This kind of code will do it for the usual Weirestrass form :

a = 1
b = 3141592653589793238462643383279502884197169399375105820974944592307816406665
p = 2^251 + 17*2^192 +1

E = EllipticCurve(GF(p), [0,0,0,a,b])
print(E)
print(E.abelian_group())

card = E.cardinality()
print("cardinality =",card)
factor(card)

G = E(874739451078007766457464989774322083649278607533249481151382481072868806602,152666792071518830868575557812948353041420400780739481342941381225525861407)
print("Generator order q=", G.order())

But how to do it for a curve in the twisted Edwards form ? Because I suppose converting the curve and the point to the Weirestrass form would change the resulting order being computed right ? 2A01:E0A:401:A7C0:DD6F:EA1B:CCA4:2633 (talk) 21:12, 7 July 2024 (UTC)[reply]

I'm not an expert, but I'd think that the group is isomorphic to the Weierstrass group by which it is induced.  --Lambiam 11:07, 8 July 2024 (UTC)[reply]

If 0 and 1 are counted as perfect powers, can every sufficiently large number be written as the sum of 3 perfect powers? 1.165.223.46 (talk) 12:09, 8 July 2024 (UTC)[reply]

Are you including perfect powers of negative numbers?

example: the first problematic number is 7, which cannot be made from 3 powers of positive numbers, but using negative numbers, 7 = 2^3 + (-1)^3 + 0^3. Dhrm77 (talk) 14:40, 8 July 2024 (UTC)[reply]

Every sufficiently large number, of course I know that 7 and 15 cannot be written as such. 220.132.216.52 (talk) 17:09, 8 July 2024 (UTC)[reply]

[edited: This comment addresses a different problem]: A necessary condition for an integer ${\displaystyle n}$ to equal such a sum a sum of three cubes is that ${\displaystyle n}$ does not equal 4 or 5 modulo 9, because the cubes modulo 9 are 0, 1, and −1, and no three of these numbers can sum to 4 or 5 modulo 9. For the remaining set of integers it is an open problem; see Sums of three cubes.  --Lambiam 16:10, 8 July 2024 (UTC)[reply]

No, I only consider nonnegative numbers. 220.132.216.52 (talk) 17:08, 8 July 2024 (UTC)[reply]

It would seem that no one knows, see OEIS:A113505. GalacticShoe (talk) 17:32, 8 July 2024 (UTC)[reply]

I'm confused. 32 is congruent to 5 mod 9, but it's the sum of 1 perfect power. All numbers not congruent to 7 mod 8 are a sum of three squares so you only have to consider 7, 15, 23, 31, ... The OEIS entry does not cite a source, other than just letting the program run to 10⁸ (which seems feasible). But in general if OEIS doesn't know then it's probably unknown. --RDBury (talk) 01:28, 9 July 2024 (UTC)[reply]

There's also the slightly-more restrictive OEIS:A135393 that doesn't use 0 or 1 as perfect powers, and even then the list seems to probably be finite. This makes me wonder what would happen if you removed the nonnegativity constraint from the base of the power. Certainly many terms would disappear (like ${\displaystyle 335=3^{5}+10^{2}+(-2)^{3}}$), and it seems likely that every number not congruent to ${\displaystyle 4}$ or ${\displaystyle 5\!\!\!{\pmod {9}}}$ would be erased as per the sums of three cubes, but the remaining terms might be interesting. GalacticShoe (talk) 04:26, 9 July 2024 (UTC)[reply]

335 is in fact 7^3 + (-2)^3 2402:7500:943:2AC:F4A8:5238:E22:338A (talk) 06:31, 9 July 2024 (UTC)[reply]

Yes, but I wanted to give an example without using -1, 0, or 1. Although there are easy examples like ${\displaystyle 167=13^{2}+(-1)^{3}+(-1)^{3}}$, I'd like to know if there are more without using the more trivial perfect powers. GalacticShoe (talk) 07:38, 9 July 2024 (UTC)[reply]

