tetrahedral number test












1














Tetrahedral numbers (https://oeis.org/A000292) are:



1, 4, 10, 20, 35, 56, 84, etc



WHERE Tetrahedral(n) = (n * (n+1) * (n+2)) / 6



But what is the test for whether a given number is a tetrahedral number?
I also need to know, if it is a tetrahedral number, then which term in the tetrahedral sequence is it? I am looking for a formula or something that would enable me to get these results:



input   result ("is tetrahedral")
1 1
2 false
3 false
4 2
5 false
6 false
7 false
8 false
9 false
10 3
11 false
12 false


I can't find any formula on the Internet for this. Any pointers in the right direction would be great - thx










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  • 1




    Hi & welcome to MSE. You could assume your number to test is a constant & $n$ is a variable, plug it into the equation for a tetrahedron # and solve for $n$ from the resulting $3$rd degree (i.e., cubic) polynomial.
    – John Omielan
    2 hours ago


















1














Tetrahedral numbers (https://oeis.org/A000292) are:



1, 4, 10, 20, 35, 56, 84, etc



WHERE Tetrahedral(n) = (n * (n+1) * (n+2)) / 6



But what is the test for whether a given number is a tetrahedral number?
I also need to know, if it is a tetrahedral number, then which term in the tetrahedral sequence is it? I am looking for a formula or something that would enable me to get these results:



input   result ("is tetrahedral")
1 1
2 false
3 false
4 2
5 false
6 false
7 false
8 false
9 false
10 3
11 false
12 false


I can't find any formula on the Internet for this. Any pointers in the right direction would be great - thx










share|cite|improve this question







New contributor




danday74 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    Hi & welcome to MSE. You could assume your number to test is a constant & $n$ is a variable, plug it into the equation for a tetrahedron # and solve for $n$ from the resulting $3$rd degree (i.e., cubic) polynomial.
    – John Omielan
    2 hours ago
















1












1








1







Tetrahedral numbers (https://oeis.org/A000292) are:



1, 4, 10, 20, 35, 56, 84, etc



WHERE Tetrahedral(n) = (n * (n+1) * (n+2)) / 6



But what is the test for whether a given number is a tetrahedral number?
I also need to know, if it is a tetrahedral number, then which term in the tetrahedral sequence is it? I am looking for a formula or something that would enable me to get these results:



input   result ("is tetrahedral")
1 1
2 false
3 false
4 2
5 false
6 false
7 false
8 false
9 false
10 3
11 false
12 false


I can't find any formula on the Internet for this. Any pointers in the right direction would be great - thx










share|cite|improve this question







New contributor




danday74 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Tetrahedral numbers (https://oeis.org/A000292) are:



1, 4, 10, 20, 35, 56, 84, etc



WHERE Tetrahedral(n) = (n * (n+1) * (n+2)) / 6



But what is the test for whether a given number is a tetrahedral number?
I also need to know, if it is a tetrahedral number, then which term in the tetrahedral sequence is it? I am looking for a formula or something that would enable me to get these results:



input   result ("is tetrahedral")
1 1
2 false
3 false
4 2
5 false
6 false
7 false
8 false
9 false
10 3
11 false
12 false


I can't find any formula on the Internet for this. Any pointers in the right direction would be great - thx







sequences-and-series






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asked 3 hours ago









danday74

1264




1264




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  • 1




    Hi & welcome to MSE. You could assume your number to test is a constant & $n$ is a variable, plug it into the equation for a tetrahedron # and solve for $n$ from the resulting $3$rd degree (i.e., cubic) polynomial.
    – John Omielan
    2 hours ago
















  • 1




    Hi & welcome to MSE. You could assume your number to test is a constant & $n$ is a variable, plug it into the equation for a tetrahedron # and solve for $n$ from the resulting $3$rd degree (i.e., cubic) polynomial.
    – John Omielan
    2 hours ago










1




1




Hi & welcome to MSE. You could assume your number to test is a constant & $n$ is a variable, plug it into the equation for a tetrahedron # and solve for $n$ from the resulting $3$rd degree (i.e., cubic) polynomial.
– John Omielan
2 hours ago






Hi & welcome to MSE. You could assume your number to test is a constant & $n$ is a variable, plug it into the equation for a tetrahedron # and solve for $n$ from the resulting $3$rd degree (i.e., cubic) polynomial.
– John Omielan
2 hours ago












3 Answers
3






active

oldest

votes


















3














Notice that $n(n+1)(n+2)/6approx n^3/6$ for large values of $n$. Thus, if you have a suspected tetrahedral number $k$, and you compute $n_0=lfloorsqrt[3]{6k}rfloor$ you will have a pretty good guess as to which index you need to check to confirm or refute your suspicions.



This of course won't work for small $k$, and your initial guess $n_0$ need not be correct, but you have at least vastly reduced the number of operations you need to finish your problem. The full algorithm now would look something like this:



1) Take the input number $k$ and compute $n_0=lfloorsqrt[3]{6k}rfloor$.



2) If $text{Tetra}(n_0)>k$, begin testing all naturals $n<n_0$ (in decreasing order) until $text{Tetra}(n)leq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.



