How to test whether a number is of the form (n(n+1)(n+2))/ 6












0












$begingroup$


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



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



WHERE:



$$mathrm{Tetrahedral}(n) = frac{(n(n+1)(n+2))}{6}$$



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.










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$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – John Omielan
    Jan 3 at 4:22


















0












$begingroup$


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



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



WHERE:



$$mathrm{Tetrahedral}(n) = frac{(n(n+1)(n+2))}{6}$$



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.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – John Omielan
    Jan 3 at 4:22
















0












0








0





$begingroup$


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



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



WHERE:



$$mathrm{Tetrahedral}(n) = frac{(n(n+1)(n+2))}{6}$$



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.










share|cite|improve this question











$endgroup$




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



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



WHERE:



$$mathrm{Tetrahedral}(n) = frac{(n(n+1)(n+2))}{6}$$



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.







sequences-and-series






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edited Jan 3 at 11:08









Mutantoe

623513




623513










asked Jan 3 at 4:14









danday74danday74

1296




1296








  • 1




    $begingroup$
    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.
    $endgroup$
    – John Omielan
    Jan 3 at 4:22
















  • 1




    $begingroup$
    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.
    $endgroup$
    – John Omielan
    Jan 3 at 4:22










1




1




$begingroup$
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.
$endgroup$
– John Omielan
Jan 3 at 4:22






$begingroup$
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.
$endgroup$
– John Omielan
Jan 3 at 4:22












4 Answers
4






active

oldest

votes


















4












$begingroup$

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









$endgroup$













  • $begingroup$
    implemented, works well in all test cases thx, but only returns a boolean (or am I doing something wrong?) - still very useful +1
    $endgroup$
    – danday74
    Jan 4 at 1:37








  • 1




    $begingroup$
    So, I take it you didn't implement the part in the parenthetical comment that says which one it is?
    $endgroup$
    – jmerry
    Jan 4 at 3:26










  • $begingroup$
    with you now, I kinda follow what's going on - soooo clever! many thx - switching to this approach shaved 100ms off my code execution time
    $endgroup$
    – danday74
    Jan 4 at 4:12



















3












$begingroup$

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









$endgroup$













  • $begingroup$
    nice answer thx, works in all cases, very few iterations needed
    $endgroup$
    – danday74
    Jan 4 at 2:27










  • $begingroup$
    other guy pipped you to post, his works without any iterations - but a superb answer, many thx - and a great generic approach to reducing iterations
    $endgroup$
    – danday74
    Jan 4 at 4:02



















3












$begingroup$

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.



If you prefer "simpler" analytical formulae, you also have
$$n=-1+t_1+t_2$$ where
$$t_1^3=frac{1}{3 left(sqrt{3} sqrt{243 k^2-1}+27 kright)}qquad text{and} qquad t_2^3=frac{1}{9} left(sqrt{3} sqrt{243 k^2-1}+27 kright)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    really appreciate your help, accepted other answer on the basis that usually only 2 iterations are needed and that this MAY be computationally more expensive - but I haven't tested performance and this MAY be faster - indeed, where performance is not an issue, this is undoubtedly a great answer
    $endgroup$
    – danday74
    Jan 4 at 3:34



















1












$begingroup$

Assume that $n(n+1)(n+2)/6=k$. Then $n(n+1)(n+2)-6k=0$, which has exactly one positive root for any positive $k$ (the left side increases monotonically and without bound from a negative intercept for positive $n$). If this is to be an integer then the rational root theorem guarantees that it will be a divisor of $6k$. So test such divisors in increasing order until you hit $n(n+1)(n+2)=6k$ or overshoot with $n(n+1)(n+2)>6k$.



We can cut down on trials by using the property that a tetrahedral number us the sum of alternating squares. With an odd argument the sum is $1^2+3^2+5^2+...$ with the sum truncated at the argument, thus the fifth nonzero terrahedral number is $1^2+3^2+5^2$. With an even argument we would use $2^2+4^2+6^2+...$.



If the argument $n$ is odd then, because all odd squares are $equiv 1bmod 8$, we must have $nequiv 2k-1bmod 8$. An even $n$ gives squares alternating between $4$ and $0bmod 8$, so if $k$ is divisible by $8$ then $nin{0,6}bmod 8$. If $k$ is instead four times an odd number then then $nin{2,4}bmod 8$.



