Finding a pattern, I'm stuck












3












$begingroup$


I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?






binomial-coefficients pattern-recognition pattern-matching







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Karæthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
    $endgroup$
    – Ak19
    7 hours ago










  • $begingroup$
    Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
    $endgroup$
    – Karæthon
    7 hours ago
















3












$begingroup$


I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?






binomial-coefficients pattern-recognition pattern-matching







share|cite|improve this question







New contributor




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







$endgroup$












  • $begingroup$
    I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
    $endgroup$
    – Ak19
    7 hours ago










  • $begingroup$
    Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
    $endgroup$
    – Karæthon
    7 hours ago














3












3








3





$begingroup$


I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?






binomial-coefficients pattern-recognition pattern-matching







share|cite|improve this question







New contributor




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







$endgroup$




I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?






binomial-coefficients pattern-recognition pattern-matching




binomial-coefficients pattern-recognition pattern-matching






share|cite|improve this question







New contributor




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











share|cite|improve this question







New contributor




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









share|cite|improve this question




share|cite|improve this question






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asked 7 hours ago









KaræthonKaræthon

183




183




New contributor




Karæthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





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






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












  • $begingroup$
    I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
    $endgroup$
    – Ak19
    7 hours ago










  • $begingroup$
    Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
    $endgroup$
    – Karæthon
    7 hours ago


















  • $begingroup$
    I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
    $endgroup$
    – Ak19
    7 hours ago










  • $begingroup$
    Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
    $endgroup$
    – Karæthon
    7 hours ago
















$begingroup$
I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
$endgroup$
– Ak19
7 hours ago




$begingroup$
I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
$endgroup$
– Ak19
7 hours ago












$begingroup$
Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
$endgroup$
– Karæthon
7 hours ago




$begingroup$
Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
$endgroup$
– Karæthon
7 hours ago










2 Answers
2






active

oldest

votes


















6












$begingroup$

If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.



Considering it has to do with ellipses, the sequence of denominators should be A056982.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
    $endgroup$
    – Karæthon
    7 hours ago










  • $begingroup$
    The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
    $endgroup$
    – Fimpellizieri
    6 hours ago










  • $begingroup$
    Ok, I think I understand, it works on my calculator so...
    $endgroup$
    – Karæthon
    6 hours ago



















0












$begingroup$

The $n^text{th}$ denominator appears to be



$$4^{large{2n-text{(the number of 1s in the base-2 representation of }n)}}$$



starting with $n=0.$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.



    Considering it has to do with ellipses, the sequence of denominators should be A056982.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
      $endgroup$
      – Karæthon
      7 hours ago










    • $begingroup$
      The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
      $endgroup$
      – Fimpellizieri
      6 hours ago










    • $begingroup$
      Ok, I think I understand, it works on my calculator so...
      $endgroup$
      – Karæthon
      6 hours ago
















    6












    $begingroup$

    If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.



    Considering it has to do with ellipses, the sequence of denominators should be A056982.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
      $endgroup$
      – Karæthon
      7 hours ago










    • $begingroup$
      The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
      $endgroup$
      – Fimpellizieri
      6 hours ago










    • $begingroup$
      Ok, I think I understand, it works on my calculator so...
      $endgroup$
      – Karæthon
      6 hours ago














    6












    6








    6





    $begingroup$

    If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.



    Considering it has to do with ellipses, the sequence of denominators should be A056982.






    share|cite|improve this answer









    $endgroup$



    If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.



    Considering it has to do with ellipses, the sequence of denominators should be A056982.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 7 hours ago









    FimpellizieriFimpellizieri

    17.3k11836




    17.3k11836












    • $begingroup$
      Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
      $endgroup$
      – Karæthon
      7 hours ago










    • $begingroup$
      The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
      $endgroup$
      – Fimpellizieri
      6 hours ago










    • $begingroup$
      Ok, I think I understand, it works on my calculator so...
      $endgroup$
      – Karæthon
      6 hours ago


















    • $begingroup$
      Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
      $endgroup$
      – Karæthon
      7 hours ago










    • $begingroup$
      The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
      $endgroup$
      – Fimpellizieri
      6 hours ago










    • $begingroup$
      Ok, I think I understand, it works on my calculator so...
      $endgroup$
      – Karæthon
      6 hours ago
















    $begingroup$
    Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
    $endgroup$
    – Karæthon
    7 hours ago




    $begingroup$
    Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
    $endgroup$
    – Karæthon
    7 hours ago












    $begingroup$
    The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
    $endgroup$
    – Fimpellizieri
    6 hours ago




    $begingroup$
    The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
    $endgroup$
    – Fimpellizieri
    6 hours ago












    $begingroup$
    Ok, I think I understand, it works on my calculator so...
    $endgroup$
    – Karæthon
    6 hours ago




    $begingroup$
    Ok, I think I understand, it works on my calculator so...
    $endgroup$
    – Karæthon
    6 hours ago











    0












    $begingroup$

    The $n^text{th}$ denominator appears to be



    $$4^{large{2n-text{(the number of 1s in the base-2 representation of }n)}}$$



    starting with $n=0.$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The $n^text{th}$ denominator appears to be



      $$4^{large{2n-text{(the number of 1s in the base-2 representation of }n)}}$$



      starting with $n=0.$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The $n^text{th}$ denominator appears to be



        $$4^{large{2n-text{(the number of 1s in the base-2 representation of }n)}}$$



        starting with $n=0.$






        share|cite|improve this answer









        $endgroup$



        The $n^text{th}$ denominator appears to be



        $$4^{large{2n-text{(the number of 1s in the base-2 representation of }n)}}$$



        starting with $n=0.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 6 hours ago









        Mitchell SpectorMitchell Spector

        7,1572824




        7,1572824






















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