Does this pattern of summing polygonal numbers to get a square repeat indefinitely?












5












$begingroup$


I am using the table of polygonal numbers on this site:



http://oeis.org/wiki/Polygonal_numbers



The first row of the table gives the value of $n=0,1,2,3,...$. The first column gives the polygonal numbers $P_{N}(n)$ starting with $N=3$.



Here's the pattern, the sum is done by adding elements of the same column starting with the triangular numbers.
$$1+1+1+1=4=2^2$$
$$3+4+5+6+7=25=5^2$$
$$6+9+12+15+18+21=81=9^2$$
$$10+16+22+28+34+40+46=196=14^2$$
...
The squares on the right hand side are given by $n+T_{n}$. The sum of index $(n+1)$ has one more element than that of $n$.



1-Does this pattern repeat indefinitely? (I suspect the answer is yes but I can't prove it)

2-Do we know why we have such a pattern?






elementary-number-theory pattern-recognition







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    I am using the table of polygonal numbers on this site:



    http://oeis.org/wiki/Polygonal_numbers



    The first row of the table gives the value of $n=0,1,2,3,...$. The first column gives the polygonal numbers $P_{N}(n)$ starting with $N=3$.



    Here's the pattern, the sum is done by adding elements of the same column starting with the triangular numbers.
    $$1+1+1+1=4=2^2$$
    $$3+4+5+6+7=25=5^2$$
    $$6+9+12+15+18+21=81=9^2$$
    $$10+16+22+28+34+40+46=196=14^2$$
    ...
    The squares on the right hand side are given by $n+T_{n}$. The sum of index $(n+1)$ has one more element than that of $n$.



    1-Does this pattern repeat indefinitely? (I suspect the answer is yes but I can't prove it)

    2-Do we know why we have such a pattern?






    elementary-number-theory pattern-recognition







    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      I am using the table of polygonal numbers on this site:



      http://oeis.org/wiki/Polygonal_numbers



      The first row of the table gives the value of $n=0,1,2,3,...$. The first column gives the polygonal numbers $P_{N}(n)$ starting with $N=3$.



      Here's the pattern, the sum is done by adding elements of the same column starting with the triangular numbers.
      $$1+1+1+1=4=2^2$$
      $$3+4+5+6+7=25=5^2$$
      $$6+9+12+15+18+21=81=9^2$$
      $$10+16+22+28+34+40+46=196=14^2$$
      ...
      The squares on the right hand side are given by $n+T_{n}$. The sum of index $(n+1)$ has one more element than that of $n$.



      1-Does this pattern repeat indefinitely? (I suspect the answer is yes but I can't prove it)

      2-Do we know why we have such a pattern?






      elementary-number-theory pattern-recognition







      share|cite|improve this question











      $endgroup$




      I am using the table of polygonal numbers on this site:



      http://oeis.org/wiki/Polygonal_numbers



      The first row of the table gives the value of $n=0,1,2,3,...$. The first column gives the polygonal numbers $P_{N}(n)$ starting with $N=3$.



      Here's the pattern, the sum is done by adding elements of the same column starting with the triangular numbers.
      $$1+1+1+1=4=2^2$$
      $$3+4+5+6+7=25=5^2$$
      $$6+9+12+15+18+21=81=9^2$$
      $$10+16+22+28+34+40+46=196=14^2$$
      ...
      The squares on the right hand side are given by $n+T_{n}$. The sum of index $(n+1)$ has one more element than that of $n$.



      1-Does this pattern repeat indefinitely? (I suspect the answer is yes but I can't prove it)

      2-Do we know why we have such a pattern?






      elementary-number-theory pattern-recognition




      elementary-number-theory pattern-recognition






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 12 hours ago









      Bernard

      122k740116




      122k740116










      asked 12 hours ago









      user25406user25406

      3571413




      3571413






















          2 Answers
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          active

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          5












          $begingroup$

          Let the first term be a triangular number $n(n+1)/2$, the common difference between successive terms be $n(n-1)/2$, and the number of terms be $n+3$. Then the average of all terms is