Let's apply these ideas to $84$. This is four times an odd number, so an odd argument must be $equiv 7bmod 8$ and an even argument must be $in{2,4}bmod 8$. The divisors of $6×84=504$ satisfying these modular requirements are



$2,4,7,12,18,...,252$



We begin testing these in ascending order. We might see that $(8)(8+1)(8+2)>8^3>504$ (the cube root test in another answer), so either $2,4$ or $7$ works or else nothing does.






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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      implemented, works well in all test cases thx, but only returns a boolean (or am I doing something wrong?) - still very useful +1
      $endgroup$
      – danday74
      Jan 4 at 1:37








    • 1




      $begingroup$
      So, I take it you didn't implement the part in the parenthetical comment that says which one it is?
      $endgroup$
      – jmerry
      Jan 4 at 3:26










    • $begingroup$
      with you now, I kinda follow what's going on - soooo clever! many thx - switching to this approach shaved 100ms off my code execution time
      $endgroup$
      – danday74
      Jan 4 at 4:12
















    4












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      implemented, works well in all test cases thx, but only returns a boolean (or am I doing something wrong?) - still very useful +1
      $endgroup$
      – danday74
      Jan 4 at 1:37








    • 1




      $begingroup$
      So, I take it you didn't implement the part in the parenthetical comment that says which one it is?
      $endgroup$
      – jmerry
      Jan 4 at 3:26










    • $begingroup$
      with you now, I kinda follow what's going on - soooo clever! many thx - switching to this approach shaved 100ms off my code execution time
      $endgroup$
      – danday74
      Jan 4 at 4:12














    4












    4








    4





    $begingroup$

    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









    $endgroup$



    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 Jan 3 at 4:28









    jmerryjmerry

    15.8k1632




    15.8k1632












    • $begingroup$
      implemented, works well in all test cases thx, but only returns a boolean (or am I doing something wrong?) - still very useful +1
      $endgroup$
      – danday74
      Jan 4 at 1:37








    • 1




      $begingroup$
      So, I take it you didn't implement the part in the parenthetical comment that says which one it is?
      $endgroup$
      – jmerry
      Jan 4 at 3:26










    • $begingroup$
      with you now, I kinda follow what's going on - soooo clever! many thx - switching to this approach shaved 100ms off my code execution time
      $endgroup$
      – danday74
      Jan 4 at 4:12


















    • $begingroup$
      implemented, works well in all test cases thx, but only returns a boolean (or am I doing something wrong?) - still very useful +1
      $endgroup$
      – danday74
      Jan 4 at 1:37








    • 1




      $begingroup$
      So, I take it you didn't implement the part in the parenthetical comment that says which one it is?
      $endgroup$
      – jmerry
      Jan 4 at 3:26










    • $begingroup$
      with you now, I kinda follow what's going on - soooo clever! many thx - switching to this approach shaved 100ms off my code execution time
      $endgroup$
      – danday74
      Jan 4 at 4:12
















    $begingroup$
    implemented, works well in all test cases thx, but only returns a boolean (or am I doing something wrong?) - still very useful +1
    $endgroup$
    – danday74
    Jan 4 at 1:37






    $begingroup$
    implemented, works well in all test cases thx, but only returns a boolean (or am I doing something wrong?) - still very useful +1
    $endgroup$
    – danday74
    Jan 4 at 1:37






    1




    1




    $begingroup$
    So, I take it you didn't implement the part in the parenthetical comment that says which one it is?
    $endgroup$
    – jmerry
    Jan 4 at 3:26




    $begingroup$
    So, I take it you didn't implement the part in the parenthetical comment that says which one it is?
    $endgroup$
    – jmerry
    Jan 4 at 3:26












    $begingroup$
    with you now, I kinda follow what's going on - soooo clever! many thx - switching to this approach shaved 100ms off my code execution time
    $endgroup$
    – danday74
    Jan 4 at 4:12




    $begingroup$
    with you now, I kinda follow what's going on - soooo clever! many thx - switching to this approach shaved 100ms off my code execution time
    $endgroup$
    – danday74
    Jan 4 at 4:12