Are there infinitely many positive integers that are not the sum of two perfect powers? (If 0 and 1 are counted as perfect powers) 2402:7500:943:2AC:F4A8:5238:E22:338A (talk) 06:31, 9 July 2024 (UTC)[reply]

The number of perfect powers up to ${\displaystyle N}$ is (if I'm not mistaken) asymptotically equal to ${\displaystyle {\sqrt {N}}.}$ Then there are ${\displaystyle \Omega (N^{\frac {3}{2}})}$ triples of perfect powers whose sum is at most ${\displaystyle N.}$ Thus, unless there is some number-theoretic conspiracy that makes many triple sums unexpectedly often have the same value, one expects, by a naive counting argument, saturation: not only can eventually all numbers be expected to be the sum of three perfect powers, but to be so in many ways.  --Lambiam 12:03, 9 July 2024 (UTC)[reply]

For N congruent to 7 mod 8 you need at least one odd power. So I think the number of triples you can use for these N is more like ${\displaystyle \Omega (N^{\frac {4}{3}})}$. The exponent is still greater than 1 though. The asymptotic density of numbers which are the sum of three squares is 7/8 and for two squares it's 0. In these cases your tuple counting argument would estimate densities of ${\displaystyle \Omega (N^{\frac {3}{2}})}$ and ${\displaystyle \Omega (N)}$ respectively, but there are indeed "number-theoretic conspiracies" in both cases. --RDBury (talk) 03:56, 10 July 2024 (UTC)[reply]

This additional question is “are there infinitely many positive integers that are not the sum of two perfect powers?” 49.217.131.145 (talk) 12:42, 10 July 2024 (UTC)[reply]

I think Lambiam's tuple counting argument with a tweak or two settles this. If N is congruent to 3 mod four then there must be an odd power. The number of perfect powers less than N is asymptotically ${\displaystyle {\sqrt {N}}}$ and the number of odd powers is asymptotically ${\displaystyle {\sqrt[{3}]{N}}.}$ The number of combinations is then ${\displaystyle \Omega (N^{\frac {5}{6}})}$ which is asymptotically less than N. I didn't see an OEIS entry for this, but that may be because I was trying to work out the first few terms in my head. Sequence A075434 comes close, but they're not counting 0 as a perfect power so it includes 4, 27, ... . --RDBury (talk) 17:26, 10 July 2024 (UTC)[reply]

I haven't proved this, but it seems that no number of the form ${\displaystyle 2^{n}-1,n\geq 18,}$ can be written as the sum of two perfect powers. The largest base-2 repunit that is such a sum may be ${\displaystyle 2^{17}-1=3^{3}+362^{2}.}$  --Lambiam 17:50, 10 July 2024 (UTC)[reply]

It seems to be true if Fermat–Catalan conjecture is true, since numbers == 3 mod 4 (as well as numbers divisible by 3 but not 9, numbers divisible by 7 but not 49, numbers divisible by 11 but not 121, etc.) are not sum of two squares. 220.132.216.52 (talk) 06:03, 11 July 2024 (UTC)[reply]

If ${\displaystyle 0}$ is not considered a perfect power, truth of the Fermat–Catalan conjecture implies a positive answer to the additional question: only a finite number of prime powers of the form ${\displaystyle p^{p}}$ are the sum of two non-zero perfect powers. I don't see how to use it here. The relevance of non-expressibility as sums of squares escapes me.  --Lambiam 05:56, 12 July 2024 (UTC)[reply]

If p == 3 mod 4 and q is odd prime, then p^q is not the sum of two squares, and truth of the Fermat–Catalan conjecture implies that only a finite number of numbers of the form p^q are the sum of two perfect powers such that at least one of them is cube or higher power. 220.132.216.52 (talk) 06:44, 13 July 2024 (UTC)[reply]

This assumes not counting ${\displaystyle 0}$ as a perfect power. The premise of the question is that ${\displaystyle 0}$ and ${\displaystyle 1}$ are both counted as perfect powers.  --Lambiam 10:13, 13 July 2024 (UTC)[reply]

