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March 4

For which natural number n, 1^k+2^k+3^k+…+n^k can be prime?

I saw the sequence (sequence A164307 in the OEIS), and now I have a problem: for For which natural number n, 1^k+2^k+3^k+…+n^k can be prime?

n=2, this is exactly the Fermat primes

n=3, numbers cannot be prime since they are divisible by 2

n=4, numbers cannot be prime since they are divisible by 2

n=5, no fixed small divisor of the numbers, but k must be divisible by 20 since they are divisible by 5 if k is not divisible by 4 and they are divisible by 11 if k == 4, 8, 12, 16 mod 20, and indeed the term k=1440 is prime, but are there infinitely many such primes?

n=6, numbers cannot be prime since they are divisible by 7 if k is not divisible by 6, while they are divisible by 13 if k is divisible by 6

n=7, numbers cannot be prime since they are divisible by 2

n=8, numbers cannot be prime since they are divisible by 2

n=9, numbers cannot be prime since they are divisible by 3

n=10, no fixed small divisor of the numbers, but k must be divisible by 20 since they are divisible by 11 if k is not divisible by 10 and they are divisible by 5 if k == 10 mod 20, but is there any such primes?

n=11, numbers cannot be prime since they are divisible by 2

n=12, numbers cannot be prime since they are divisible by 2

n=13, numbers cannot be prime since they are divisible by 13 if k is not divisible by 12, while they are divisible by 3 if k is divisible by 12

n=14, numbers cannot be prime since they are divisible by 7 if k is not divisible by 6, while they are divisible by 5 if k == 6 mod 12, while they are divisible by 13 if k is divisible by 12

n=15, numbers cannot be prime since they are divisible by 2

n=16, numbers cannot be prime since they are divisible by 2 218.187.65.97 (talk) 15:05, 4 March 2024 (UTC)

These sums have values given by Faulhaber's formula. That article has quite a bit about it but does not touch this question. I'm not sure it'll help much but it's worth a good look. NadVolum (talk) 21:43, 4 March 2024 (UTC)

Faulhaber should definitely be looked at here. The problem is that it involves those pesky Bernoulli numbers which aren't integers. If a prime is not a factor of the corresponding denominators then you can say something about whether it divides the sum, but otherwise the situation is more complicated. Let S(k, n) denote the sum. There are two variations on the problem here: First, given k, determine n so that S(k, n) is prime, and second, given n, determine k so that S(k, n) is prime. Faulhaber seems more relevant to the first problem, but the problem as stated in the question is more like the second version. For example, in the n=5 case the problem becomes: For which k is 1^k+2^k+3^k+4^k+5^k prime? Faulhaber doesn't seem to be very useful here since it gives a different formula for each k; you might as well just compute the sum manually. Fermat's little theorem seems more relevant since it implies that, for a fixed n, S(k, n) has period p-1 mod p. So if p|S(k, n) then p|S(j, n) for any j congruent to k mod p-1 (Edit: Assuming p>n). Unfortunately, unless p is very small, this only eliminates a fraction of possible k's (if any), This is the point where I would consult the literature, for which OEIS would be the starting point. But that's where this whole thing began. It would be nice to get some kind of result as in the n=2 case, that k must have a particular form such as 2^a, but that seems unlikely for n>2 unless you can eliminate all k as in the n=3 case. --RDBury (talk) 15:26, 5 March 2024 (UTC)

Note: There is a small error above in the n=6 case: S(12, 6) ≡ 2438235715 is not divisible by 7 or 13. All you can state from divisibility by p=7 and 13 these is that 12|k. But you can eliminate 4|k using p=5. Between p=5, 7 & 13 you can eliminate n=6. --RDBury (talk) 16:18, 5 March 2024 (UTC)

Also note: You can show that if S(k, 10) is prime then k is a multiple of 60; consider also p=7. I didn't see any other easy improvements though. S(60, 10) is composite having a factor of 137. S(120, 10) exceeded the amount of computation time I'm allowed on WolframAlpha; a primality test on a 120 digit number is not infeasible though. --RDBury (talk) 16:55, 5 March 2024 (UTC)

