Square triangular number: Difference between revisions

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[[File:TriSquare36.svg|200px|thumb|Square triangular number 36 depicted as a triangular number and as a square number.]]

In [[mathematics]], a '''square triangular number''' (or '''triangular square number''') is a number which is both a [[triangular number]] and a [[square number|perfect square]]. There are [[Infinity|infinitely many]] square triangular numbers; the first few are:

There are [[Infinity|infinitely many]] square triangular numbers; the first few are :0, 1, 36, {{val|1225|fmt=gaps}}, {{val|41616}}, {{val|1413721}}, {{val|48024900}}, {{val|1631432881}}, {{val|55420693056}}, {{val|1882672131025}} {{OEIS|id=A001110}}.

==Explicit formulas==

Write {{math|''N''_''k''}} for the ''{{mvar|k''}}th square triangular number, and write {{math|''s''_''k''}} and {{math|''t''_''k''}} for the sides of the corresponding square and triangle, so that

:<math>N_k = s_k^2 = \frac{t_k(t_k+1)}{2}.</math>

Define the ''triangular root'' of a triangular number {{math|''N'' {{=}} {{sfrac|''n''(''n'' + 1)|2}}}} to be {{mvar|n}}. From this definition and the quadratic formula,

:<math>n = \frac{\sqrt{8N + 1} - 1}{2}.</math>

Therefore, {{mvar|N}} is triangular ({{mvar|n}} is an integer) [[if and only if]] {{math|8''N'' + 1}} is square. Consequently, a square number {{math|''M''²}} is also triangular if and only if {{math|8''M''² + 1}} is square, that is, there are numbers {{mvar|x}} and {{mvar|y}} such that {{math|''x''² − 8''y''² {{=}} 1}}. This is an instance of the [[Pell equation]] with {{math|''n'' {{=}} 8}}. All Pell equations have the trivial solution {{math|''x'' {{=}} 1, ''y'' {{=}} 0}} for any {{mvar|n}}; this is called the zeroth solution, and indexed as {{math|(''x''₀, ''y''₀) {{=}} (1,0)}}. If {{math|(''x_k'', ''y_k'')}} denotes the {{mvar|k}}th nontrivial solution to any Pell equation for a particular {{mvar|n}}, it can be shown by the method of descent that

:<math>\begin{align}

x_{k+1} &= 2x_k x_1 - x_{k-1}, \\

y_{k+1} &= 2y_k x_1 - y_{k-1}.

\end{align}</math>

Hence there are an infinity of solutions to any Pell equation for which there is one non-trivial one, which holds whenever {{mvar|n}} is not a square. The first non-trivial solution when {{math|''n'' {{=}} 8}} is easy to find: it is (3,1). A solution {{math|(''x_k'', ''y_k'')}} to the Pell equation for {{math|''n'' {{=}} 8}} yields a square triangular number and its square and triangular roots as follows:

:<math>s_k = y_k , \quad t_k = \frac{x_k - 1}{2}, \quad N_k = y_k^2.</math>

Hence, the first square triangular number, derived from (3,1), is 1, and the next, derived from {{nowrap|6 × (3,1) − (1,0) {{=}} (17,6)}}, is 36.

The sequences {{math|''N''_''k''}}, {{math|''s''_''k''}} and {{math|''t''_''k''}} are the [[OEIS]] sequences {{OEIS2C|id=A001110}}, {{OEIS2C|id=A001109}}, and {{OEIS2C|id=A001108}} respectively.

