Square triangular number: Difference between revisions

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:''For{{short squaresdescription|Integer ofthat triangularis both a perfect numbers,square seeand [[squareda triangular number]]''}}

{{for|squares of triangular numbers|squared triangular number}}

A '''square triangular number''' (or '''triangular square number''') is a number which is both a [[triangular number]] and a [[square number|perfect square]]. There is an [[Infinity|infinite]] number of triangular squares, given by the formula

[[File:square_triangular_number_36.svg|234px|thumb|Square triangular number 36 depicted as a triangular number and as a square number.]]

:<math> N_k = {1 \over 32} \left( \left( 1 + \sqrt{2} \right)^{2k} - \left( 1 - \sqrt{2} \right)^{2k} \right)^2 . </math>

or by the linear [[recursion]]

:<math>N_k = 34N_{k-1} - N_{k-2} + 2</math> with <math>N_0 = 0</math> and <math>N_1 = 1</math>

In [[mathematics]], a '''square triangular number''' (or '''triangular square number''') is a number which is both a [[triangular number]] and a [[square number]], in other words, the sum of all integers from <math>1</math> to <math>n</math> has a square root that is an integer. There are [[Infinity|infinitely many]] square triangular numbers; the first few are:

The first few square triangular numbers are [[1 (number)|1]], [[36 (number)|36]], 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, ... {{OEIS|id=A001110}}

{{bi|left=1.6|0, 1, 36, {{val|1225}}, {{val|41616}}, {{val|1413721}}, {{val|48024900}}, {{val|1631432881}}, {{val|55420693056}}, {{val|1882672131025}} {{OEIS|id=A001110}}}}

==Solution as a Pell equation==

The problem of finding square triangular numbers reduces to [[Pell's equation]] in the following way. Every triangular number is of the form ''n''(''n'' + 1)/2. Therefore we seek integers ''n'', ''m'' such that

Write <math>N_k</math> for the <math>k</math>th square triangular number, and write <math>s_k</math> and <math>t_k</math> for the sides of the corresponding square and triangle, so that

:<math>n(n+1)/2 = m^2.</math>

{{bi|left=1.6|<math>\displaystyle N_k = s_k^2 = \frac{t_k(t_k+1)}{2}.</math>}}

With a bit of algebra this becomes

Define the ''triangular root'' of a triangular number <math>N=\tfrac{n(n+1)}{2}</math> to be <math>n</math>. From this definition and the quadratic formula,

:<math>(2n+1)^2=8m^2+1,</math>

{{bi|left=1.6|<math>\displaystyle n = \frac{\sqrt{8N + 1} - 1}{2}.</math>}}

and then letting ''k'' = 2''n'' + 1 and ''h'' = 2''m'', we get the [[Diophantine equation]]

Therefore, <math>N</math> is triangular (<math>n</math> is an integer) [[if and only if]] <math>8N+1</math> is square. Consequently, a square number <math>M^2</math> is also triangular if and only if <math>8M^2+1</math> is square, that is, 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]] <math>x^2-ny^2=1</math> with <math>n=8</math>. All Pell equations have the trivial solution <math>x=1,y=0</math> for any <math>n</math>; this is called the zeroth solution, and indexed as <math>(x_0,y_0)=(1,0)</math>. If <math>(x_k,y_k)</math> denotes the <math>k</math>th nontrivial solution to any Pell equation for a particular <math>n</math>, it can be shown by the method of descent that the next solution is

:<math>k^2=2h^2+1</math>

{{bi|left=1.6|<math>\displaystyle \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 infinitely many solutions to any Pell equation for which there is one non-trivial one, which is true whenever <math>n</math> is not a square. The first non-trivial solution when <math>n=8</math> is easy to find: it is <math>(3,1)</math>. A solution <math>(x_k,y_k)</math> to the Pell equation for <math>n=8</math> yields a square triangular number and its square and triangular roots as follows:

{{bi|left=1.6|<math>\displaystyle s_k = y_k , \quad t_k = \frac{x_k - 1}{2}, \quad N_k = y_k^2.</math>}}

which is an instance of Pell's equation and is solved by the [[Pell number]]s.