3) If $text{Tetra}(n_0)<k$, begin testing all naturals $n>n_0$ until $text{Tetra}(n)geq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.



4) If $text{Tetra}(n_0)=k$, then you are done, and the index is $n_0$.






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    2














    Another way to look at that: $frac{n(n+1)(n+2)}{6}=frac{(n+1)^3-(n+1)}{6}$.



    So then, if we have some $m$ we want to test? Multiply it by $6$. Add its cube root, rounded up - $k^3-k > (k-1)^3$ for all $k > 1$. If we get a perfect cube ($6m+lceilsqrt[3]{6m}rceil=k^3$ for some integer $k$), it's a tetrahedral number (Specifically, for $n=k-1$). If not, it isn't.






    share|cite|improve this answer





























      0














      For a given $k$, you want to know if it exists an integer $n$ such that
      $$frac{n(n+1)(n+2)}{6}=k$$ A simple way is to look at the cubic equation
      $$n^3+3n^2+2n-6k=0$$ which has only one real root. Using the hyperbolic solution for this case, the solution is given by
      $$n=-1+frac{2 }{sqrt{3}}cosh left(frac{1}{3} cosh ^{-1}left(9 sqrt{3},
      kright)right)$$
      If this $n$ is an integer (or very very close to because of possible rounding errors), then you are done.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

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        active

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        active

        oldest

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        3














        Notice that $n(n+1)(n+2)/6approx n^3/6$ for large values of $n$. Thus, if you have a suspected tetrahedral number $k$, and you compute $n_0=lfloorsqrt[3]{6k}rfloor$ you will have a pretty good guess as to which index you need to check to confirm or refute your suspicions.



        This of course won't work for small $k$, and your initial guess $n_0$ need not be correct, but you have at least vastly reduced the number of operations you need to finish your problem. The full algorithm now would look something like this:



        1) Take the input number $k$ and compute $n_0=lfloorsqrt[3]{6k}rfloor$.



        2) If $text{Tetra}(n_0)>k$, begin testing all naturals $n<n_0$ (in decreasing order) until $text{Tetra}(n)leq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.



        3) If $text{Tetra}(n_0)<k$, begin testing all naturals $n>n_0$ until $text{Tetra}(n)geq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.



        4) If $text{Tetra}(n_0)=k$, then you are done, and the index is $n_0$.






        share|cite|improve this answer








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          3














          Notice that $n(n+1)(n+2)/6approx n^3/6$ for large values of $n$. Thus, if you have a suspected tetrahedral number $k$, and you compute $n_0=lfloorsqrt[3]{6k}rfloor$ you will have a pretty good guess as to which index you need to check to confirm or refute your suspicions.



          This of course won't work for small $k$, and your initial guess $n_0$ need not be correct, but you have at least vastly reduced the number of operations you need to finish your problem. The full algorithm now would look something like this:



          1) Take the input number $k$ and compute $n_0=lfloorsqrt[3]{6k}rfloor$.



          2) If $text{Tetra}(n_0)>k$, begin testing all naturals $n<n_0$ (in decreasing order) until $text{Tetra}(n)leq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.



          3) If $text{Tetra}(n_0)<k$, begin testing all naturals $n>n_0$ until $text{Tetra}(n)geq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.



          4) If $text{Tetra}(n_0)=k$, then you are done, and the index is $n_0$.






          share|cite|improve this answer








          New contributor




          ItsJustASeriesBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





















            3












            3








            3






            Notice that $n(n+1)(n+2)/6approx n^3/6$ for large values of $n$. Thus, if you have a suspected tetrahedral number $k$, and you compute $n_0=lfloorsqrt[3]{6k}rfloor$ you will have a pretty good guess as to which index you need to check to confirm or refute your suspicions.



            This of course won't work for small $k$, and your initial guess $n_0$ need not be correct, but you have at least vastly reduced the number of operations you need to finish your problem. The full algorithm now would look something like this:



            1) Take the input number $k$ and compute $n_0=lfloorsqrt[3]{6k}rfloor$.



            2) If $text{Tetra}(n_0)>k$, begin testing all naturals $n<n_0$ (in decreasing order) until $text{Tetra}(n)leq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.



            3) If $text{Tetra}(n_0)<k$, begin testing all naturals $n>n_0$ until $text{Tetra}(n)geq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.



            4) If $text{Tetra}(n_0)=k$, then you are done, and the index is $n_0$.






            share|cite|improve this answer








            New contributor




            ItsJustASeriesBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            Notice that $n(n+1)(n+2)/6approx n^3/6$ for large values of $n$. Thus, if you have a suspected tetrahedral number $k$, and you compute $n_0=lfloorsqrt[3]{6k}rfloor$ you will have a pretty good guess as to which index you need to check to confirm or refute your suspicions.



            This of course won't work for small $k$, and your initial guess $n_0$ need not be correct, but you have at least vastly reduced the number of operations you need to finish your problem. The full algorithm now would look something like this:



            1) Take the input number $k$ and compute $n_0=lfloorsqrt[3]{6k}rfloor$.