Define the ''triangular root'' of a triangular number <math>N = \frac{n(n+1)}{2}</math> to be <math>n</math>. From this definition and the quadratic formula, <math>n = \frac{\sqrt{8N + 1} - 1}{2}.</math> Therefore, <math>N</math> is triangular if and only if <math>8N + 1</math> is square. Consequently, a number <math>M^2</math> is square and triangular if and only if <math>8M^2 + 1</math> is square, i. e., there are numbers <math>x</math> and <math>y</math> such that <math>x^2 - 8y^2 = 1</math>. This is an instance of the Pell equation, with <math>n=8</math>. All Pell equations have the trivial solution (1,0), for any n; this solution is called the zeroth, and indexed as <math>(x_0,y_0)</math>. If <math> (x_k,y_k)</math> denotes the k'th non-trivial solution to any Pell equation for a particular n, it can be shown by the method of descent that <math>x_{k+1} = 2x_k x_1 - x_{k-1}</math> and <math>y_{k+1} = 2y_k x_1 - y_{k-1}</math>. Hence there are an infinity of solutions to any Pell equation for which there is one non-trivial one, which holds whenever n is not a square. The first non-trivial solution when n=8 is easy to find: it is (3,1). A solution <math>(x_k,y_k)</math> to the Pell equation for n=8 yields a square triangular number and its square and triangular roots as follows: <math>s_k = y_k , t_k = \frac{x_k - 1}{2},</math> and <math>N_k = y_k^2.</math> Hence, the first square triangular number, derived from (3,1), is 1, and the next, derived from (17,6) (=6×(3,1)-(1,0)), is 36.

The sequences ''N''_''k'', ''s''_''k'' and ''t''_''k'' are the [[OEIS]] sequences {{OEIS2C|id=A001110}}, {{OEIS2C|id=A001109}}, and {{OEIS2C|id=A001108}} respectively.

In 1778 [[Leonhard Euler]] determined the explicit formula<ref name=Dickson>

Line 19 ⟶ 27:

{{cite journal |last=Euler |first=Leonhard |authorlink=Leonhard Euler |year=1813 |title=Regula facilis problemata Diophantea per numeros integros expedite resolvendi (An easy rule for Diophantine problems which are to be resolved quickly by integral numbers) |journal=Mémoires de l'Académie des Sciences de St.-Pétersbourg |volume= 4 |pages=3–17 |url=http://math.dartmouth.edu/~euler/pages/E739.html |language=Latin |accessdate=2009-05-11 |quote=According to the records, it was presented to the St. Petersburg Academy on May 4, 1778.}}

</ref>{{Rp|12–13}}

:<math>N_k = \left( \frac{\left(3 + 2\sqrt{2}\right)^k - \left(3 - 2\sqrt{2}\right)^k}{4\sqrt{2}} \right)^2.

</math>

Other equivalent formulas (obtained by expanding this formula) that may be convenient include

: <math>\begin{align}

N_k &= \tfrac{1 \over }{32} \left( \left( 1 + \sqrt{2} \right)^{2k} - \left( 1 - \sqrt{2} \right)^{2k} \right)^2 \\
&= \tfrac{1 \over }{32} \left( \left( 1 + \sqrt{2} \right)^{4k}-2 + \left( 1 - \sqrt{2} \right)^{4k} \right) \\

&= \tfrac{1 \over }{32} \left( \left( 17 + 12\sqrt{2} \right)^k -2 + \left( 17 - 12\sqrt{2} \right)^k \right).

\end{align}</math>

The corresponding explicit formulas for {{math|''s''_''k''}} and {{math|''t''_''k''}} are:<ref name=Euler />{{Rp|13}}

:<math>\begin{align}

s_k &= \frac{\left(3 + 2\sqrt{2}\right)^k - \left(3 - 2\sqrt{2}\right)^k}{4\sqrt{2}}, \\

t_k &= \frac{\left(3 + 2\sqrt{2}\right)^k + \left(3 - 2\sqrt{2}\right)^k - 2}{4}.

\end{align}</math>

The corresponding explicit formulas for ''s''_''k'' and ''t''_''k'' are <ref name=Euler />{{Rp|13}}

:<math> s_k = \frac{(3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^k}{4\sqrt{2}} </math>

and

:<math> t_k = \frac{(3 + 2\sqrt{2})^k + (3 - 2\sqrt{2})^k - 2}{4}. </math>

==Pell's equation==

Line 35 ⟶ 45:

{{cite book | last1 = Barbeau | first1 = Edward | title = Pell's Equation | pages = 16–17 | url=https://books.google.com/books?id=FtoFImV5BKMC&pg=PA16 | accessdate = 2009-05-10 |series = Problem Books in Mathematics | publisher = Springer | ___location = New York | year = 2003 | isbn = 978-0-387-95529-2 }}

