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?










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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

















up vote
6
down vote

favorite
3












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?










share|cite|improve this question
























  • 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















up vote
6
down vote

favorite
3









up vote
6
down vote

favorite
3






3





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?










share|cite|improve this question















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






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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




















  • 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












2 Answers
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2
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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.






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    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





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      2 Answers
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      2 Answers
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      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.






      share|cite|improve this answer

























        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.






        share|cite|improve this answer























          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.






          share|cite|improve this answer












          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.







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          share|cite|improve this answer










          answered 6 hours ago









          René Gy

          1,009613




          1,009613






















              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





              share|cite|improve this answer



























                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





                share|cite|improve this answer

























                  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





                  share|cite|improve this answer














                  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






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                  edited 10 hours ago

























                  answered 10 hours ago









                  Will Jagy

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