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Jul 9 at 17:34 vote accept Notamathematician
Jul 9 at 15:12 answer added Max Alekseyev timeline score: 4
Jul 7 at 13:54 comment added მამუკა ჯიბლაძე Also above million, there are $a(2498)=a(3014)=a(17414)=a(20462)=1526775$ and $a(13140)=a(13644)=a(267678)=a(275134)=184481855$. Most probably there will be larger and larger $a$-preimages further
Jul 7 at 13:49 comment added მამუკა ჯიბლაძე Oh sorry! Indeed $a(1208)=a(1814)=a(131072)=a(262142)=262143$
Jul 7 at 13:02 comment added Notamathematician @მამუკაჯიბლაძე, please double check your result.
Jul 7 at 12:41 comment added მამუკა ჯიბლაძე Not sure about that, I can only confirm that $a(2n)=a(\tau(2n))$ where binary digits of $\tau(x)$ are $1-\text{reversed binary digits of $x$}$
Jul 7 at 12:34 comment added Notamathematician @მამუკაჯიბლაძე, it looks like we can permute $a(2n)$ with A059894 and it gives no changes. So A035928 is the double of its fixed points (A290254).
Jul 7 at 9:55 comment added მამუკა ჯიბლაძე Some fun facts. Up to $n=$ one million, for most values of $a(n)$ there are exactly two even $n$ with this value. Only for the value $31$ there are three $n$ with this value of $a(n)$, namely $a(12)=a(16)=a(30)=31$. Those even $n$ for which no other even $n$ attains the same value are $2,10,38,42,52,56,142,...$; except for one term, namely 12, this is A035928. These are the only possibilities, i. e. up to one million, number of the even $n$ with the same value of $a(n)$ is either 3 (only for $a(n)=31$), 1 (for $n$ in A035928), or 2 (in all remaining cases)
Jul 7 at 6:13 history asked Notamathematician CC BY-SA 4.0