    3












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      nice answer thx, works in all cases, very few iterations needed
      $endgroup$
      – danday74
      Jan 4 at 2:27










    • $begingroup$
      other guy pipped you to post, his works without any iterations - but a superb answer, many thx - and a great generic approach to reducing iterations
      $endgroup$
      – danday74
      Jan 4 at 4:02
















    3












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      nice answer thx, works in all cases, very few iterations needed
      $endgroup$
      – danday74
      Jan 4 at 2:27










    • $begingroup$
      other guy pipped you to post, his works without any iterations - but a superb answer, many thx - and a great generic approach to reducing iterations
      $endgroup$
      – danday74
      Jan 4 at 4:02














    3












    3








    3





    $begingroup$

    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









    $endgroup$



    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












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 3 at 4:27









    ItsJustASeriesBroItsJustASeriesBro

    1563




    1563












    • $begingroup$
      nice answer thx, works in all cases, very few iterations needed
      $endgroup$
      – danday74
      Jan 4 at 2:27










    • $begingroup$
      other guy pipped you to post, his works without any iterations - but a superb answer, many thx - and a great generic approach to reducing iterations
      $endgroup$
      – danday74
      Jan 4 at 4:02


















    • $begingroup$
      nice answer thx, works in all cases, very few iterations needed
      $endgroup$
      – danday74
      Jan 4 at 2:27










    • $begingroup$
      other guy pipped you to post, his works without any iterations - but a superb answer, many thx - and a great generic approach to reducing iterations
      $endgroup$
      – danday74
      Jan 4 at 4:02
















    $begingroup$
    nice answer thx, works in all cases, very few iterations needed
    $endgroup$
    – danday74
    Jan 4 at 2:27




    $begingroup$
    nice answer thx, works in all cases, very few iterations needed
    $endgroup$
    – danday74
    Jan 4 at 2:27












    $begingroup$
    other guy pipped you to post, his works without any iterations - but a superb answer, many thx - and a great generic approach to reducing iterations
    $endgroup$
    – danday74
    Jan 4 at 4:02




    $begingroup$
    other guy pipped you to post, his works without any iterations - but a superb answer, many thx - and a great generic approach to reducing iterations
    $endgroup$
    – danday74
    Jan 4 at 4:02











    3












    $begingroup$

    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.



    If you prefer "simpler" analytical formulae, you also have
    $$n=-1+t_1+t_2$$ where
    $$t_1^3=frac{1}{3 left(sqrt{3} sqrt{243 k^2-1}+27 kright)}qquad text{and} qquad t_2^3=frac{1}{9} left(sqrt{3} sqrt{243 k^2-1}+27 kright)$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      really appreciate your help, accepted other answer on the basis that usually only 2 iterations are needed and that this MAY be computationally more expensive - but I haven't tested performance and this MAY be faster - indeed, where performance is not an issue, this is undoubtedly a great answer
      $endgroup$
      – danday74
      Jan 4 at 3:34
















    3












    $begingroup$

    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.



    If you prefer "simpler" analytical formulae, you also have
    $$n=-1+t_1+t_2$$ where
    $$t_1^3=frac{1}{3 left(sqrt{3} sqrt{243 k^2-1}+27 kright)}qquad text{and} qquad t_2^3=frac{1}{9} left(sqrt{3} sqrt{243 k^2-1}+27 kright)$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      really appreciate your help, accepted other answer on the basis that usually only 2 iterations are needed and that this MAY be computationally more expensive - but I haven't tested performance and this MAY be faster - indeed, where performance is not an issue, this is undoubtedly a great answer
      $endgroup$
      – danday74
      Jan 4 at 3:34














    3












    3








    3





    $begingroup$

    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.



    If you prefer "simpler" analytical formulae, you also have
    $$n=-1+t_1+t_2$$ where
    $$t_1^3=frac{1}{3 left(sqrt{3} sqrt{243 k^2-1}+27 kright)}qquad text{and} qquad t_2^3=frac{1}{9} left(sqrt{3} sqrt{243 k^2-1}+27 kright)$$






    share|cite|improve this answer











    $endgroup$



    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.



