Sequence in which adding 2 produces a square
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Consider the sequence defined by $x_1=2$, $x_2=x_3=7$, $x_{n+1}=x_{n}x_{n-1}-x_{n-2}$.
Then $x_4=7 cdot 7 - 2 = 47, x_5=47 cdot 7 - 7=322, x_6=322 cdot 47 - 7 =15127, x_7=15127 cdot 322 -47=4870847,x_8=4870847 cdot 15127-322=73681302247.$
My spreadsheet cannot calculate any more, but notice that $x_n+2$ is always square:
$sqrt{x_4+2}=7$, $sqrt{x_5+2}=18$, $sqrt{x_6+2}=123$, $sqrt{x_7+2}=2207$,$sqrt{x_8+2}=271443.$ Is this a coincidence?
sequences-and-series
|
show 2 more comments
up vote
6
down vote
favorite
Consider the sequence defined by $x_1=2$, $x_2=x_3=7$, $x_{n+1}=x_{n}x_{n-1}-x_{n-2}$.
Then $x_4=7 cdot 7 - 2 = 47, x_5=47 cdot 7 - 7=322, x_6=322 cdot 47 - 7 =15127, x_7=15127 cdot 322 -47=4870847,x_8=4870847 cdot 15127-322=73681302247.$
My spreadsheet cannot calculate any more, but notice that $x_n+2$ is always square:
$sqrt{x_4+2}=7$, $sqrt{x_5+2}=18$, $sqrt{x_6+2}=123$, $sqrt{x_7+2}=2207$,$sqrt{x_8+2}=271443.$ Is this a coincidence?
sequences-and-series
First tought in my mind: did you try induction?
– Will M.
13 hours ago
It's not that easy.
– Craig Feinstein
13 hours ago
It's true up to $n=35$, according to Mathematica, so it's probably true in general.
– Patrick Stevens
10 hours ago
@PatrickStevens evidently the indicated square roots satisfy the same type of recurrence; I put a brief output as answer, including the square roots
– Will Jagy
10 hours ago
$sqrt frac{x_{n+1}-2}{5})$ seems to be the sequence A101361
– René Gy
10 hours ago
|
show 2 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Consider the sequence defined by $x_1=2$, $x_2=x_3=7$, $x_{n+1}=x_{n}x_{n-1}-x_{n-2}$.
Then $x_4=7 cdot 7 - 2 = 47, x_5=47 cdot 7 - 7=322, x_6=322 cdot 47 - 7 =15127, x_7=15127 cdot 322 -47=4870847,x_8=4870847 cdot 15127-322=73681302247.$
My spreadsheet cannot calculate any more, but notice that $x_n+2$ is always square:
$sqrt{x_4+2}=7$, $sqrt{x_5+2}=18$, $sqrt{x_6+2}=123$, $sqrt{x_7+2}=2207$,$sqrt{x_8+2}=271443.$ Is this a coincidence?
sequences-and-series
Consider the sequence defined by $x_1=2$, $x_2=x_3=7$, $x_{n+1}=x_{n}x_{n-1}-x_{n-2}$.
Then $x_4=7 cdot 7 - 2 = 47, x_5=47 cdot 7 - 7=322, x_6=322 cdot 47 - 7 =15127, x_7=15127 cdot 322 -47=4870847,x_8=4870847 cdot 15127-322=73681302247.$
My spreadsheet cannot calculate any more, but notice that $x_n+2$ is always square:
$sqrt{x_4+2}=7$, $sqrt{x_5+2}=18$, $sqrt{x_6+2}=123$, $sqrt{x_7+2}=2207$,$sqrt{x_8+2}=271443.$ Is this a coincidence?
sequences-and-series
sequences-and-series
edited 13 hours ago
asked 14 hours ago
Craig Feinstein
380320
380320
First tought in my mind: did you try induction?
– Will M.
13 hours ago
It's not that easy.
– Craig Feinstein
13 hours ago
It's true up to $n=35$, according to Mathematica, so it's probably true in general.
– Patrick Stevens
10 hours ago
@PatrickStevens evidently the indicated square roots satisfy the same type of recurrence; I put a brief output as answer, including the square roots
– Will Jagy
10 hours ago
$sqrt frac{x_{n+1}-2}{5})$ seems to be the sequence A101361
– René Gy
10 hours ago
|
show 2 more comments
First tought in my mind: did you try induction?