          $dfrac{n(n+1)}{2}+dfrac{n+2}{2}×dfrac{n(n-1)}{2}=dfrac{n^2(n+3)}{4}$



          Multiplying this by the number of terms $n+3$ then gives a sum of $(n(n+3)/2)^2$.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            so this pattern is just a property of the triangular numbers though we are summing over polygonal numbers (including of course the triangular number).
            $endgroup$
            – user25406
            11 hours ago






          • 2




            $begingroup$
            Yes. Polygonal numbers with a different polygon but the same order (e.g. triangular of 3, square of 3, pentagonal of 3, etc) form an arithmetic sequence with a triangular number difference.
            $endgroup$
            – Oscar Lanzi
            11 hours ago



















          2












          $begingroup$

          This is a much more drawn out approach using summations. It doesn't have the elegance of Oscar's solutions, but it still works. You have that with $T_k$ being the $k$-th Triangular number (here $T_0=0$)



          $$sum_{k=1}^{4+0}{[T_1+(k-1)T_0]}=(T_1+1)^2$$



          $$sum_{k=1}^{4+1}{[T_2+(k-1)T_1]}=(T_2+2)^2$$



          $$sum_{k=0}^{4+2}{[T_3+(k-1)T_2]}=(T_3+3)^2$$



          This suggests that



          $$sum_{k=1}^{m+4}{[T_{m+1}+(k-1)T_{m}]}=(T_{m+1}+(m+1))^2$$



          Since summation is linear, we have



          $$sum_{k=1}^{m+4}{[T_m+(k-1)kT_{m-1}]}=T_{m+1}sum_{k=1}^{m+4}1+T_msum_{k=1}^{m+4}{(k-1)}=T_{m+1}(m+4)+T_msum_{k=1}^{m+3}{k}$$



          Using the formula for the $m$-th Triangular number, we have



          $$=left[frac{(m+1)(m+2)}{2}right](m+4)+left[frac{m(m+1)}{2}right]left[frac{(m+3)(m+4)}{2}right]$$



          Factoring and simplifying gives



          $$frac{(m+1)(m+4)}{2}left[(m+2)+frac{m(m+3)}{2}right]=frac{(m+1)^2(m+4)^2}{4}$$



          Now, the right hand side



          $$[T_{m+1}+(m+1)]^2=T_{m+1}^2+2(m+1)T_{m+1}+(m+1)^2$$



          $$=frac{(m+1)^2(m+2)^2}{4}+2(m+1)^2frac{(m+2)}{2}+(m+1)^2$$



          Factoring out an $(m+1)^2$,



          $$=(m+1)^2left[frac{(m+2)^2}{4}+frac{4(m+2)}{4}+frac{4}{4}right]=frac{(m+1)^2}{4}left[(m^2+4m+4)+(4m+8)+4right]$$



          $$=frac{(m+1)^2}{4}(m^2+8m+16)=frac{(m+1)^2(m+4)^2}{4}$$



          Since both expressions equal $frac{(m+1)^2(m+4)^2}{4}$, we are done.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Impressive. I noticed that the square were of the form $(n+T_{n})^2$ but I couldn't do the math.
            $endgroup$
            – user25406
            10 hours ago






          • 2




            $begingroup$
            These types of patterns are not unique to triangular numbers, they pop up a bit. For example, $$1=1^3$$ $$3+5=2^3$$ $$7+9+11=3^3$$ $$13+15+17+19=4^3$$ I can see if there is a link to other problems like this...
            $endgroup$
            – Eleven-Eleven
            10 hours ago











          Your Answer





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          2 Answers
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          2 Answers
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          active

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          active

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          active

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          5












          $begingroup$

          Let the first term be a triangular number $n(n+1)/2$, the common difference between successive terms be $n(n-1)/2$, and the number of terms be $n+3$. Then the average of all terms is



          $dfrac{n(n+1)}{2}+dfrac{n+2}{2}×dfrac{n(n-1)}{2}=dfrac{n^2(n+3)}{4}$



          Multiplying this by the number of terms $n+3$ then gives a sum of $(n(n+3)/2)^2$.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            so this pattern is just a property of the triangular numbers though we are summing over polygonal numbers (including of course the triangular number).
            $endgroup$
            – user25406
            11 hours ago