Our article on Faulhaber's formula points out that for odd $k>1$, $S(k,n)$ is a polynomial which has $n^{2}$ and $(n+1)^{2}$ terms. In general, we can write $S(k,n)={\frac {n^{2}(n+1)^{2}P(n)}{N}}$ where $P(n)$ is a polynomial with integer coefficients and $N$ is a constant positive integer. We can see that if $P(x)$ and $Q(x)$ are integer-valued polynomials, then if $P(n)Q(n)/N$ is an integer, then it must be composite if $P(n)>N$ and $Q(n)>N$, since some factor must remain of both $P(n)$ and $Q(n)$ after division by $N$. So if $n^{2}(n+1)^{2}>N$ and $P(n)>N$, then $S(k,n)$ cannot be prime. Since $P(n)>N$ if and only if ${\frac {n^{2}(n+1)^{2}P(n)}{N}}>n^{2}(n+1)^{2}$, we see that $S(k,n)$ cannot be prime for odd $k$ if $S(k,n)>n^{2}(n+1)^{2}>N$. For even $k$ an analogous result exists, except replacing $n^{2}(n+1)^{2}$ with $n(n+1)(2n+1)$. Of course, this leaves the problem of finding out what $N$ actually is. GalacticShoe (talk) 19:19, 5 March 2024 (UTC)

$N$ is given by the sequence OEIS:A064538. I couldn't find an upper bound in the references. It appears to be extraordinarily fast growing. GalacticShoe (talk) 04:25, 6 March 2024 (UTC)

For odd $k\geq 7$, $S(k,n)>n^{2}(n+1)^{2}$ when $n\geq 2$, and for even $k\geq 6$, $S(k,n)>n(n+1)(2n+1)$ when $n\geq 2$. So really, since $S(k,1)=1$ is never prime, the only inequality that matters is $n^{2}(n+1)^{2}>N$ for odd $k\geq 7$ and $n(n+1)(2n+1)>N$ for even $k\geq 6$. GalacticShoe (talk) 05:16, 6 March 2024 (UTC)

For $k=5$, all $n\geq 3$ cannot produce primes, and by direct testing $S(5,2)$ is not. So there are no $S(5,n)$ primes.

For $k=4$, all $n\geq 3$ cannot produce primes, but again by direct testing, $S(4,2)$ is prime (17).

For $k=3$ we can no longer use the limits, since $S(3,n)={\frac {n^{2}(n+1)^{2}}{4}}<n^{2}(n+1)^{2}$, but for obvious reasons there are no $S(3,n)$ primes.

For $k=2$ we again run into a problem with the limits as $S(3,n)={\frac {n(n+1)(2n+1)}{6}}<n(n+1)(2n+1)$, but again it can be seen that $S(2,2)=5$ is the only prime here.

For $k=1$ we see that $S(1,2)=3$ is the only prime.

In sum (pun intended), for $k\leq 5$ the only primes are $S(1,2),S(2,2),S(4,2)$. The set of primes for any fixed value of $k>5$ can be established with a finite but apparently quickly-growing computation. GalacticShoe (talk) 05:30, 6 March 2024 (UTC)

Coming back to this, for $k$ from $1$ through $48$ (the last term that A064538 has listed), the only primes are $S(1,2),S(2,2),S(4,2),S(8,2),S(16,2)$, all Fermat primes as the OP pointed out. GalacticShoe (talk) 07:48, 9 March 2024 (UTC)

March 6

Simple math, or semantics?

