Timeline for Count sums of two squares
Current License: CC BY-SA 3.0
7 events
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| Nov 27, 2015 at 18:44 | comment | added | Alex A. | @GlenO That simplification is seriously genius. I recommend you submit that as an alternative formula for the sequence on OEIS! | |
| Nov 27, 2015 at 18:43 | history | edited | Alex A. | CC BY-SA 3.0 |
This simplification is incredible
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| Nov 27, 2015 at 3:11 | comment | added | Glen O |
So, it can be recovered - if you let the i=0 case happen in the sum, you can switch the sign on 4g(n). So (n==0)-4g(n)-4g(n/2)+8sum(i->g(i)g(n-i),0:n/2), which will not run into the error. But you can do even better, by noting the symmetries - (n==0)+4sum([g(i)g(n-i)for i=1:n])
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| Nov 26, 2015 at 3:45 | comment | added | Alex A. |
@GlenO Great suggestions, thanks. The sum trick fails for n=0 though, so I kept array comprehension.
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| Nov 26, 2015 at 3:43 | history | edited | Alex A. | CC BY-SA 3.0 |
saved 10
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| Nov 26, 2015 at 3:12 | comment | added | Glen O |
Use x^.5 rather than sqrt(x) to save 3 bytes. And (n==0) saves 2 bytes over 1÷(n+1). And you can save 4 more characters by using cos(π*sqrt(x))^2÷1 rather than floor(cos(π*sqrt(x))^2). Also, use 1:n/2 rather than 1:n÷2, because there's no harm using a float in g(x) and it'll be locked to the integers for i anyway. And sum(i->g(i)g(n-i),1:n/2) will shave some more characters, too.
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| Nov 26, 2015 at 1:25 | history | answered | Alex A. | CC BY-SA 3.0 |