          • 2




            $begingroup$
            Yes. Polygonal numbers with a different polygon but the same order (e.g. triangular of 3, square of 3, pentagonal of 3, etc) form an arithmetic sequence with a triangular number difference.
            $endgroup$
            – Oscar Lanzi
            11 hours ago
















          5












          $begingroup$

          Let the first term be a triangular number $n(n+1)/2$, the common difference between successive terms be $n(n-1)/2$, and the number of terms be $n+3$. Then the average of all terms is



          $dfrac{n(n+1)}{2}+dfrac{n+2}{2}×dfrac{n(n-1)}{2}=dfrac{n^2(n+3)}{4}$



          Multiplying this by the number of terms $n+3$ then gives a sum of $(n(n+3)/2)^2$.






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            so this pattern is just a property of the triangular numbers though we are summing over polygonal numbers (including of course the triangular number).
            $endgroup$
            – user25406
            11 hours ago






          • 2




            $begingroup$
            Yes. Polygonal numbers with a different polygon but the same order (e.g. triangular of 3, square of 3, pentagonal of 3, etc) form an arithmetic sequence with a triangular number difference.
            $endgroup$
            – Oscar Lanzi
            11 hours ago














          5












          5








          5





          $begingroup$

          Let the first term be a triangular number $n(n+1)/2$, the common difference between successive terms be $n(n-1)/2$, and the number of terms be $n+3$. Then the average of all terms is



          $dfrac{n(n+1)}{2}+dfrac{n+2}{2}×dfrac{n(n-1)}{2}=dfrac{n^2(n+3)}{4}$



          Multiplying this by the number of terms $n+3$ then gives a sum of $(n(n+3)/2)^2$.






          share|cite|improve this answer









          $endgroup$



          Let the first term be a triangular number $n(n+1)/2$, the common difference between successive terms be $n(n-1)/2$, and the number of terms be $n+3$. Then the average of all terms is



          $dfrac{n(n+1)}{2}+dfrac{n+2}{2}×dfrac{n(n-1)}{2}=dfrac{n^2(n+3)}{4}$



          Multiplying this by the number of terms $n+3$ then gives a sum of $(n(n+3)/2)^2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 11 hours ago









          Oscar LanziOscar Lanzi

          13k12136




          13k12136








          • 2




            $begingroup$
            so this pattern is just a property of the triangular numbers though we are summing over polygonal numbers (including of course the triangular number).
            $endgroup$
            – user25406
            11 hours ago






          • 2




            $begingroup$
            Yes. Polygonal numbers with a different polygon but the same order (e.g. triangular of 3, square of 3, pentagonal of 3, etc) form an arithmetic sequence with a triangular number difference.
            $endgroup$
            – Oscar Lanzi
            11 hours ago














          • 2




            $begingroup$
            so this pattern is just a property of the triangular numbers though we are summing over polygonal numbers (including of course the triangular number).
            $endgroup$
            – user25406
            11 hours ago






          • 2




            $begingroup$
            Yes. Polygonal numbers with a different polygon but the same order (e.g. triangular of 3, square of 3, pentagonal of 3, etc) form an arithmetic sequence with a triangular number difference.
            $endgroup$
            – Oscar Lanzi
            11 hours ago








          2




          2




          $begingroup$
          so this pattern is just a property of the triangular numbers though we are summing over polygonal numbers (including of course the triangular number).
          $endgroup$
          – user25406
          11 hours ago




          $begingroup$
          so this pattern is just a property of the triangular numbers though we are summing over polygonal numbers (including of course the triangular number).
          $endgroup$
          – user25406
          11 hours ago




          2




          2




          $begingroup$
          Yes. Polygonal numbers with a different polygon but the same order (e.g. triangular of 3, square of 3, pentagonal of 3, etc) form an arithmetic sequence with a triangular number difference.
          $endgroup$
          – Oscar Lanzi
          11 hours ago




          $begingroup$
          Yes. Polygonal numbers with a different polygon but the same order (e.g. triangular of 3, square of 3, pentagonal of 3, etc) form an arithmetic sequence with a triangular number difference.
          $endgroup$
          – Oscar Lanzi
          11 hours ago