    If you prefer "simpler" analytical formulae, you also have
    $$n=-1+t_1+t_2$$ where
    $$t_1^3=frac{1}{3 left(sqrt{3} sqrt{243 k^2-1}+27 kright)}qquad text{and} qquad t_2^3=frac{1}{9} left(sqrt{3} sqrt{243 k^2-1}+27 kright)$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 3 at 7:21

























    answered Jan 3 at 6:39









    Claude LeiboviciClaude Leibovici

    124k1158135




    124k1158135












    • $begingroup$
      really appreciate your help, accepted other answer on the basis that usually only 2 iterations are needed and that this MAY be computationally more expensive - but I haven't tested performance and this MAY be faster - indeed, where performance is not an issue, this is undoubtedly a great answer
      $endgroup$
      – danday74
      Jan 4 at 3:34


















    • $begingroup$
      really appreciate your help, accepted other answer on the basis that usually only 2 iterations are needed and that this MAY be computationally more expensive - but I haven't tested performance and this MAY be faster - indeed, where performance is not an issue, this is undoubtedly a great answer
      $endgroup$
      – danday74
      Jan 4 at 3:34
















    $begingroup$
    really appreciate your help, accepted other answer on the basis that usually only 2 iterations are needed and that this MAY be computationally more expensive - but I haven't tested performance and this MAY be faster - indeed, where performance is not an issue, this is undoubtedly a great answer
    $endgroup$
    – danday74
    Jan 4 at 3:34




    $begingroup$
    really appreciate your help, accepted other answer on the basis that usually only 2 iterations are needed and that this MAY be computationally more expensive - but I haven't tested performance and this MAY be faster - indeed, where performance is not an issue, this is undoubtedly a great answer
    $endgroup$
    – danday74
    Jan 4 at 3:34











    1












    $begingroup$

    Assume that $n(n+1)(n+2)/6=k$. Then $n(n+1)(n+2)-6k=0$, which has exactly one positive root for any positive $k$ (the left side increases monotonically and without bound from a negative intercept for positive $n$). If this is to be an integer then the rational root theorem guarantees that it will be a divisor of $6k$. So test such divisors in increasing order until you hit $n(n+1)(n+2)=6k$ or overshoot with $n(n+1)(n+2)>6k$.



    We can cut down on trials by using the property that a tetrahedral number us the sum of alternating squares. With an odd argument the sum is $1^2+3^2+5^2+...$ with the sum truncated at the argument, thus the fifth nonzero terrahedral number is $1^2+3^2+5^2$. With an even argument we would use $2^2+4^2+6^2+...$.



    If the argument $n$ is odd then, because all odd squares are $equiv 1bmod 8$, we must have $nequiv 2k-1bmod 8$. An even $n$ gives squares alternating between $4$ and $0bmod 8$, so if $k$ is divisible by $8$ then $nin{0,6}bmod 8$. If $k$ is instead four times an odd number then then $nin{2,4}bmod 8$.



    Let's apply these ideas to $84$. This is four times an odd number, so an odd argument must be $equiv 7bmod 8$ and an even argument must be $in{2,4}bmod 8$. The divisors of $6×84=504$ satisfying these modular requirements are



    $2,4,7,12,18,...,252$



    We begin testing these in ascending order. We might see that $(8)(8+1)(8+2)>8^3>504$ (the cube root test in another answer), so either $2,4$ or $7$ works or else nothing does.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Assume that $n(n+1)(n+2)/6=k$. Then $n(n+1)(n+2)-6k=0$, which has exactly one positive root for any positive $k$ (the left side increases monotonically and without bound from a negative intercept for positive $n$). If this is to be an integer then the rational root theorem guarantees that it will be a divisor of $6k$. So test such divisors in increasing order until you hit $n(n+1)(n+2)=6k$ or overshoot with $n(n+1)(n+2)>6k$.



      We can cut down on trials by using the property that a tetrahedral number us the sum of alternating squares. With an odd argument the sum is $1^2+3^2+5^2+...$ with the sum truncated at the argument, thus the fifth nonzero terrahedral number is $1^2+3^2+5^2$. With an even argument we would use $2^2+4^2+6^2+...$.