Hence, the first square triangular number, derived from <math>(3,1)</math>, is <math>1</math>, and the next, derived from <math>6\cdot (3,1)-(1,0)=(17,6)</math>, is <math>36</math>.

We get the [[recursion]]

The sequences <math>N_k</math>, <math>s_k</math> and <math>t_k</math> are the [[OEIS]] sequences {{OEIS2C|id=A001110}}, {{OEIS2C|id=A001109}}, and {{OEIS2C|id=A001108}} respectively.

:<math>m_{k}=6m_{k-1}-m_{k-2}.</math>

==Explicit formula==

Also, note that

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

{{cite book | last1 = Dickson | first1 = Leonard Eugene | author-link1 = Leonard Eugene Dickson |title = [[History of the Theory of Numbers]] | volume = 2 | publisher = American Mathematical Society | ___location = Providence | year = 1999 |orig-year = 1920 | page = 16 | isbn = 978-0-8218-1935-7 }}

</ref><ref name=Euler>

{{cite journal |last=Euler |first=Leonhard |author-link=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=la |access-date=2009-05-11 |quote=According to the records, it was presented to the St. Petersburg Academy on May 4, 1778.}}

</ref>{{Rp|12–13}}

{{bi|left=1.6|<math>\displaystyle N_k = \left( \frac{\left(3 + 2\sqrt{2}\right)^k - \left(3 - 2\sqrt{2}\right)^k}{4\sqrt{2}} \right)^2.

:<math>m^2_{k}-1=m_{k+1}m_{k-1}</math>

</math>}}

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

since <math>m_{0}=1</math> and <math>m_{1}=6</math>.

{{bi|left=1.6|<math>\displaystyle \begin{align}

The ''k^th'' triangular square ''N_k'' is equal to the ''s^th'' perfect square and the ''t^th'' triangular number, such that

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

:<math> s(N) = \sqrt{N}, </math>

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

:<math> t(N) = \lfloor \sqrt{2 N} \rfloor. </math>

&= \tfrac{1}{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</math> and <math>t_k</math> are:<ref name=Euler />{{Rp|13}}

''t'' is given by the formula

:<math> t(N_k) = {1 \over 4} \left[ \left( \left( 1 + \sqrt{2} \right)^k + \left( 1 - \sqrt{2} \right)^k \right)^2 - \left( 1 + (-1)^k \right)^2 \right]. </math>

{{bi|left=1.6|<math>\displaystyle \begin{align}

or by the recursion

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

:<math> t_k = 2\sqrt{2t_{k-1}(t_{k-1}+1)} + 3t_{k-1} + 1</math>

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

\end{align}</math>}}

==Recurrence relations==

As ''k'' becomes larger, the ratio ''t/s'' approaches the square root of two: Also ratio of successive square triangulars converges to 17+12(sqrt(2))

The solution to the Pell equation can be expressed as a [[recurrence relation]] for the equation's solutions. This can be translated into recurrence equations that directly express the square triangular numbers, as well as the sides of the square and triangle involved. We have<ref>{{MathWorld|title=Square Triangular Number|urlname=SquareTriangularNumber}}</ref>{{Rp|(12)}}

{{bi|left=1.6|<math>\displaystyle \begin{matrixalign} N=1 & s=1 & t=1 & t/s=1

N_k &= 34N_{k-1} - N_{k-2} + 2,& \text{with }N_0 &= 0\text{ and }N_1 = 1; \\

\\ N=36 & s=6 & t=8 & t/s = 1.3333333

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

\\ N=1225 & s=35 & t=49 & t/s = 1.4

\end{align}</math>}}

\\ N=41616 & s=204 & t=288 & t/s = 1.4117647

\\ N=1,413,721 & s=1189 & t=1681 & t/s = 1.4137931

\\ N=48,024,900 & s=6930 & t=9800 & t/s = 1.4141414

\\ N=1,631,432,881 & s=40391 & t=57121 & t/s = 1.4142011

\end{matrix} </math>

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

== References ==

*{{cite journal

| author = Sesskin, Sam

| title = A "converse" to Fermat's last theorem?

