(x=scan())%in%(2*4^(n=0:x)-2^n-1)

March 27, 2019
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(This article was first published on R – Xi'an's Og, and kindly contributed to R-bloggers)

One challenge on code golf is to find the shortest possible code to identify whether or not an integer belongs to the binary cyclops numbers which binary expansion is 0, 101, 11011, 1110111, 111101111, &tc. The n-th such number being

a(n) = 2^{2n + 1} - 2^n - 1 = 2\,4^n - 2^n - 1 = (2^n - 1)(2\,2^n + 1)

this leads to the above solution in R (26 bits). The same length as the C solution [which I do not get]

f(n){n=~n==(n^=-~n)*~n/2;}

And with shorter versions in many esoteric languages I had never heard of, like the 8 bits Brachylog code

ḃD↔Dḍ×ᵐ≠

or the 7 bits Jelly

B¬ŒḂ⁼SƊ

As a side remark, since this was not the purpose of the game, the R code is most inefficient in creating a set of size (x+1), with most terms being Inf.

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