– Will M.
13 hours ago
It's not that easy.
– Craig Feinstein
13 hours ago
It's true up to $n=35$, according to Mathematica, so it's probably true in general.
– Patrick Stevens
10 hours ago
@PatrickStevens evidently the indicated square roots satisfy the same type of recurrence; I put a brief output as answer, including the square roots
– Will Jagy
10 hours ago
$sqrt frac{x_{n+1}-2}{5})$ seems to be the sequence A101361
– René Gy
10 hours ago
First tought in my mind: did you try induction?
– Will M.
13 hours ago
First tought in my mind: did you try induction?
– Will M.
13 hours ago
It's not that easy.
– Craig Feinstein
13 hours ago
It's not that easy.
– Craig Feinstein
13 hours ago
It's true up to $n=35$, according to Mathematica, so it's probably true in general.
– Patrick Stevens
10 hours ago
It's true up to $n=35$, according to Mathematica, so it's probably true in general.
– Patrick Stevens
10 hours ago
@PatrickStevens evidently the indicated square roots satisfy the same type of recurrence; I put a brief output as answer, including the square roots
– Will Jagy
10 hours ago
@PatrickStevens evidently the indicated square roots satisfy the same type of recurrence; I put a brief output as answer, including the square roots
– Will Jagy
10 hours ago
$sqrt frac{x_{n+1}-2}{5})$ seems to be the sequence A101361
– René Gy
10 hours ago
$sqrt frac{x_{n+1}-2}{5})$ seems to be the sequence A101361
– René Gy
10 hours ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
We have
begin{align*} x_{n+1}+x_{n-2}&=x_{n}x_{n-1}\
end{align*}
then
begin{align*} x_{n+1}x_{n-2}&=x_{n}x_{n-1}x_{n-2}-x_{n-2}^2\
&=x_n(x_{n}+x_{n-3})-x_{n-2}^2\
&=x_nx_{n-3}+x_{n}^2-x_{n-2}^2\
end{align*}
then by telescoping
begin{align*} x_{n+1}x_{n-2}&=x_4x_1+x_n^2+x_{n-1}^2-x_3^2-x_2^2\
x_{n+1}x_{n-2}&=x_n^2+x_{n-1}^2-4\
end{align*}
Now we let $x_n=z_n-2$. We then have
begin{align*}
z_{n+1}z_{n-2}&=z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+4+2z_{n+1}-4+2z_{n-2}-4\
&= z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+2x_{n+1}+2z_{n-2}+4
end{align*}
But begin{align*}
x_{n+1}&=x_nx_{n-1}-x_{n-2}\
&=(z_n-2)(z_{n-1}-2)-z_{n-2}+2
end{align*}
then by substitution and reduction, we find
begin{align*}
z_{n+1}z_{n-2}&=(z_n+z_{n-1}-4)^2
end{align*}
from which a proof by induction easily follows.
add a comment |
up vote
1
down vote
evidently the square root $s_n = sqrt {x_n + 2}$ satisfies
$$ s_{n+1} = s_n s_{n-1} - s_{n-2} $$
This tends to mean that a proof will be possible.
