How to prove these two derangement number formulas are equivalent
In helping someone with a coding challenge for calculating the derangement number, I learned that the formula I came up with:
$$d(n)=ncdot d(n-1)+(-1)^n$$
is not the one that most references quote: both Wikipedia and Wolfram list:
$$d(n)=(n-1)[(d(n-1)+d(n-2)]$$
instead. I can see how the second formulation is easier to arrive at intuitively, but I can't think how these two can be proven to be the same. Wolfram lists both, and indeed the series of $d(n)$ are identical empirically, but how would one go about proving they are the same?
combinatorics
asked Nov 24 at 20:08
jbeldock
1064
add a comment |
In helping someone with a coding challenge for calculating the derangement number, I learned that the formula I came up with:
$$d(n)=ncdot d(n-1)+(-1)^n$$
is not the one that most references quote: both Wikipedia and Wolfram list:
$$d(n)=(n-1)[(d(n-1)+d(n-2)]$$
instead. I can see how the second formulation is easier to arrive at intuitively, but I can't think how these two can be proven to be the same. Wolfram lists both, and indeed the series of $d(n)$ are identical empirically, but how would one go about proving they are the same?
combinatorics
asked Nov 24 at 20:08
jbeldock
1064
Wolfram lists both recursions with just the word “and” between them.
– David K
Nov 24 at 22:06
add a comment |
In helping someone with a coding challenge for calculating the derangement number, I learned that the formula I came up with:
$$d(n)=ncdot d(n-1)+(-1)^n$$
is not the one that most references quote: both Wikipedia and Wolfram list:
$$d(n)=(n-1)[(d(n-1)+d(n-2)]$$
instead. I can see how the second formulation is easier to arrive at intuitively, but I can't think how these two can be proven to be the same. Wolfram lists both, and indeed the series of $d(n)$ are identical empirically, but how would one go about proving they are the same?
combinatorics
asked Nov 24 at 20:08
jbeldock
1064
In helping someone with a coding challenge for calculating the derangement number, I learned that the formula I came up with:
$$d(n)=ncdot d(n-1)+(-1)^n$$
is not the one that most references quote: both Wikipedia and Wolfram list:
$$d(n)=(n-1)[(d(n-1)+d(n-2)]$$
instead. I can see how the second formulation is easier to arrive at intuitively, but I can't think how these two can be proven to be the same. Wolfram lists both, and indeed the series of $d(n)$ are identical empirically, but how would one go about proving they are the same?
combinatorics
combinatorics
asked Nov 24 at 20:08
jbeldock
1064
asked Nov 24 at 20:08
jbeldock
1064
asked Nov 24 at 20:08
jbeldock
1064
asked Nov 24 at 20:08
jbeldock
1064
asked Nov 24 at 20:08
jbeldock
1064
1064
Wolfram lists both recursions with just the word “and” between them.
– David K
Nov 24 at 22:06
add a comment |
Wolfram lists both recursions with just the word “and” between them.
– David K
Nov 24 at 22:06
Wolfram lists both recursions with just the word “and” between them.
– David K
Nov 24 at 22:06
Wolfram lists both recursions with just the word “and” between them.
– David K
Nov 24 at 22:06
add a comment |
1 Answer
1
active
oldest
votes
The first formula can easily be used to derive the second formula ...
begin{eqnarray*}
d_{n-1}&=&(n-1)d_{n-2}+(-1)^{n-1} \
d_{n}&=&nd_{n-1}+(-1)^{n} \
&=&(n-1)d_{n-1}+d_{n-1}+(-1)^{n} \
&=&(n-1)d_{n-1}+(n-1)d_{n-2}+underbrace{(-1)^{n-1}+(-1)^{n}}_{=0}. \
end{eqnarray*}
answered Nov 24 at 20:26
Donald Splutterwit
22.3k21244
Thank you! Too me a minute to figure out what you did from line 2 to 3, but get it now! $nd_{n-1} = (n-1)d_{n-1} + d_{n-1}$
– jbeldock
Nov 24 at 21:15
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The first formula can easily be used to derive the second formula ...