          2












          $begingroup$

          This is a much more drawn out approach using summations. It doesn't have the elegance of Oscar's solutions, but it still works. You have that with $T_k$ being the $k$-th Triangular number (here $T_0=0$)



          $$sum_{k=1}^{4+0}{[T_1+(k-1)T_0]}=(T_1+1)^2$$



          $$sum_{k=1}^{4+1}{[T_2+(k-1)T_1]}=(T_2+2)^2$$



          $$sum_{k=0}^{4+2}{[T_3+(k-1)T_2]}=(T_3+3)^2$$



          This suggests that



          $$sum_{k=1}^{m+4}{[T_{m+1}+(k-1)T_{m}]}=(T_{m+1}+(m+1))^2$$



          Since summation is linear, we have



          $$sum_{k=1}^{m+4}{[T_m+(k-1)kT_{m-1}]}=T_{m+1}sum_{k=1}^{m+4}1+T_msum_{k=1}^{m+4}{(k-1)}=T_{m+1}(m+4)+T_msum_{k=1}^{m+3}{k}$$



          Using the formula for the $m$-th Triangular number, we have



          $$=left[frac{(m+1)(m+2)}{2}right](m+4)+left[frac{m(m+1)}{2}right]left[frac{(m+3)(m+4)}{2}right]$$



          Factoring and simplifying gives



          $$frac{(m+1)(m+4)}{2}left[(m+2)+frac{m(m+3)}{2}right]=frac{(m+1)^2(m+4)^2}{4}$$



          Now, the right hand side



          $$[T_{m+1}+(m+1)]^2=T_{m+1}^2+2(m+1)T_{m+1}+(m+1)^2$$



          $$=frac{(m+1)^2(m+2)^2}{4}+2(m+1)^2frac{(m+2)}{2}+(m+1)^2$$



          Factoring out an $(m+1)^2$,



          $$=(m+1)^2left[frac{(m+2)^2}{4}+frac{4(m+2)}{4}+frac{4}{4}right]=frac{(m+1)^2}{4}left[(m^2+4m+4)+(4m+8)+4right]$$



          $$=frac{(m+1)^2}{4}(m^2+8m+16)=frac{(m+1)^2(m+4)^2}{4}$$



          Since both expressions equal $frac{(m+1)^2(m+4)^2}{4}$, we are done.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Impressive. I noticed that the square were of the form $(n+T_{n})^2$ but I couldn't do the math.
            $endgroup$
            – user25406
            10 hours ago






          • 2




            $begingroup$
            These types of patterns are not unique to triangular numbers, they pop up a bit. For example, $$1=1^3$$ $$3+5=2^3$$ $$7+9+11=3^3$$ $$13+15+17+19=4^3$$ I can see if there is a link to other problems like this...
            $endgroup$
            – Eleven-Eleven
            10 hours ago
















          2












          $begingroup$

          This is a much more drawn out approach using summations. It doesn't have the elegance of Oscar's solutions, but it still works. You have that with $T_k$ being the $k$-th Triangular number (here $T_0=0$)



          $$sum_{k=1}^{4+0}{[T_1+(k-1)T_0]}=(T_1+1)^2$$



          $$sum_{k=1}^{4+1}{[T_2+(k-1)T_1]}=(T_2+2)^2$$



          $$sum_{k=0}^{4+2}{[T_3+(k-1)T_2]}=(T_3+3)^2$$



          This suggests that



          $$sum_{k=1}^{m+4}{[T_{m+1}+(k-1)T_{m}]}=(T_{m+1}+(m+1))^2$$



          Since summation is linear, we have



          $$sum_{k=1}^{m+4}{[T_m+(k-1)kT_{m-1}]}=T_{m+1}sum_{k=1}^{m+4}1+T_msum_{k=1}^{m+4}{(k-1)}=T_{m+1}(m+4)+T_msum_{k=1}^{m+3}{k}$$



          Using the formula for the $m$-th Triangular number, we have



          $$=left[frac{(m+1)(m+2)}{2}right](m+4)+left[frac{m(m+1)}{2}right]left[frac{(m+3)(m+4)}{2}right]$$