I think I screwed up the beginning labels for the square roots, but the sequence $2,3,3,7, 18, 123, 2207,...$ gives $3 cdot 3 - 2 = 7,$ $3 cdot 7 - 3 = 18,$ $7 cdot 18 - 3 = 123,$ $18 cdot 123 - 7 = 2207$
4 47 49 int sqrt 7 sdiff -1
5 322 324 int sqrt 18 sdiff 3
6 15127 15129 int sqrt 123 sdiff 3
7 4870847 4870849 int sqrt 2207 sdiff 7
8 73681302247 73681302249 int sqrt 271443 sdiff 18
9 358890350005878082 358890350005878084 int sqrt 599074578 sdiff 123
10 26443508352314721186469779407 26443508352314721186469779409 int sqrt 162614600673847 sdiff 2207
11 int sqrt 97418273275323406890123 sdiff 271443
12 int sqrt 15841633607002416873831447357889638603 sdiff 599074578
13 int sqrt 1543264591854508694059691789796980188767738307671225999544322 sdiff 162614600673847
14 int sqrt 24447832222819253258096747170722821932737551721781484698278226571759551365824711139513519853772043 sdiff 97418273275323406890123
15 int sqrt 37729473817076660931593825922523825105418958136330582571231905937822606847169782819188349018726882837649307587607345212765019953877746594614121187432973351243 sdiff 15841633607002416873831447357889638603
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
We have
begin{align*} x_{n+1}+x_{n-2}&=x_{n}x_{n-1}\
end{align*}
then
begin{align*} x_{n+1}x_{n-2}&=x_{n}x_{n-1}x_{n-2}-x_{n-2}^2\
&=x_n(x_{n}+x_{n-3})-x_{n-2}^2\
&=x_nx_{n-3}+x_{n}^2-x_{n-2}^2\
end{align*}
then by telescoping
begin{align*} x_{n+1}x_{n-2}&=x_4x_1+x_n^2+x_{n-1}^2-x_3^2-x_2^2\
x_{n+1}x_{n-2}&=x_n^2+x_{n-1}^2-4\
end{align*}
Now we let $x_n=z_n-2$. We then have
begin{align*}
z_{n+1}z_{n-2}&=z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+4+2z_{n+1}-4+2z_{n-2}-4\
&= z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+2x_{n+1}+2z_{n-2}+4
end{align*}
But begin{align*}
x_{n+1}&=x_nx_{n-1}-x_{n-2}\
&=(z_n-2)(z_{n-1}-2)-z_{n-2}+2
end{align*}
then by substitution and reduction, we find
begin{align*}
z_{n+1}z_{n-2}&=(z_n+z_{n-1}-4)^2
end{align*}
from which a proof by induction easily follows.
add a comment |
up vote
2
down vote
We have
begin{align*} x_{n+1}+x_{n-2}&=x_{n}x_{n-1}\
end{align*}
then
begin{align*} x_{n+1}x_{n-2}&=x_{n}x_{n-1}x_{n-2}-x_{n-2}^2\
&=x_n(x_{n}+x_{n-3})-x_{n-2}^2\
&=x_nx_{n-3}+x_{n}^2-x_{n-2}^2\
end{align*}
then by telescoping
begin{align*} x_{n+1}x_{n-2}&=x_4x_1+x_n^2+x_{n-1}^2-x_3^2-x_2^2\
x_{n+1}x_{n-2}&=x_n^2+x_{n-1}^2-4\
end{align*}
Now we let $x_n=z_n-2$. We then have
begin{align*}
z_{n+1}z_{n-2}&=z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+4+2z_{n+1}-4+2z_{n-2}-4\
&= z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+2x_{n+1}+2z_{n-2}+4
end{align*}
But begin{align*}
x_{n+1}&=x_nx_{n-1}-x_{n-2}\
&=(z_n-2)(z_{n-1}-2)-z_{n-2}+2
end{align*}
then by substitution and reduction, we find
begin{align*}
z_{n+1}z_{n-2}&=(z_n+z_{n-1}-4)^2
end{align*}
from which a proof by induction easily follows.
add a comment |
up vote
2
down vote
up vote
2
down vote
We have
begin{align*} x_{n+1}+x_{n-2}&=x_{n}x_{n-1}\
end{align*}
then
begin{align*} x_{n+1}x_{n-2}&=x_{n}x_{n-1}x_{n-2}-x_{n-2}^2\
&=x_n(x_{n}+x_{n-3})-x_{n-2}^2\
&=x_nx_{n-3}+x_{n}^2-x_{n-2}^2\
end{align*}
then by telescoping
begin{align*} x_{n+1}x_{n-2}&=x_4x_1+x_n^2+x_{n-1}^2-x_3^2-x_2^2\
x_{n+1}x_{n-2}&=x_n^2+x_{n-1}^2-4\
end{align*}
Now we let $x_n=z_n-2$. We then have
begin{align*}
z_{n+1}z_{n-2}&=z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+4+2z_{n+1}-4+2z_{n-2}-4\
&= z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+2x_{n+1}+2z_{n-2}+4
end{align*}
But begin{align*}
x_{n+1}&=x_nx_{n-1}-x_{n-2}\
&=(z_n-2)(z_{n-1}-2)-z_{n-2}+2
end{align*}
then by substitution and reduction, we find
begin{align*}
z_{n+1}z_{n-2}&=(z_n+z_{n-1}-4)^2
end{align*}
from which a proof by induction easily follows.
