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

Extract a Pattern of Iterated continued fractions from convergentsA triangular array of numbers.Is there a $3times 3$ magic square adding up to $7$.Which basis orders [for the natural numbers] have been proven?Seeking for an examples of non-trivial sets that can be used to generate all the natural numbersSelf-Similar Reverse-Sum SequencesTriangular pyramid. Find the sum of certain numbers on pyramid.Collatz conjecture pattern (3n + 1 problem).Standalone proof of a conditional part of Lagrange’s Four-Square Theorem?AMC 2000, Problem 16

4

$begingroup$

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

edited 8 hours ago

Bernard

122k740116

asked 8 hours ago

user25406user25406

3521413

$endgroup$

add a comment | 

4

$begingroup$

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

edited 8 hours ago

Bernard

122k740116

asked 8 hours ago

user25406user25406

3521413

$endgroup$

add a comment | 

$begingroup$

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

edited 8 hours ago

Bernard

122k740116

asked 8 hours ago

user25406user25406

3521413

$endgroup$

share|cite|improve this question

edited 8 hours ago

Bernard

122k740116

asked 8 hours ago

user25406user25406

3521413

share|cite|improve this question

edited 8 hours ago

Bernard

122k740116

asked 8 hours ago

user25406user25406

3521413

edited 8 hours ago

Bernard

122k740116

edited 8 hours ago

Bernard

122k740116

asked 8 hours ago

user25406user25406

3521413

asked 8 hours ago

user25406user25406

3521413

2 Answers
2

active

oldest

votes

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

answered 7 hours ago

Oscar LanziOscar Lanzi

13k12136

$endgroup$

• 
  
  
  

  
  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
  7 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
  6 hours ago
  

  
  
  
  

add a comment | 

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

answered 6 hours ago

Eleven-ElevenEleven-Eleven

5,75572759

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Impressive. I noticed that the square were of the form $(n+T_{n})^2$ but I couldn't do the math.
  $endgroup$
  – user25406
  6 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
  6 hours ago
  

  
  
  
  

add a comment | 

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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

answered 7 hours ago

Oscar LanziOscar Lanzi

13k12136

$endgroup$

• 
  
  
  

  
  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
  7 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
  6 hours ago
  

  
  
  
  

add a comment | 

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

answered 7 hours ago

Oscar LanziOscar Lanzi

13k12136

$endgroup$

• 
  
  
  

  
  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
  7 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
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 7 hours ago

Oscar LanziOscar Lanzi

13k12136

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Oscar LanziOscar Lanzi

13k12136

answered 7 hours ago

Oscar LanziOscar Lanzi

13k12136

answered 7 hours ago

Oscar LanziOscar Lanzi

13k12136

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

answered 6 hours ago

Eleven-ElevenEleven-Eleven

5,75572759

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Impressive. I noticed that the square were of the form $(n+T_{n})^2$ but I couldn't do the math.
  $endgroup$
  – user25406
  6 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
  6 hours ago
  

  
  
  
  

add a comment | 

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

answered 6 hours ago

Eleven-ElevenEleven-Eleven

5,75572759

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Impressive. I noticed that the square were of the form $(n+T_{n})^2$ but I couldn't do the math.
  $endgroup$
  – user25406
  6 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
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 6 hours ago

Eleven-ElevenEleven-Eleven

5,75572759

$endgroup$

share|cite|improve this answer

answered 6 hours ago

Eleven-ElevenEleven-Eleven

5,75572759

answered 6 hours ago

Eleven-ElevenEleven-Eleven

5,75572759

answered 6 hours ago

Eleven-ElevenEleven-Eleven

5,75572759