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Apr 30, 2018 at 19:15 vote accept David S. Newman
Apr 30, 2018 at 5:20 answer added Gjergji Zaimi timeline score: 4
Apr 30, 2018 at 4:35 comment added David S. Newman @FedorPetrov I finally understood your question and corrected the problem to read as you have indicated. I would like to think that those extra commas were not put there by me, but by some other well meaning editor.
Apr 30, 2018 at 4:31 history edited David S. Newman CC BY-SA 3.0
I removed some commas which didn't belong. I wish that I could see previous versions of this problem to see if those commas were put there by me or by someone else.
Apr 29, 2018 at 1:23 comment added Gjergji Zaimi Yes, almost verbatim when $k$ is even.
Apr 29, 2018 at 0:56 comment added David S. Newman @GjergjiZaimi Will similar arguments work if f(n) is taken to be 1+x^n+x^2n+x^3n+...+x^kn . ?
Apr 28, 2018 at 15:10 comment added David S. Newman @Fedor Petrov I've made a correction in response to your comment , but I'm not sure if it answers your question.
Apr 28, 2018 at 15:06 history edited David S. Newman CC BY-SA 3.0
corrected question 2
Apr 28, 2018 at 12:57 comment added Fedor Petrov You mean $f(a(1))\cdot f(a(2))\cdot f(a(3))...$?
Apr 27, 2018 at 22:05 comment added Gjergji Zaimi You can actually describe exactly which numbers appear as a value of $a(n)$. For a positive integer $m$ relatively prime to $6$, and nonnegative integers $a,b$, the integer $2^a3^bm$ is a member of the $a(n)$ sequence if and only if $2^a$ appears in the binary expansion of $b+1$. This implies a positive answer to question 1 and a negative answer to question 2. (Apologies for leaving a comment instead of a detailed answer. If no one has expanded on this I can write a proof when I have access to a keyboard.)
S Apr 27, 2018 at 21:57 history suggested Vincent CC BY-SA 3.0
cleared up some of the mess I created with my previous edit...
Apr 27, 2018 at 20:39 review Suggested edits
S Apr 27, 2018 at 21:57
Apr 27, 2018 at 20:03 history asked David S. Newman CC BY-SA 3.0