| journal = Mathematics Magazine

| volume = 35

| issue = 4

| year = 1962

| pages = 215–217

| url = http://links.jstor.org/sici?sici=0025-570X(196209)35%3A4%3C215%3AA%22TFLT%3E2.0.CO%3B2-6}}

{{bi|left=1.6|<math>\displaystyle \begin{align}

== External links ==

s_k &= 6s_{k-1} - s_{k-2},& \text{with }s_0 &= 0\text{ and }s_1 = 1; \\

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

\end{align}</math>}}

==Other characterizations==

All square triangular numbers have the form <math>b^2c^2</math>, where <math>\tfrac{b}{c}</math> is a [[Convergent (continued fraction)|convergent]] to the [[simple continued fraction|continued fraction expansion]] of <math>\sqrt2</math>, the [[square root of 2]].<ref name=Ball>

{{cite book | last1 = Ball | first1 = W. W. Rouse |author-link1 = W. W. Rouse Ball | last2 = Coxeter | first2 = H. S. M. |author-link2 = Harold Scott MacDonald Coxeter | title = Mathematical Recreations and Essays | url = https://archive.org/details/mathematicalrecr00coxe | url-access = limited | publisher = Dover Publications | ___location = New York | year = 1987 | page = [https://archive.org/details/mathematicalrecr00coxe/page/n72 59]| isbn = 978-0-486-25357-2 }}

</ref>

A. V. Sylwester gave a short proof that there are infinitely many square triangular numbers: If the <math>n</math>th triangular number <math>\tfrac{n(n+1)}{2}</math> is square, then so is the larger <math>4n(n+1)</math>th triangular number, since:

{{bi|left=1.6|<math>\displaystyle\frac{\bigl( 4n(n+1) \bigr) \bigl( 4n(n+1)+1 \bigr)}{2} = 4 \, \frac{n(n+1)}{2} \,\left(2n+1\right)^2.</math>}}

The left hand side of this equation is in the form of a triangular number, and as the product of three squares, the right hand side is square.<ref name=Sylwester>

{{cite journal |last1=Pietenpol |first1=J. L. |first2=A. V. |last2=Sylwester |first3=Erwin |last3=Just |first4=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>

The [[generating function]] for the square triangular numbers is:<ref>{{cite web |first=Simon |last=Plouffe |author-link=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 |date=August 1992 |access-date=2009-05-11 |archive-date=2012-08-20 |archive-url=https://web.archive.org/web/20120820012535/http://www.plouffe.fr/simon/articles/FonctionsGeneratrices.pdf |url-status=dead }}</ref>

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

==See also==

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

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

==Notes==

{{reflist}}

==External links==

* [http://www.cut-the-knot.org/do_you_know/triSquare.shtml Triangular numbers that are also square] at [[cut-the-knot]]

* {{MathWorld|title=Square Triangular Number|urlname=SquareTriangularNumber}}

* [http://www.research.att.com/projects/OEIS?Anum=A001110 Sequence A001110] from the [[On-Line Encyclopedia of Integer Sequences]].

* [http://opinionator.blogs.nytimes.com/2012/01/04/remembering-michael-dummett/ Michael Dummett's solution]

{{Classes of natural numbers|collapsed}}

[[Category:Figurate numbers]]

[[Category:Figurate numbers]]

[[fr:Nombre carré triangulaire]]

[[Category:Integer sequences]]

[[it:Numero quadrato triangolare]]

[[he:מספר משולשי ריבועי]]

[[ja:平方三角数]]

[[sl:Trikotniško kvadratno število]]

[[zh:三角平方數]]