Timeline for Generate Recamán's sequence
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 31, 2018 at 7:24 | comment | added | don bright | "m+=k+2*p*(k*(k>0)in m" < that is very interesting trick! | |
| Sep 12, 2014 at 15:17 | comment | added | Markuz |
@flornquake Thank you. I did not know that [-1] returns the last element. I learned something today.
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| Sep 12, 2014 at 15:12 | history | edited | Markuz | CC BY-SA 3.0 |
added 97 characters in body
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| Sep 12, 2014 at 13:23 | comment | added | flornquake |
You can then change the first line to m=p,=0, and increment p at the start of the loop to save another character.
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| Sep 12, 2014 at 13:14 | comment | added | flornquake |
You can write -1 instead of p-1. Edit: You can also make m a tuple and write m=0, and m+=k+2*p*(k*(k>0)in m),.
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| Sep 11, 2014 at 19:35 | history | edited | Markuz | CC BY-SA 3.0 |
added 8 characters in body
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| Sep 11, 2014 at 19:22 | comment | added | xnor |
Also, k in m or k<0 can be k*(k>=0)in m since if k<0, the product is 0, which is in m.
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| Sep 11, 2014 at 19:15 | history | edited | Markuz | CC BY-SA 3.0 |
added 7 characters in body
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| Sep 11, 2014 at 19:04 | comment | added | xnor |
You can shorten [k,k+2*p][bool] to k+2*p*(bool).
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| Sep 11, 2014 at 18:53 | review | Low quality posts | |||
| Sep 11, 2014 at 20:56 | |||||
| Sep 11, 2014 at 18:35 | history | answered | Markuz | CC BY-SA 3.0 |