 at 5:39
1
1
$begingroup$
Nice thanks. I got it it's $$ binomn-k+1k-1 + binomn-k+1k-2 $$
$endgroup$
– user1792605
Mar 22 at 2:44
$begingroup$
Nice thanks. I got it it's $$ binomn-k+1k-1 + binomn-k+1k-2 $$
$endgroup$
– user1792605
Mar 22 at 2:44
$begingroup$
Nice work! NB using the first recurrence you mention (the one relating different binomial coefficients) you can simplify this to $$n - k + 2choosek - 1 .$$
$endgroup$
– Travis
Mar 22 at 5:39
$begingroup$
Nice work! NB using the first recurrence you mention (the one relating different binomial coefficients) you can simplify this to $$n - k + 2choosek - 1 .$$
$endgroup$
– Travis
Mar 22 at 5:39
add a comment |
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