</ref>

Every triangular number is of the form ''t''(''t'' + 1)/2. Therefore we seek integers ''t'', ''s'' such that

Every triangular number is of the form {{math|{{sfrac|''t''(''t'' + 1)|2}}}}. Therefore we seek integers {{mvar|t}}, {{mvar|s}} such that

:<math>\frac{t(t+1)}{2} = s^2.</math>

With a bit of algebraRearranging, this becomes

:<math>\left(2t+1\right)^2=8s^2+1,</math>

and then letting {{math|''x'' {{=}} 2''t'' + 1}} and {{math|''y'' {{=}} 2''s''}}, we get the [[Diophantine equation]]

:<math>x^2 - 2y^2 =1,</math>

which is an instance of [[Pell's equation]]. This particular equation is solved by the [[Pell number]]s {{math|''P''_''k''}} as<ref>

{{cite book |last1=Hardy |first1=G. H. |authorlink1=G. H. Hardy |last2=Wright |first2=E. M. |authorlink2 = E. M. Wright |title=An Introduction to the Theory of Numbers |edition=5th |year=1979 |publisher=Oxford University Press |isbn=0-19-853171-0 |page=210|quote= Theorem 244 }}

</ref>

Line 63 ⟶ 74:

There are [[recurrence relation]]s for the square triangular numbers, as well as for the sides of the square and triangle involved. We have<ref>{{MathWorld|title=Square Triangular Number|urlname=SquareTriangularNumber}}</ref>{{Rp|(12)}}

:<math>\begin{align}
N_k &= 34N_{k-1} - N_{k-2} + 2,& \text{ with }N_0 &= 0\text{ and }N_1 = 1.</math>; \\

:<math>N_k &= \left(6\sqrt{N_{k-1}} - \sqrt{N_{k-2}}\right)^2,& \text{ with }N_0 &= 0\text{ and }N_1 = 1.</math>

\end{align}</math>

We have<ref name=Dickson /><ref name=Euler />{{Rp|13}}

:<math>\begin{align}
s_k &= 6s_{k-1} - s_{k-2},& \text{ with }s_0 &= 0\text{ and }s_1 = 1;</math> \\

t_k &= 6t_{k-1} - t_{k-2} + 2,& \text{with }t_0 &= 0\text{ and }t_1 = 1.

\end{align}</math>

:<math>t_k = 6t_{k-1} - t_{k-2} + 2,\text{ with }t_0 = 0\text{ and }t_1 = 1.</math>

==Other characterizations==

All square triangular numbers have the form {{math|''b''²''c''²}}, where {{math|{{sfrac|''b'' / |''c''}}}} is a [[Convergent (continued fraction)|convergent]] to the [[continued fraction]] for the [[square root of 2|{{sqrt|2}}]].<ref name=Ball>

{{cite book | last1 = Ball | first1 = W. W. Rouse |authorlink1 = W. W. Rouse Ball | last2 = Coxeter | first2 = H. S. M. |authorlink2 = Harold Scott MacDonald Coxeter | title = Mathematical Recreations and Essays | publisher = Dover Publications | ___location = New York | year = 1987 | page = 59| isbn = 978-0-486-25357-2 }}

</ref>

A. V. Sylwester gave a short proof that there are an infinity of square triangular numbers, to wit:<ref name=Sylwester>

{{cite journal |last=Pietenpol |first=J. L. |author2first2=A. V. |last2=Sylwester |author3first3=Erwin |last3=Just |author4first4=R. M. |last4=Warten |date=February 1962 |title=Elementary Problems and Solutions: E 1473, Square Triangular Numbers |journal=American Mathematical Monthly |volume=69 |issue=2 |pages=168–169 |ISSN=0002-9890 |jstor=2312558|publisher=Mathematical Association of America | doi = 10.2307/2312558}}

</ref>

If the {{mvar|n}}th triangular number {{math|{{sfrac|''n''(''n'' + 1)/|2}}}} is square, then so is the larger {{math|4''n''(''n'' + 1)}}th triangular number, since:

:<math>\frac{\bigl( 4n(n+1) \bigr) \bigl( 4n(n+1)+1 \bigr)}{2} = 2^24 \, \frac{n(n+1)}{2} \,\left(2n+1\right)^2.</math>

We know this result has to be a square, because it is a product of three squares: 2^2 (by the exponent)4, ({{math|{{sfrac|''n''(''n'' + 1))/|2}}}} (the n'thoriginal square triangular number, by proof assumption), and the{{math|(2''n'' (2n+ 1)^^{2 (by the exponent). The product of any numbers that are squares is naturally going to result in another square. This can be seen from the fact that a necessary and sufficient condition for a number to be square is that there should be only even powers of primes in its prime factorisation, and multiplying two square numbers preserves this property in the product}}}.

The triangular roots <{{math|''t<sub>t_kk</mathsub>''}} are alternately simultaneously one less than a square and twice a square, if {{mvar|k}} is even, and simultaneously a square and one less than twice a square, if {{mvar|k}} is odd. Thus, <math>
:49 = 7^² = 2* × 5^² -− 1,
:288 = 17^² -− 1 = 2 *× 12^<sup>2</mathsup>, and <math>
:1681 = 41^² = 2 *× 29^<sup>2 - 1.</mathsup> − 1.
In each case, the two square roots involved multiply to give <{{math|''s_ks_k''}}: {{nowrap|1=5 *× 7 = 35}}, {{nowrap|1=12 *× 17 = 204}}, </math> and <math>{{nowrap|1=29 *× 41 = 1189}}.</math>{{citation needed|date=December 2014}}

Additionally:

<math>N_k - N_{k-1}=s_{2k-1}: 36 - 1 = 35, 1225 - 36 = 1189,</math> and <math>41616 - 1225 = 40391.</math> In other words, the difference between two consecutive square triangular numbers is the square root of another square triangular number.{{citation needed|date=December 2014}}

:<math>N_k - N_{k-1}=s_{2k-1};</math>

{{nowrap|1=36 − 1 = 35}}, {{nowrap|1=1225 − 36 = 1189}}, and {{nowrap|1=41616 − 1225 = 40391}}. In other words, the difference between two consecutive square triangular numbers is the square root of another square triangular number.{{citation needed|date=December 2014}}

The generating function for the square triangular numbers is:<ref>

{{cite web |first=Simon |last=Plouffe |authorlink=Simon Plouffe |title=1031 Generating Functions |url=http://www.lacim.uqam.ca/%7Eplouffe/articles/FonctionsGeneratrices.pdf |publisher=University of Quebec, Laboratoire de combinatoire et d'informatique mathématique |page=A.129 |format=PDF |date=August 1992 |accessdate=2009-05-11 }}

</ref>

:<math>\frac{1+z}{(1-z)\left(z^2 - 34z + 1\right)} = 1 + 36z + 1225 z^2 + \cdots.</math>

==Numerical data==

As <math>{{mvar|k</math>}} becomes larger, the ratio <{{math|{{sfrac|''t<sub>t_kk</s_ksub>''|''s<sub>k</mathsub>''}}}} approaches <math>\sqrt[[square root of 2|{{sqrt|2}}]] \approx ≈&nbsp;{{val|1.41421356</math>}}, and the ratio of successive square triangular numbers approaches <math> {{nowrap|(1 +\sqrt {{sqrt|2}})^⁴}} {{nowrap|{{=}} 17 + 12\sqrt{{sqrt|2}}}} \approx ≈&nbsp;{{val|33.970562748</math>}}. The table below shows values of <math>{{mvar|k</math>}} between 0 and 11, which comprehend all square triangular numbers up to <math>100\,000\,000</math>{{val|e=16}}.