      If the argument $n$ is odd then, because all odd squares are $equiv 1bmod 8$, we must have $nequiv 2k-1bmod 8$. An even $n$ gives squares alternating between $4$ and $0bmod 8$, so if $k$ is divisible by $8$ then $nin{0,6}bmod 8$. If $k$ is instead four times an odd number then then $nin{2,4}bmod 8$.



      Let's apply these ideas to $84$. This is four times an odd number, so an odd argument must be $equiv 7bmod 8$ and an even argument must be $in{2,4}bmod 8$. The divisors of $6×84=504$ satisfying these modular requirements are



      $2,4,7,12,18,...,252$



      We begin testing these in ascending order. We might see that $(8)(8+1)(8+2)>8^3>504$ (the cube root test in another answer), so either $2,4$ or $7$ works or else nothing does.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Assume that $n(n+1)(n+2)/6=k$. Then $n(n+1)(n+2)-6k=0$, which has exactly one positive root for any positive $k$ (the left side increases monotonically and without bound from a negative intercept for positive $n$). If this is to be an integer then the rational root theorem guarantees that it will be a divisor of $6k$. So test such divisors in increasing order until you hit $n(n+1)(n+2)=6k$ or overshoot with $n(n+1)(n+2)>6k$.



        We can cut down on trials by using the property that a tetrahedral number us the sum of alternating squares. With an odd argument the sum is $1^2+3^2+5^2+...$ with the sum truncated at the argument, thus the fifth nonzero terrahedral number is $1^2+3^2+5^2$. With an even argument we would use $2^2+4^2+6^2+...$.



        If the argument $n$ is odd then, because all odd squares are $equiv 1bmod 8$, we must have $nequiv 2k-1bmod 8$. An even $n$ gives squares alternating between $4$ and $0bmod 8$, so if $k$ is divisible by $8$ then $nin{0,6}bmod 8$. If $k$ is instead four times an odd number then then $nin{2,4}bmod 8$.



        Let's apply these ideas to $84$. This is four times an odd number, so an odd argument must be $equiv 7bmod 8$ and an even argument must be $in{2,4}bmod 8$. The divisors of $6×84=504$ satisfying these modular requirements are



        $2,4,7,12,18,...,252$



        We begin testing these in ascending order. We might see that $(8)(8+1)(8+2)>8^3>504$ (the cube root test in another answer), so either $2,4$ or $7$ works or else nothing does.






        share|cite|improve this answer











        $endgroup$



        Assume that $n(n+1)(n+2)/6=k$. Then $n(n+1)(n+2)-6k=0$, which has exactly one positive root for any positive $k$ (the left side increases monotonically and without bound from a negative intercept for positive $n$). If this is to be an integer then the rational root theorem guarantees that it will be a divisor of $6k$. So test such divisors in increasing order until you hit $n(n+1)(n+2)=6k$ or overshoot with $n(n+1)(n+2)>6k$.



        We can cut down on trials by using the property that a tetrahedral number us the sum of alternating squares. With an odd argument the sum is $1^2+3^2+5^2+...$ with the sum truncated at the argument, thus the fifth nonzero terrahedral number is $1^2+3^2+5^2$. With an even argument we would use $2^2+4^2+6^2+...$.



        If the argument $n$ is odd then, because all odd squares are $equiv 1bmod 8$, we must have $nequiv 2k-1bmod 8$. An even $n$ gives squares alternating between $4$ and $0bmod 8$, so if $k$ is divisible by $8$ then $nin{0,6}bmod 8$. If $k$ is instead four times an odd number then then $nin{2,4}bmod 8$.



        Let's apply these ideas to $84$. This is four times an odd number, so an odd argument must be $equiv 7bmod 8$ and an even argument must be $in{2,4}bmod 8$. The divisors of $6×84=504$ satisfying these modular requirements are



        $2,4,7,12,18,...,252$



        We begin testing these in ascending order. We might see that $(8)(8+1)(8+2)>8^3>504$ (the cube root test in another answer), so either $2,4$ or $7$ works or else nothing does.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 14:46

























        answered Jan 3 at 12:59









        Oscar LanziOscar Lanzi

        13.2k12136




        13.2k12136






























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