Timeline for Find the number of tuples $(x_0,x_1,\dots)$ such that $n=\sum\limits_{j=0}^\infty x_j2^{j}$ and $m=\sum\limits_{j=0}^\infty x_j(2^{j}-1)$
Current License: CC BY-SA 4.0
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Sep 13, 2023 at 1:02 | comment | added | Brian Hopkins | Following up on both Dave and BillyJoe's comments, see if the following values through $x^6y^6$ make sense: $$1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + y(x^2 + x^3 + x^4 + x^5 + x^6) \\+ y^2(x^4 + x^5 + x^6) + y^3(x^4+x^5+2x^6) + y^4x^6.$$ | |
Sep 11, 2023 at 17:02 | comment | added | Marcos | @BillyJoe I understand that you can take the generating function, but I dont know how this is useful to get any information of the problem. Maybe there is some standard arguments, but Im not used to work with generating functions. | |
Sep 11, 2023 at 14:01 | comment | added | Fabius Wiesner | Take it just as a hint, it should be the coefficient $x^ny^m$ of the generating function: $$\prod_{j=0}^\infty \frac{1}{1-x^{2^j}y^{2^j-1}}$$ and I think it should be possible to get a recursion. See here for a similar problem. | |
Sep 11, 2023 at 11:07 | comment | added | Dave Benson | Have you tried tabulating the numbers for small $n$ and $m$, and looking up the answer in Sloane? | |
Sep 11, 2023 at 10:21 | history | edited | Marcos | CC BY-SA 4.0 |
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Sep 11, 2023 at 9:27 | history | asked | Marcos | CC BY-SA 4.0 |