          Factoring and simplifying gives



          $$frac{(m+1)(m+4)}{2}left[(m+2)+frac{m(m+3)}{2}right]=frac{(m+1)^2(m+4)^2}{4}$$



          Now, the right hand side



          $$[T_{m+1}+(m+1)]^2=T_{m+1}^2+2(m+1)T_{m+1}+(m+1)^2$$



          $$=frac{(m+1)^2(m+2)^2}{4}+2(m+1)^2frac{(m+2)}{2}+(m+1)^2$$



          Factoring out an $(m+1)^2$,



          $$=(m+1)^2left[frac{(m+2)^2}{4}+frac{4(m+2)}{4}+frac{4}{4}right]=frac{(m+1)^2}{4}left[(m^2+4m+4)+(4m+8)+4right]$$



          $$=frac{(m+1)^2}{4}(m^2+8m+16)=frac{(m+1)^2(m+4)^2}{4}$$



          Since both expressions equal $frac{(m+1)^2(m+4)^2}{4}$, we are done.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Impressive. I noticed that the square were of the form $(n+T_{n})^2$ but I couldn't do the math.
            $endgroup$
            – user25406
            10 hours ago






          • 2




            $begingroup$
            These types of patterns are not unique to triangular numbers, they pop up a bit. For example, $$1=1^3$$ $$3+5=2^3$$ $$7+9+11=3^3$$ $$13+15+17+19=4^3$$ I can see if there is a link to other problems like this...
            $endgroup$
            – Eleven-Eleven
            10 hours ago














          2












          2








          2





          $begingroup$

          This is a much more drawn out approach using summations. It doesn't have the elegance of Oscar's solutions, but it still works. You have that with $T_k$ being the $k$-th Triangular number (here $T_0=0$)



          $$sum_{k=1}^{4+0}{[T_1+(k-1)T_0]}=(T_1+1)^2$$



          $$sum_{k=1}^{4+1}{[T_2+(k-1)T_1]}=(T_2+2)^2$$



          $$sum_{k=0}^{4+2}{[T_3+(k-1)T_2]}=(T_3+3)^2$$



          This suggests that



          $$sum_{k=1}^{m+4}{[T_{m+1}+(k-1)T_{m}]}=(T_{m+1}+(m+1))^2$$



          Since summation is linear, we have



          $$sum_{k=1}^{m+4}{[T_m+(k-1)kT_{m-1}]}=T_{m+1}sum_{k=1}^{m+4}1+T_msum_{k=1}^{m+4}{(k-1)}=T_{m+1}(m+4)+T_msum_{k=1}^{m+3}{k}$$



          Using the formula for the $m$-th Triangular number, we have



          $$=left[frac{(m+1)(m+2)}{2}right](m+4)+left[frac{m(m+1)}{2}right]left[frac{(m+3)(m+4)}{2}right]$$



          Factoring and simplifying gives



          $$frac{(m+1)(m+4)}{2}left[(m+2)+frac{m(m+3)}{2}right]=frac{(m+1)^2(m+4)^2}{4}$$



          Now, the right hand side



          $$[T_{m+1}+(m+1)]^2=T_{m+1}^2+2(m+1)T_{m+1}+(m+1)^2$$



          $$=frac{(m+1)^2(m+2)^2}{4}+2(m+1)^2frac{(m+2)}{2}+(m+1)^2$$



          Factoring out an $(m+1)^2$,



          $$=(m+1)^2left[frac{(m+2)^2}{4}+frac{4(m+2)}{4}+frac{4}{4}right]=frac{(m+1)^2}{4}left[(m^2+4m+4)+(4m+8)+4right]$$



          $$=frac{(m+1)^2}{4}(m^2+8m+16)=frac{(m+1)^2(m+4)^2}{4}$$



          Since both expressions equal $frac{(m+1)^2(m+4)^2}{4}$, we are done.






