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{{for|squares of triangular numbers|squared triangular number}}
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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 an [[Infinity|infinite]] number of square triangular numbers; the first few are 0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025 {{OEIS|id=A001110}}.
 
==Explicit formulas==
 
Write ''N''<sub>''k''</sub> for the ''k''th square triangular number, and write ''s''<sub>''k''</sub> and ''t''<sub>''k''</sub> 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>
 
The sequences ''N''<sub>''k''</sub>, ''s''<sub>''k''</sub> and ''t''<sub>''k''</sub> 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>
{{cite book | last1 = Dickson | first1 = Leonard Eugene | authorlink1 = Leonard Eugene Dickson |title = [[History of the Theory of Numbers]] | volume = 2 | publisher = American Mathematical Society | location = Providence | year = 1999 |origyear = 1920 | page = 16 | isbn = 978-0-8218-1935-7 }}
</ref><ref name=Euler>
{{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=Memoires de l'academie des sciences de St.-Petersbourg |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{(3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^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 &= {1 \over 32} \left( ( 1 + \sqrt{2} )^{2k} - ( 1 - \sqrt{2} )^{2k} \right)^2 = {1 \over 32} \left( ( 1 + \sqrt{2} )^{4k}-2 + ( 1 - \sqrt{2} )^{4k} \right) \\
&= {1 \over 32} \left( ( 17 + 12\sqrt{2} )^k -2 + ( 17 - 12\sqrt{2} )^k \right).
\end{align}</math>
The corresponding explicit formulas for ''s''<sub>''k''</sub> and ''t''<sub>''k''</sub> 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==
The problem of finding square triangular numbers reduces to [[Pell's equation]] in the following way.<ref>
{{cite book | last1 = Barbeau | first1 = Edward | title = Pell's Equation | pages = 16–17 | url=http://books.google.com/?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
 
:<math>\frac{t(t+1)}{2} = s^2.</math>
 
With a bit of algebra this becomes
 
:<math>(2t+1)^2=8s^2+1,</math>
 
and then letting ''x'' = 2''t'' + 1 and ''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 ''P''<sub>''k''</sub> 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>
 
:<math>x = P_{2k} + P_{2k-1}, \quad y = P_{2k};</math>
 
and therefore all solutions are given by
 
:<math> s_k = \frac{P_{2k}}{2}, \quad t_k = \frac{P_{2k} + P_{2k-1} -1}{2}, \quad N_k = \left( \frac{P_{2k}}{2} \right)^2.</math>
 
There are many identities about the Pell numbers, and these translate into identities about the square triangular numbers.
 
==Recurrence relations==
 
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>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 = 1\text{ and }N_1 = 36.</math>
 
We have<ref name=Dickson /><ref name=Euler />{{Rp|13}}
 
:<math>s_k = 6s_{k-1} - s_{k-2},\text{ with }s_0 = 0\text{ and }s_1 = 1;</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 ''b''<sup>2</sup>''c''<sup>2</sup>, where ''b'' / ''c'' is a [[Convergent (continued fraction)|convergent]] to the [[continued fraction]] for the square root of 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. |coauthors=A. V. Sylwester, Erwin Just, R. M 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=00029890  |jstor=2312558|publisher=Mathematical Association of America }}
</ref>
 
If the triangular number ''n''(''n''+1)/2 is square, then so is the larger triangular number
:<math>\frac{\bigl( 4n(n+1) \bigr) \bigl( 4n(n+1)+1 \bigr)}{2} = 2^2 \, \frac{n(n+1)}{2} \,(2n+1)^2.</math>
 
We know this result has to be a square, because it is a product of three squares: 2^2 (by the exponent), (n(n+1))/2 (the n'th triangular number, by proof assumption), and the (2n+1)^2 (by the exponent). The product of any numbers that are squares is naturally going to result in another square, which can best be proven by geometrically visualizing the multiplication as the multiplying of a NxN box by an MxM box, which is done by placing one MxM box inside each cell of the NxN box, naturally producing another square result.
 
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)(z^2 - 34z + 1)} = 1 + 36z + 1225 z^2 + \cdots.</math>
 
==Numerical data==
 
As ''k'' becomes larger, the ratio ''t''<sub>''k''</sub> / ''s''<sub>''k''</sub> approaches <math>\sqrt{2} \approx 1.41421</math> and the ratio of successive square triangular numbers approaches <math> (1+\sqrt{2})^4 = 17+12\sqrt{2} \approx 33.97056</math>.
 
:<math> \begin{array}{rrrrll}
k & N_k & s_k & t_k & t_k/s_k & N_k/N_{k-1} \\
0 & 0 & 0 & 0 & & \\
1 & 1 & 1 & 1 & 1 & \\
2 & 36 & 6 & 8 &  1.33333 & 36\\
3 & 1\,225 & 35 & 49 &  1.4 & 34.02778\\
4 & 41\,616 & 204 & 288 &  1.41176 & 33.97224\\
5 & 1\,413\,721 & 1\,189 & 1\,681 &  1.41379 & 33.97061\\
6 & 48\,024\,900 & 6\,930 & 9\,800 &  1.41414 & 33.97056\\
7 & 1\,631\,432\,881 & 40\,391 & 57\,121 &  1.41420 & 33.97056\\
\end{array} </math>
 
==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://opinionator.blogs.nytimes.com/2012/01/04/remembering-michael-dummett/ Michael Dummett's solution]
 
[[Category:Figurate numbers]]
[[Category:Integer sequences]]

Latest revision as of 16:00, 27 February 2014

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Document files allow the user to input information, images, tables and additional elements to improve the presentation. The only issue with this structure compared to additional file kinds including .pdf for example is its ability to be readily editable. This signifies which anybody viewing the file may change it by accident. Additionally, this file formatting can be opened by different programs but it refuses to guarantee which what you see inside the Microsoft Word application can still become the same whenever you see it utilizing another system. However, it happens to be nonetheless preferred by many computer users for its ease of use and features.

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Whenever you are trying to find the greatest tuneup utilities 2014 program, be sure to look for 1 which defragments the registry. It should also scan for assorted items, including invalid paths plus invalid shortcuts plus programs. It could equally identify invalid fonts, check for device driver issues plus repair files. Also, be sure that it has a scheduler. That method, you are able to set it to scan the system at certain occasions on certain days. It sounds like a lot, nevertheless it's completely vital.

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