We have
begin{align*} x_{n+1}+x_{n-2}&=x_{n}x_{n-1}\
end{align*}
then
begin{align*} x_{n+1}x_{n-2}&=x_{n}x_{n-1}x_{n-2}-x_{n-2}^2\
&=x_n(x_{n}+x_{n-3})-x_{n-2}^2\
&=x_nx_{n-3}+x_{n}^2-x_{n-2}^2\
end{align*}
then by telescoping
begin{align*} x_{n+1}x_{n-2}&=x_4x_1+x_n^2+x_{n-1}^2-x_3^2-x_2^2\
x_{n+1}x_{n-2}&=x_n^2+x_{n-1}^2-4\
end{align*}
Now we let $x_n=z_n-2$. We then have
begin{align*}
z_{n+1}z_{n-2}&=z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+4+2z_{n+1}-4+2z_{n-2}-4\
&= z_n^2-4z_n+4+z_{n-1}^2-4z_{n-1}+2x_{n+1}+2z_{n-2}+4
end{align*}
But begin{align*}
x_{n+1}&=x_nx_{n-1}-x_{n-2}\
&=(z_n-2)(z_{n-1}-2)-z_{n-2}+2
end{align*}
then by substitution and reduction, we find
begin{align*}
z_{n+1}z_{n-2}&=(z_n+z_{n-1}-4)^2
end{align*}
from which a proof by induction easily follows.
answered 6 hours ago
René Gy
1,009613
1,009613
add a comment |
add a comment |
up vote
1
down vote
evidently the square root $s_n = sqrt {x_n + 2}$ satisfies
$$ s_{n+1} = s_n s_{n-1} - s_{n-2} $$
This tends to mean that a proof will be possible.
I think I screwed up the beginning labels for the square roots, but the sequence $2,3,3,7, 18, 123, 2207,...$ gives $3 cdot 3 - 2 = 7,$ $3 cdot 7 - 3 = 18,$ $7 cdot 18 - 3 = 123,$ $18 cdot 123 - 7 = 2207$
4 47 49 int sqrt 7 sdiff -1
5 322 324 int sqrt 18 sdiff 3
6 15127 15129 int sqrt 123 sdiff 3
7 4870847 4870849 int sqrt 2207 sdiff 7
8 73681302247 73681302249 int sqrt 271443 sdiff 18
9 358890350005878082 358890350005878084 int sqrt 599074578 sdiff 123
10 26443508352314721186469779407 26443508352314721186469779409 int sqrt 162614600673847 sdiff 2207
11 int sqrt 97418273275323406890123 sdiff 271443
12 int sqrt 15841633607002416873831447357889638603 sdiff 599074578
13 int sqrt 1543264591854508694059691789796980188767738307671225999544322 sdiff 162614600673847
14 int sqrt 24447832222819253258096747170722821932737551721781484698278226571759551365824711139513519853772043 sdiff 97418273275323406890123
15 int sqrt 37729473817076660931593825922523825105418958136330582571231905937822606847169782819188349018726882837649307587607345212765019953877746594614121187432973351243 sdiff 15841633607002416873831447357889638603
add a comment |
up vote
1
down vote
evidently the square root $s_n = sqrt {x_n + 2}$ satisfies
$$ s_{n+1} = s_n s_{n-1} - s_{n-2} $$
This tends to mean that a proof will be possible.
I think I screwed up the beginning labels for the square roots, but the sequence $2,3,3,7, 18, 123, 2207,...$ gives $3 cdot 3 - 2 = 7,$ $3 cdot 7 - 3 = 18,$ $7 cdot 18 - 3 = 123,$ $18 cdot 123 - 7 = 2207$
4 47 49 int sqrt 7 sdiff -1
5 322 324 int sqrt 18 sdiff 3
6 15127 15129 int sqrt 123 sdiff 3
7 4870847 4870849 int sqrt 2207 sdiff 7
8 73681302247 73681302249 int sqrt 271443 sdiff 18
9 358890350005878082 358890350005878084 int sqrt 599074578 sdiff 123
10 26443508352314721186469779407 26443508352314721186469779409 int sqrt 162614600673847 sdiff 2207
11 int sqrt 97418273275323406890123 sdiff 271443
12 int sqrt 15841633607002416873831447357889638603 sdiff 599074578
13 int sqrt 1543264591854508694059691789796980188767738307671225999544322 sdiff 162614600673847
14 int sqrt 24447832222819253258096747170722821932737551721781484698278226571759551365824711139513519853772043 sdiff 97418273275323406890123
15 int sqrt 37729473817076660931593825922523825105418958136330582571231905937822606847169782819188349018726882837649307587607345212765019953877746594614121187432973351243 sdiff 15841633607002416873831447357889638603
add a comment |
up vote
1
down vote
up vote
1
down vote
evidently the square root $s_n = sqrt {x_n + 2}$ satisfies
$$ s_{n+1} = s_n s_{n-1} - s_{n-2} $$
This tends to mean that a proof will be possible.