I've asked a couple of AIs the following, resulting in either 12 or 13 18: Add together the first five integers, excluding the number three. Perhaps this is more of a semantics problem, but what is the "correct" answer, or is it ambiguous? -- 136.54.106.120 (talk) 00:30, 6 March 2024 (UTC)

Well, there is no "first integer". But I don't see how they get 13. Bubba73 ^{You talkin' to me?} 00:56, 6 March 2024 (UTC)

Oops, my mistake (corrected: 18, strike 13); per AI response: To add together the first five integers, excluding the number three, you would sum 1 + 2 + 4 + 5 + 6, which equals 18. --136.54.106.120 (talk) 01:34, 6 March 2024 (UTC)

Besides there not being any first integer, it is ambiguous because it isn't clear if the third integer should be excluded from the sum or excluded from the count of the first five integers. Bubba73 ^{You talkin' to me?} 01:41, 6 March 2024 (UTC)

So short answer, semantics. The integers are ... -3, -2, -1, 0, 1, 2, 3, ... . So there is no "first" one. To be clear you have to specify "positive integer" meaning 1, 2, 3, ... . Another interpretation would be "natural number" which, depending on who you ask is either the same as "positive integer" or "non-negative integer", meaning 0, 1, 2, 3, ... . In that case the answer would be either 8, 9, 12, or 13, depending on whether you meant to exclude the third in the sequence or the actual number 3, and whether you meant the excluded number was to be counted as one of the five. The upshot is that the question has to be clear and unambiguous to have any hope of getting a meaningful answer. In any case, it's generally not a good idea to ask AI's math questions; they tend to make stuff up and at best only repeat what is said the most often, without questioning if it's actually correct. --RDBury (talk) 02:44, 6 March 2024 (UTC)

The ambiguity of this ill-stated problem reminds me of stupid postings circulating on social media like "Can you solve this: 8 ÷ 4 × 2 = ? Most people get this wrong!", ignoring that there is no universal convention for interpreting a one-line term involving division followed by multiplication.  --Lambiam 09:32, 6 March 2024 (UTC)

March 8

Bouncing ball confinement

When a ball is dropped in vacuum in a uniform gravitational field from a height $h$ onto an immovable surface whose lowest point is at height $0,$ obeying an idealized version of Newton's laws and bouncing back totally elastically, it can theoretically bounce back to its release height, possibly only after many bounces, but will never get higher than $h.$ This is an easy consequence of the preservation of energy, itself a consequence of Newton's laws, combined with the fact that kinetic energy, initially zero, is a nonnegative quantity.

Watching a video illustrating the butterfly effect in a 2D setting, in which 100 balls with imperceptibly different initial positions all very close to $(x_{0},y_{0})$ are simultaneously dropped on the curve given by $y=|x|,$ it seemed to me that the balls basically remained confined to the rectangle with vertices

$(0,0),\left({\frac {x_{0}{+}y_{0}}{2}},{\frac {y_{0}{+}x_{0}}{2}}\right),(x_{0},y_{0}),\left({\frac {x_{0}{-}y_{0}}{2}},{\frac {y_{0}{-}x_{0}}{2}}\right).$

This is a much stronger confinement than $y\leq y_{0}.$ Is there some simple mathematical argument supporting my observation?  --Lambiam 19:58, 8 March 2024 (UTC)

Don't have an answer, but someone else has a video which documents essentially the same phenomenon, this time tracing a path. The video title says "square" but I'm fairly sure that's just an artifact of the ball being dropped close to the apex. GalacticShoe (talk) 07:07, 9 March 2024 (UTC)

Indeed, one of the replies to the video is just the three words, "not a square". When $x_{0}=0$ the trajectory is either undefined or confined to the line $x=0$, but the limit case of the hypothesized rectangle for $x_{0}\to 0$ is a square. (The best way to define the bounce-back when the trajectory hits $(0,0)$ is to have the ball move back in the direction it came from, the limit case of hitting $(\pm \varepsilon ,\varepsilon ).$)  --Lambiam 10:06, 9 March 2024 (UTC)

A few thoughts: The second video shows that the boundary is the envelope of all the possible trajectories. The possible trajectories are parabolas $y=y_{\mathrm {max} }-b(x-x_{0})^{2}$, with the apex at $x_{0}$ (to be shown that this is always so), height given by $y_{\mathrm {max} }=y_{0}-{\tfrac {1}{2}}{\dot {x}}_{0}^{2}$, and $b$ to be expressed in terms of the transverse velocity at the apex ${\dot {x}}_{0}$ as well. This should give the one-parameter family of parabolas, and somebody should be able to determine the envelope. --Wrongfilter (talk) 07:42, 9 March 2024 (UTC)