How does Gödel defined a proof in his version of Intermediate logic? 2A02:8071:60A0:92E0:78B6:4D3A:774B:E50C (talk) 18:34, 10 July 2024 (UTC)[reply]

Can you give us a pointer to a source defining "Gödel's version" of intermediate logic? Also, is there evidence that Gödel defined the notion of proof for this logic?  --Lambiam 18:46, 10 July 2024 (UTC)[reply]

Moved to the Computing section of the Reference desk —  --Lambiam 13:36, 16 July 2024 (UTC)[reply]

Moved to the Science section of the Reference desk —  --Lambiam 13:24, 16 July 2024 (UTC).[reply]

Moved to the Computing section of the Reference desk —  --Lambiam 13:37, 16 July 2024 (UTC)[reply]

Is there any smooth function with the following two properties:

${\displaystyle f^{(n)}(x)>0}$, i.e. the nth derivative of f is strictly positive for every x and n.

${\displaystyle \lim _{x\rightarrow \infty }{\frac {f(x)}{b^{x}}}=0}$ for every b > 1. The hard case is when b is small.

Functions like ${\displaystyle a^{x}}$ (for a > 1) are the only ones I can think of with the first property, but none of them has the second property because you can always choose b < a. So I am asking whether there is any function with the first property that grows slower than exponential.

120.21.218.123 (talk) 10:09, 18 July 2024 (UTC)[reply]

Wouldn't any power series with positive coefficients that decrease compared to the coefficients of the exponential do? The exponential is ${\displaystyle 1+\sum _{k=1}^{\infty }{\frac {1}{k!}}x^{k}}$, so e.g. ${\displaystyle f(x)=1+\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {1}{k!}}x^{k}}$ should do the trick. The next question is whether you can find a closed-form expression for this or a similar power series. --Wrongfilter (talk) 13:02, 18 July 2024 (UTC)[reply]

Good thinking. It is of course the case that the first property holds for any power series where all coefficients are positive. Plotting on a graph, I think your specific example doesn't satisfy the second property, but others where the coefficients decrease more rapidly do. 120.21.218.123 (talk) 13:26, 18 July 2024 (UTC)[reply]

${\displaystyle 1+\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {1}{k!}}x^{k}>1+\sum _{k=1}^{\infty }{\frac {1}{k{+}1}}{\frac {1}{k!}}x^{k}=}$ ${\displaystyle 1+\sum _{k=1}^{\infty }{\frac {1}{(k{+}1)!}}x^{k}=}$ ${\displaystyle 1+{\frac {1}{x}}\sum _{k=1}^{\infty }{\frac {1}{(k{+}1)!}}x^{k{+}1}=}$ ${\displaystyle 1+{\frac {1}{x}}\sum _{k=2}^{\infty }{\frac {1}{k}}x^{k}={\frac {\exp x-1}{x}}=}$ ${\displaystyle \Omega (2^{x}).}$  --Lambiam 13:45, 18 July 2024 (UTC)[reply]

A half-exponential function will satisfy your requirements. Hellmuth Kneser famously defined an analytic function that is the functional square root of the exponential function.^[1]  --Lambiam 14:04, 18 July 2024 (UTC)[reply]

References

A variant of Wrongfilter's idea that I think does work:

${\displaystyle f(x)=\sum _{k=0}^{\infty }{\frac {x^{k}}{(k!)^{2}}}}$

(taking ${\displaystyle 0^{0}}$ to be ${\displaystyle 1}$).

Numerical evidence suggests that ${\displaystyle \log f(x)\sim 2{\sqrt {x}}.}$ One might herefore hope that ${\displaystyle e^{2{\sqrt {x}}}}$ would also work. However, its second derivative is negative for ${\displaystyle x<{\tfrac {1}{4}}.}$  --Lambiam 22:20, 20 July 2024 (UTC)[reply]