            2) If $text{Tetra}(n_0)>k$, begin testing all naturals $n<n_0$ (in decreasing order) until $text{Tetra}(n)leq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.



            3) If $text{Tetra}(n_0)<k$, begin testing all naturals $n>n_0$ until $text{Tetra}(n)geq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.



            4) If $text{Tetra}(n_0)=k$, then you are done, and the index is $n_0$.







            share|cite|improve this answer








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            answered 2 hours ago









            ItsJustASeriesBro

            1313




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                2














                Another way to look at that: $frac{n(n+1)(n+2)}{6}=frac{(n+1)^3-(n+1)}{6}$.



                So then, if we have some $m$ we want to test? Multiply it by $6$. Add its cube root, rounded up - $k^3-k > (k-1)^3$ for all $k > 1$. If we get a perfect cube ($6m+lceilsqrt[3]{6m}rceil=k^3$ for some integer $k$), it's a tetrahedral number (Specifically, for $n=k-1$). If not, it isn't.






                share|cite|improve this answer


























                  2














                  Another way to look at that: $frac{n(n+1)(n+2)}{6}=frac{(n+1)^3-(n+1)}{6}$.



                  So then, if we have some $m$ we want to test? Multiply it by $6$. Add its cube root, rounded up - $k^3-k > (k-1)^3$ for all $k > 1$. If we get a perfect cube ($6m+lceilsqrt[3]{6m}rceil=k^3$ for some integer $k$), it's a tetrahedral number (Specifically, for $n=k-1$). If not, it isn't.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    Another way to look at that: $frac{n(n+1)(n+2)}{6}=frac{(n+1)^3-(n+1)}{6}$.



                    So then, if we have some $m$ we want to test? Multiply it by $6$. Add its cube root, rounded up - $k^3-k > (k-1)^3$ for all $k > 1$. If we get a perfect cube ($6m+lceilsqrt[3]{6m}rceil=k^3$ for some integer $k$), it's a tetrahedral number (Specifically, for $n=k-1$). If not, it isn't.






                    share|cite|improve this answer












                    Another way to look at that: $frac{n(n+1)(n+2)}{6}=frac{(n+1)^3-(n+1)}{6}$.



                    So then, if we have some $m$ we want to test? Multiply it by $6$. Add its cube root, rounded up - $k^3-k > (k-1)^3$ for all $k > 1$. If we get a perfect cube ($6m+lceilsqrt[3]{6m}rceil=k^3$ for some integer $k$), it's a tetrahedral number (Specifically, for $n=k-1$). If not, it isn't.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 hours ago









                    jmerry

                    1,87229




                    1,87229























                        0














                        For a given $k$, you want to know if it exists an integer $n$ such that
                        $$frac{n(n+1)(n+2)}{6}=k$$ A simple way is to look at the cubic equation
                        $$n^3+3n^2+2n-6k=0$$ which has only one real root. Using the hyperbolic solution for this case, the solution is given by
                        $$n=-1+frac{2 }{sqrt{3}}cosh left(frac{1}{3} cosh ^{-1}left(9 sqrt{3},
                        kright)right)$$
                        If this $n$ is an integer (or very very close to because of possible rounding errors), then you are done.






                        share|cite|improve this answer


























                          0














                          For a given $k$, you want to know if it exists an integer $n$ such that
                          $$frac{n(n+1)(n+2)}{6}=k$$ A simple way is to look at the cubic equation
                          $$n^3+3n^2+2n-6k=0$$ which has only one real root. Using the hyperbolic solution for this case, the solution is given by
                          $$n=-1+frac{2 }{sqrt{3}}cosh left(frac{1}{3} cosh ^{-1}left(9 sqrt{3},
                          kright)right)$$
                          If this $n$ is an integer (or very very close to because of possible rounding errors), then you are done.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            For a given $k$, you want to know if it exists an integer $n$ such that
                            $$frac{n(n+1)(n+2)}{6}=k$$ A simple way is to look at the cubic equation
                            $$n^3+3n^2+2n-6k=0$$ which has only one real root. Using the hyperbolic solution for this case, the solution is given by
                            $$n=-1+frac{2 }{sqrt{3}}cosh left(frac{1}{3} cosh ^{-1}left(9 sqrt{3},
                            kright)right)$$
                            If this $n$ is an integer (or very very close to because of possible rounding errors), then you are done.






                            share|cite|improve this answer












                            For a given $k$, you want to know if it exists an integer $n$ such that
                            $$frac{n(n+1)(n+2)}{6}=k$$ A simple way is to look at the cubic equation
                            $$n^3+3n^2+2n-6k=0$$ which has only one real root. Using the hyperbolic solution for this case, the solution is given by
                            $$n=-1+frac{2 }{sqrt{3}}cosh left(frac{1}{3} cosh ^{-1}left(9 sqrt{3},
                            kright)right)$$
                            If this $n$ is an integer (or very very close to because of possible rounding errors), then you are done.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 40 mins ago









                            Claude Leibovici

                            119k1157132




                            119k1157132






















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