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Mar 28, 2021 at 17:09 history edited YCor
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Mar 28, 2021 at 5:58 comment added Gerry Myerson The sum $a_n$ of the first $n$ odious numbers is tabulated at oeis.org/A173209 and the formula $a_n=n^2-(n/2)+O(1)$, attributed to Charles R Greathouse IV, is given without proof.
Mar 28, 2021 at 5:52 comment added Gerry Myerson I would hazard a guess that, for $n$ large, roughly half the positive integers up to $n$ are odious, and the sum of their $k$th powers would be roughly $(1/2)\sum_{r=1}^nr^k$.
Mar 28, 2021 at 4:48 answer added Alapan Das timeline score: 0
Mar 27, 2021 at 21:15 comment added Sylvain JULIEN Also, defining an "inner product" on odious numbers $n_{1}$ and $n_{2}$ as $\sum_{i}b_{1}(i)b_{2}(i)$ where $n_{j}=\sum_{i\geq 0}b_{j}(i)2^{i}$, one can easily figure out that the sum of an odd number of pairwise orthogonal odious numbers is an odious number.
Mar 27, 2021 at 21:07 comment added Sylvain JULIEN Perhaps one can use the fact that any permutation of the binary digits of an odious number gives rise to the binary expansion of another odious number, so group theoretic considerations may be useful.
Mar 27, 2021 at 20:20 answer added Max Alekseyev timeline score: 6
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Mar 27, 2021 at 7:44
Mar 27, 2021 at 1:57 history asked MathNoob CC BY-SA 4.0