Thanks. This probably gives enough hints to prove the confinement hypothesis, after which it may not be a trivial exercise to reduce the proof to a much more elementary one based on invariants following from Newton's laws plus elasticity.  --Lambiam 10:06, 9 March 2024 (UTC)

I think the first thing to do is apply the conservation of energy, if the apex of the parabola is at $y_{1}$ then we have:

${\frac {Energy}{m}}=const.=y_{0}g=y_{1}g+{\frac {{\dot {x}}^{2}}{2}}.$

Assume for now that this occurs at $x=x_{0}$; we'll verify this later. Call the value of ${\dot {x}}$ for a specific path $v_{x}$. Parametric equations for the parabola are given by:

$x=x_{0}+v_{x}t,y=y_{1}-{\frac {gt^{2}}{2}}=y_{0}-{\frac {{v_{x}}^{2}}{2g}}-{\frac {gt^{2}}{2}}.$

Eliminate t and write $k={\frac {{v_{x}}^{2}}{g}}$ to get an implicit form for the equation:

$y-y_{0}+{\frac {k}{2}}+{\frac {(x-x_{0})^{2}}{2k}}=0,$

where k is a constant determining the path. The next step is to verify that that if two of these curves intersect on the line $x=y$, then their angles with the line are the same. That will confirm then when the ball bounces off the line the new curve will be in the same family, which is what we assumed at the beginning. Suppose the two curves

$y-y_{0}+{\frac {k}{2}}+{\frac {(x-x_{0})^{2}}{2k}}=0$

$y-y_{0}+{\frac {l}{2}}+{\frac {(x-x_{0})^{2}}{2l}}=0$

intersect at (x, y) with k≠l. Eliminating y:

$kl=(x-x_{0})^{2}.$

The normal vectors to the two curves are given by

$({\frac {x-x_{0}}{k}},1),({\frac {x-x_{0}}{l}},1).$

The reflection of the first vector by the line x=y is

$(1,{\frac {x-x_{0}}{k}})$

and by the previous equation this is a scalar multiple of the second vector, and this proves that the directions are reflections of each other. The corresponding fact for y=-x should be similar. Note that we did not use x=y anywhere, and I think this implies that you could insert short diagonal line segments into the "field of play" and the result would be the same. The balls may bounce around as in a pinball machine but they would still be constrained to be in the family of parabolic paths. Finally, it remains to find the envelope of the family. This is done by setting the partial derivative with respect to k of the implicit equation to 0 and eliminating k. We get:

${\frac {\partial }{\partial k}}(2k(y-y_{0})+k^{2}+(x-x_{0})^{2})=0,$

which gives:

$k=-(y-y_{0}).$

Plug this back into the original equation to produce:

$(x-x_{0})^{2}=(y-y_{0})^{2},$

and this gives the two lines $y-y_{0}=\pm (x-x_{0})$ in question. As far as the original question is concerned I'm not sure if this counts as "simple" or not. There is a lot of messy algebra going on but conceptually it's not that hard. Consider too that this is from someone specializing in math and having little working knowledge of physics. I'm sure someone who has taken a few more physics courses than I have would have more tricks for tackling this sort of thing mare easily, using additional physical invariants etc. --RDBury (talk) 20:09, 9 March 2024 (UTC)

PS. If you don't like the envelope part there is a simpler approach. If

$2k(y-y_{0})+k^{2}+(x-x_{0})^{2}=0$

has a solution in k then the discriminant must be non-negative, in other words

$(y_{0}-y)^{2}\geq (x_{0}-x)^{2}.$

We know that $y_{0}-y\geq 0$ from which we can deduce $y\leq y_{0}-|x_{0}-x|.$ --RDBury (talk) 20:35, 9 March 2024 (UTC)

PPS. (Some additional random thoughts.) If you replace k with z in the above equation, the result is the equation of a surface in space. This turns out to be a cone and slicing the cone by fixing z=k gives you conic sections. These turn out to be the parabolic trajectories. When you project the cone to the x-y plane you get the region $(y_{0}-y)^{2}\geq (x_{0}-x)^{2}.$ as before.