I think I screwed up the beginning labels for the square roots, but the sequence $2,3,3,7, 18, 123, 2207,...$ gives $3 cdot 3 - 2 = 7,$ $3 cdot 7 - 3 = 18,$ $7 cdot 18 - 3 = 123,$ $18 cdot 123 - 7 = 2207$
4 47 49 int sqrt 7 sdiff -1
5 322 324 int sqrt 18 sdiff 3
6 15127 15129 int sqrt 123 sdiff 3
7 4870847 4870849 int sqrt 2207 sdiff 7
8 73681302247 73681302249 int sqrt 271443 sdiff 18
9 358890350005878082 358890350005878084 int sqrt 599074578 sdiff 123
10 26443508352314721186469779407 26443508352314721186469779409 int sqrt 162614600673847 sdiff 2207
11 int sqrt 97418273275323406890123 sdiff 271443
12 int sqrt 15841633607002416873831447357889638603 sdiff 599074578
13 int sqrt 1543264591854508694059691789796980188767738307671225999544322 sdiff 162614600673847
14 int sqrt 24447832222819253258096747170722821932737551721781484698278226571759551365824711139513519853772043 sdiff 97418273275323406890123
15 int sqrt 37729473817076660931593825922523825105418958136330582571231905937822606847169782819188349018726882837649307587607345212765019953877746594614121187432973351243 sdiff 15841633607002416873831447357889638603
evidently the square root $s_n = sqrt {x_n + 2}$ satisfies
$$ s_{n+1} = s_n s_{n-1} - s_{n-2} $$
This tends to mean that a proof will be possible.
I think I screwed up the beginning labels for the square roots, but the sequence $2,3,3,7, 18, 123, 2207,...$ gives $3 cdot 3 - 2 = 7,$ $3 cdot 7 - 3 = 18,$ $7 cdot 18 - 3 = 123,$ $18 cdot 123 - 7 = 2207$
4 47 49 int sqrt 7 sdiff -1
5 322 324 int sqrt 18 sdiff 3
6 15127 15129 int sqrt 123 sdiff 3
7 4870847 4870849 int sqrt 2207 sdiff 7
8 73681302247 73681302249 int sqrt 271443 sdiff 18
9 358890350005878082 358890350005878084 int sqrt 599074578 sdiff 123
10 26443508352314721186469779407 26443508352314721186469779409 int sqrt 162614600673847 sdiff 2207
11 int sqrt 97418273275323406890123 sdiff 271443
12 int sqrt 15841633607002416873831447357889638603 sdiff 599074578
13 int sqrt 1543264591854508694059691789796980188767738307671225999544322 sdiff 162614600673847
14 int sqrt 24447832222819253258096747170722821932737551721781484698278226571759551365824711139513519853772043 sdiff 97418273275323406890123
15 int sqrt 37729473817076660931593825922523825105418958136330582571231905937822606847169782819188349018726882837649307587607345212765019953877746594614121187432973351243 sdiff 15841633607002416873831447357889638603
edited 10 hours ago
answered 10 hours ago
Will Jagy
100k597198
100k597198
add a comment |
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First tought in my mind: did you try induction?
– Will M.
13 hours ago
It's not that easy.
– Craig Feinstein
13 hours ago
It's true up to $n=35$, according to Mathematica, so it's probably true in general.
– Patrick Stevens
10 hours ago
@PatrickStevens evidently the indicated square roots satisfy the same type of recurrence; I put a brief output as answer, including the square roots
– Will Jagy
10 hours ago
$sqrt frac{x_{n+1}-2}{5})$ seems to be the sequence A101361
– René Gy
10 hours ago