          share|cite|improve this answer









          $endgroup$



          This is a much more drawn out approach using summations. It doesn't have the elegance of Oscar's solutions, but it still works. You have that with $T_k$ being the $k$-th Triangular number (here $T_0=0$)



          $$sum_{k=1}^{4+0}{[T_1+(k-1)T_0]}=(T_1+1)^2$$



          $$sum_{k=1}^{4+1}{[T_2+(k-1)T_1]}=(T_2+2)^2$$



          $$sum_{k=0}^{4+2}{[T_3+(k-1)T_2]}=(T_3+3)^2$$



          This suggests that



          $$sum_{k=1}^{m+4}{[T_{m+1}+(k-1)T_{m}]}=(T_{m+1}+(m+1))^2$$



          Since summation is linear, we have



          $$sum_{k=1}^{m+4}{[T_m+(k-1)kT_{m-1}]}=T_{m+1}sum_{k=1}^{m+4}1+T_msum_{k=1}^{m+4}{(k-1)}=T_{m+1}(m+4)+T_msum_{k=1}^{m+3}{k}$$



          Using the formula for the $m$-th Triangular number, we have



          $$=left[frac{(m+1)(m+2)}{2}right](m+4)+left[frac{m(m+1)}{2}right]left[frac{(m+3)(m+4)}{2}right]$$



          Factoring and simplifying gives



          $$frac{(m+1)(m+4)}{2}left[(m+2)+frac{m(m+3)}{2}right]=frac{(m+1)^2(m+4)^2}{4}$$



          Now, the right hand side



          $$[T_{m+1}+(m+1)]^2=T_{m+1}^2+2(m+1)T_{m+1}+(m+1)^2$$



          $$=frac{(m+1)^2(m+2)^2}{4}+2(m+1)^2frac{(m+2)}{2}+(m+1)^2$$



          Factoring out an $(m+1)^2$,



          $$=(m+1)^2left[frac{(m+2)^2}{4}+frac{4(m+2)}{4}+frac{4}{4}right]=frac{(m+1)^2}{4}left[(m^2+4m+4)+(4m+8)+4right]$$



          $$=frac{(m+1)^2}{4}(m^2+8m+16)=frac{(m+1)^2(m+4)^2}{4}$$



          Since both expressions equal $frac{(m+1)^2(m+4)^2}{4}$, we are done.







          share|cite|improve this answer












          share|cite|improve this answer



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          answered 10 hours ago









          Eleven-ElevenEleven-Eleven

          5,75572759




          5,75572759












          • $begingroup$
            Impressive. I noticed that the square were of the form $(n+T_{n})^2$ but I couldn't do the math.
            $endgroup$
            – user25406
            10 hours ago






          • 2




            $begingroup$
            These types of patterns are not unique to triangular numbers, they pop up a bit. For example, $$1=1^3$$ $$3+5=2^3$$ $$7+9+11=3^3$$ $$13+15+17+19=4^3$$ I can see if there is a link to other problems like this...
            $endgroup$
            – Eleven-Eleven
            10 hours ago


















          • $begingroup$
            Impressive. I noticed that the square were of the form $(n+T_{n})^2$ but I couldn't do the math.
            $endgroup$
            – user25406
            10 hours ago






          • 2




            $begingroup$
            These types of patterns are not unique to triangular numbers, they pop up a bit. For example, $$1=1^3$$ $$3+5=2^3$$ $$7+9+11=3^3$$ $$13+15+17+19=4^3$$ I can see if there is a link to other problems like this...
            $endgroup$
            – Eleven-Eleven
            10 hours ago
















          $begingroup$
          Impressive. I noticed that the square were of the form $(n+T_{n})^2$ but I couldn't do the math.
          $endgroup$
          – user25406
          10 hours ago




          $begingroup$
          Impressive. I noticed that the square were of the form $(n+T_{n})^2$ but I couldn't do the math.
          $endgroup$
          – user25406
          10 hours ago




          2




          2




          $begingroup$
          These types of patterns are not unique to triangular numbers, they pop up a bit. For example, $$1=1^3$$ $$3+5=2^3$$ $$7+9+11=3^3$$ $$13+15+17+19=4^3$$ I can see if there is a link to other problems like this...
          $endgroup$
          – Eleven-Eleven
          10 hours ago




          $begingroup$
          These types of patterns are not unique to triangular numbers, they pop up a bit. For example, $$1=1^3$$ $$3+5=2^3$$ $$7+9+11=3^3$$ $$13+15+17+19=4^3$$ I can see if there is a link to other problems like this...
          $endgroup$
          – Eleven-Eleven
          10 hours ago


















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