begin{eqnarray*}
d_{n-1}&=&(n-1)d_{n-2}+(-1)^{n-1} \
d_{n}&=&nd_{n-1}+(-1)^{n} \
&=&(n-1)d_{n-1}+d_{n-1}+(-1)^{n} \
&=&(n-1)d_{n-1}+(n-1)d_{n-2}+underbrace{(-1)^{n-1}+(-1)^{n}}_{=0}. \
end{eqnarray*}
answered Nov 24 at 20:26
Donald Splutterwit
22.3k21244
Thank you! Too me a minute to figure out what you did from line 2 to 3, but get it now! $nd_{n-1} = (n-1)d_{n-1} + d_{n-1}$
– jbeldock
Nov 24 at 21:15
add a comment |
The first formula can easily be used to derive the second formula ...
begin{eqnarray*}
d_{n-1}&=&(n-1)d_{n-2}+(-1)^{n-1} \
d_{n}&=&nd_{n-1}+(-1)^{n} \
&=&(n-1)d_{n-1}+d_{n-1}+(-1)^{n} \
&=&(n-1)d_{n-1}+(n-1)d_{n-2}+underbrace{(-1)^{n-1}+(-1)^{n}}_{=0}. \
end{eqnarray*}
answered Nov 24 at 20:26
Donald Splutterwit
22.3k21244
Thank you! Too me a minute to figure out what you did from line 2 to 3, but get it now! $nd_{n-1} = (n-1)d_{n-1} + d_{n-1}$
– jbeldock
Nov 24 at 21:15
add a comment |
The first formula can easily be used to derive the second formula ...
begin{eqnarray*}
d_{n-1}&=&(n-1)d_{n-2}+(-1)^{n-1} \
d_{n}&=&nd_{n-1}+(-1)^{n} \
&=&(n-1)d_{n-1}+d_{n-1}+(-1)^{n} \
&=&(n-1)d_{n-1}+(n-1)d_{n-2}+underbrace{(-1)^{n-1}+(-1)^{n}}_{=0}. \
end{eqnarray*}
answered Nov 24 at 20:26
Donald Splutterwit
22.3k21244
The first formula can easily be used to derive the second formula ...
begin{eqnarray*}
d_{n-1}&=&(n-1)d_{n-2}+(-1)^{n-1} \
d_{n}&=&nd_{n-1}+(-1)^{n} \
&=&(n-1)d_{n-1}+d_{n-1}+(-1)^{n} \
&=&(n-1)d_{n-1}+(n-1)d_{n-2}+underbrace{(-1)^{n-1}+(-1)^{n}}_{=0}. \
end{eqnarray*}
answered Nov 24 at 20:26
Donald Splutterwit
22.3k21244
answered Nov 24 at 20:26
Donald Splutterwit
22.3k21244
answered Nov 24 at 20:26
Donald Splutterwit
22.3k21244
answered Nov 24 at 20:26
Donald Splutterwit
22.3k21244
22.3k21244
Thank you! Too me a minute to figure out what you did from line 2 to 3, but get it now! $nd_{n-1} = (n-1)d_{n-1} + d_{n-1}$
– jbeldock
Nov 24 at 21:15
add a comment |
Thank you! Too me a minute to figure out what you did from line 2 to 3, but get it now! $nd_{n-1} = (n-1)d_{n-1} + d_{n-1}$
– jbeldock
Nov 24 at 21:15
Thank you! Too me a minute to figure out what you did from line 2 to 3, but get it now! $nd_{n-1} = (n-1)d_{n-1} + d_{n-1}$
– jbeldock
Nov 24 at 21:15
Thank you! Too me a minute to figure out what you did from line 2 to 3, but get it now! $nd_{n-1} = (n-1)d_{n-1} + d_{n-1}$
– jbeldock
Nov 24 at 21:15
add a comment |
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Wolfram lists both recursions with just the word “and” between them.
– David K
Nov 24 at 22:06