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Project Euler #28

DulcetAirman Jul 20th, 2019 62 Never
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  1. https://projecteuler.net/problem=28
  2.  
  3.  
  4. The series to get the result has this pattern:
  5. 1 + [3 + 5 + 7 + 9] + [13 + 17 + 21 + 25 ] + [31 + 37 + 43 + 49] + [ 56 ...
  6. There's always a quertet of four summands.
  7. How many you skip between the 4 summands (difference of one summand to the next):
  8. 0    2                 4                      6                      8    
  9.  
  10. x :    
  11. 0    1                 2                      3                      4
  12. f(x) := x*2 :
  13. 0    2                 4                      6                      8
  14. h(x) := (2x+1)^2 :
  15. 1    (2+1)^2=9         (4+1)^2=25            49                     81
  16. Note that this is always the last of the quertet.
  17. Now we take that number 4 times because each block is:
  18. [ h(x)-3f(x) + h(x)-2f(x) + h(x)-f(x) + h(x) ] = 4h(x)-6f(x)
  19. g(h) := g(x-1) + 4*h(x) - 6*f(x)
  20. 1   25               101                    261                    537
  21.  
  22. That doesn't mean that it has to be recursive.
  23. With a bit of magic (also know as mathematics) you get this formula:
  24. g(x) := 1 + 2/3 x (13 + x (15 + 8 x))
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