:{| class="wikitable" border="1" style="text-align:right"

|-

! {{mvar|k}}

! <math>k</math>

! {{math|''N<mathsub>N_kk</mathsub>''}}

! {{math|''s<mathsub>s_kk</mathsub>''}}

! {{math|''t<mathsub>t_kk</mathsub>''}}

! {{math|{{sfrac|''t_k''|''s_k''}}}}

! <math>t_k/s_k</math>

! {{math|{{sfrac|''N_k''|''N''_{''k'' − 1}}}}}

! <math> N_k/N_{k-1}</math>

|-

|0

|<math>0</math>

|0

|<math>0</math>

|0

|<math>0</math>

|0

|<math>0</math>

|align=left{{N/A|}}

|

|align=left{{N/A|}}

|

|-

|1

|<math>1</math>

|1

|<math>1</math>

|1

|<math>1</math>

|1

|<math>1</math>

|align=left|1

|<math>1.00000000</math>

|align=left{{N/A|}}

|

|-

|2

|<math>2</math>

|36

|<math>36</math>

|6

|<math>6</math>

|8

|<math>8</math>

|<math>align=left|{{val|1.33333333</math>}}

|align=left|36

|<math>36.000000000</math>

|-

|3

|<math>3</math>

|{{val|1225|fmt=gaps}}

|<math>1\,225</math>

|35

|<math>35</math>

|49

|<math>49</math>

|align=left|1.4

|<math>1.40000000</math>

|<math>align=left|{{val|34.027777778</math>}}

|-

|4

|<math>4</math>

|{{val|41616}}

|<math>41\,616</math>

|<math>204</math>

|<math>288</math>

|<math>align=left|{{val|1.41176471</math>}}

|<math>align=left|{{val|33.972244898</math>}}

|-

|5

|<math>5</math>

|{{val|1413721}}

|<math>1\,413\,721</math>

|{{val|1189|fmt=gaps}}

|<math>1\,189</math>

|{{val|1681|fmt=gaps}}

|<math>1\,681</math>

|<math>align=left|{{val|1.41379310</math>}}

|<math>align=left|{{val|33.970612265</math>}}

|-

|6

|<math>6</math>

|{{val|48024900}}

|<math>48\,024\,900</math>

|{{val|6930|fmt=gaps}}

|<math>6\,930</math>

|{{val|9800|fmt=gaps}}

|<math>9\,800</math>

|<math>align=left|{{val|1.41414141</math>}}

|<math>align=left|{{val|33.970564206</math>}}

|-

|7

|<math>7</math>

|{{val|1631432881}}

|<math>1\,631\,432\,881</math>

|{{val|40391}}

|<math>40\,391</math>

|{{val|57121}}

|<math>57\,121</math>

|<math>align=left|{{val|1.41420118</math>}}

|<math>align=left|{{val|33.970562791</math>}}

|-

|8

|<math>8</math>

|{{val|55420693056}}

|<math>55\,420\,693\,056</math>

|{{val|235416}}

|<math>235\,416</math>

|{{val|332928}}

|<math>332\,928</math>

|<math>align=left|{{val|1.41421144</math>}}

|<math>align=left|{{val|33.970562750</math>}}

|-

|9

|<math>9</math>

|{{val|1882672131025}}

|<math>1\,882\,672\,131\,025</math>

|{{val|1372105}}

|<math>1\,372\,105</math>

|{{val|1940449}}

|<math>1\,940\,449</math>

|<math>align=left|{{val|1.41421320</math>}}

|<math>align=left|{{val|33.970562749</math>}}

|-

|10

|<math>10</math>

|{{val|63955431761796}}

|<math>63\,955\,431\,761\,796</math>

|{{val|7997214}}

|<math>7\,997\,214</math>

|{{val|11309768}}

|<math>11\,309\,768</math>

|<math>align=left|{{val|1.41421350</math>}}

|<math>align=left|{{val|33.970562748</math>}}

|-

|11

|<math>11</math>

|{{val|2172602007770041}}

|<math>2\,172\,602\,007\,770\,041</math>

|{{val|46611179}}

|<math>46\,611\,179</math>

|{{val|65918161}}

|<math>65\,918\,161</math>

|<math>align=left|{{val|1.41421355</math>}}

|<math>align=left|{{val|33.970562748</math>}}

|}<!-- The table was generated in 23-jul-2016 using a Python script available at http://pastebin.com/sWyesrR8 -->

==See also==

*[[Cannonball problem]], on numbers that are simultaneously square and square pyramidal

*[[Sixth power]], numbers that are simultaneously square and cubical