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Zack Wolske's user avatar
Zack Wolske's user avatar
Zack Wolske
  • Member for 13 years, 2 months
  • Last seen this week
  • Toronto, ON, CAN
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On A057985 and A287066
Remarkable! Thank you.
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On A057985 and A287066
No problem! 53 states seems quite nice. I suppose P4 and these fixed points are very intimately related. Is this typical? That is, suppose p and q are fixed points of the same replacement morphism which are not automatic, and p is automatic in some numeration system N with related recurrence - then are p and q connected through this recurrence? Maybe too much to ask in a comment (or in general), but there must be smaller examples or non-examples.
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On A057985 and A287066
$e(10)=6$. I think the definitions for that statement are a bit convoluted. Something equivalent is "Let $r(n)$ be the largest value less than $n$ in A005314. Then $b(n)=a(n-r)$." And similarly, "Let $s$ be the largest sum $\sum c(i)$ less than $n-1$. Then $a(n)=b(n-1-s)$."
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Are there integers $x,y,z$ such that $(x+1)y^2-xz^2=x^3+2x+2$?
I do not know how to use these arguments to solve the cases where $x$ is odd.
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Are there integers $x,y,z$ such that $(x+1)y^2-xz^2=x^3+2x+2$?
@Fedor: suppose $x=-14$, $z^2+(-14)^2=-13(k-14)$, and $-14(k-14)=y^2-2$. Then $k$ is odd and $k-14$ is $1$ mod $4$, because it divides a sum of two squares. But then $-13(k-14)\equiv 3 \pmod 4$.
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Are there integers $x,y,z$ such that $(x+1)y^2-xz^2=x^3+2x+2$?
@Fedor: following JoshuaZ's idea, write $x^2+z^2=(x+1)(x+k)$, so that $y^2-2=x(x+k)$, and using a similar observation, the only primes dividing $(x+k)$ are $1\pmod 8$ and possibly a single $2$. If $k$ is even, then $x$ is odd, else $y^2-2 \equiv 0 \pmod 4$; and if $k$ is odd, then $x$ is even, else $x^2+z^2\equiv 0 \pmod 4$ with $x$ odd. So $k+x$ is odd and $1 \pmod 8$. When $x$ is even, it is $2 \pmod 4$, which means $z^2=kx+k+x \equiv 3 \pmod 4$. Maybe someone can finish the case where $k$ is even.
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Primes above the distant prime neighbors
You are asking if there can be a prime gap larger than 2M with a start below 2p+1 when M is a maximal prime gap starting at p. Seems unlikely! Like a lot of questions about prime gaps, you'll probably be stuck with conjectures about asymptotics. See primerecords.dk/primegaps/maximal.htm for computational records.
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Minimal index of number fields of small degree
The usual index of an element will be $\sqrt{\text{disc}(\mathbb{Z}[a])/\text{disc}(\mathcal{O}_K)}$. @alpoge By Minkowski arguments, this is bounded above by $|\text{disc}(\mathcal{O}_K)|^{(d-2)/2}$, with some improvements when d=3 or 5, see "Algebraic Integers with Small Discriminant" by Thunder and Wolfskill.
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Can $n>7$ be written as $p + 2^k + (1 + (n\ \text{mod}\ 2))\times5^m$ with $p$ an odd prime and $2^k + (1 + (n\ \text{mod}\ 2))\times5^m$ squarefree?
Do you know that $S$ is infinite? That alone seems difficult enough, without the Goldbach style condition, and is necessary.
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Number theory question from Homotopy groups of spheres
Doesn't this procedure miss every prime that is $1$ mod $8$?
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An old paper of S.Chowla on unit equations
fixed a mistake in the book title
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