I think this could be generalized as follows. Start with any curve, say x=x(k), y=y(k) parameterized by k. Assume the apex of each parabola is on this curve, then the equation for each parabola can be worked out using the conservation of energy as above. This gives an equation in x, y and k. If this turns out to be quadratic in k then there is a region where each point is crossed by two of the parabolas. You can work out their tangent lines at that point, find the bisectors, and by assuming that a new curve is tangent to one of the bisectors you get a differential equation for the new curves. There are two bisectors so there are two differential equations, and the two sets of solutions would form orthogonal trajectories. In the case where the starting curve is a vertical line the resulting families of curves turn out to be diagonal lines. It's not hard to see that if the starting curve is a horizontal line then you get vertical and horizontal lines. The upshot in that case is that if you drop a ball in to a rectangular well then it will keep bouncing back to original height. I suppose the next easiest case would be when the starting curve is a diagonal line. The more complex the starting curve the more difficult it will be to work out the envelope of the parabolas and solve the differential equations. In the case where g=0 this is related to reflected caustics

Going back to the original problem, instead of saying that the ball bounces off the wall, say that it goes through but now the gravitational force is reflected through the wall. The result would be a gravitational field of the form f=(0, -g) if y>|x|, (-g, 0) if x>|y|, f=(0, g) if y<-|x|, and (g, 0) if x<-|y|. (Sort of like the taxicab metric but with gravity.) This is not a central field so the conservation of angular momentum would not hold, and indeed it's not hard to see that an "orbit" can change direction sometimes. But there is a well defined potential function, namely g⋅max(|x|, |y|).

Finally, are we sure that we're observing an actual butterfly effect here, or is it an artifact of the simulation? Looking at some of the other bouncing ball simulations on the same channel, with most of them the balls become a chaotic mess limited above only by the original height. With g=0 and the container a square, the path will eventually get arbitrarily close to any point in the interior provided that the starting angle is irrational. But that's not chaotic behavior since small changes in the starting position do not result in increasingly larger changes in future positions. Are we sure that something similar wouldn't happen if the simulation was perfectly accurate? --RDBury (talk) 18:33, 11 March 2024 (UTC)

It is hard to determine visually whether the system is chaotic. It may make a difference that we are dealing with a piecewise linear curve. Does it have a positive Lyapunov exponent? (Rather than $|\delta \mathbf {Z} (t)|\approx e^{\lambda t}|\delta \mathbf {Z} _{0}|,$ I think the criterion should be $|\delta \mathbf {Z} (t)|=O(e^{\lambda t}|\delta \mathbf {Z} _{0}|)$ for some $\lambda >0$; "the" Lyapunov exponent should then be the supremum of the range satisfying the condition.) But with a bounded phase space the supremum is necessarily zero, so we need a modified definition, perhaps something related to the notion of entropy.

Not obviously related, but perhaps there is a connection: a version with more balls shows an intriguing phenomenon at the end of 0:41.  --Lambiam 22:52, 11 March 2024 (UTC)

The fact that you can still make out the moose at the end of video makes me suspicious. There is another video comparing walls y=2|x| and y=.5|x|. Those are piecewise linear but the case for those systems being truly chaotic is a lot more believable for me. I suspect that the two lines being at right angles is the critical factor, but I'm not going to write my own simulator just to find out. Note that the rectangular well I mentioned above does not produce chaotic motion. --RDBury (talk) 06:16, 12 March 2024 (UTC)

March 16

Smallest triangular number with prime signature the same as A025487(n)

Smallest triangular number with prime signature the same as (sequence A025487 in the OEIS)(n), or 0 if no such number exists. (there is a similar sequence (sequence A081978 in the OEIS) in OEIS)

For n = 1 through n = 29, this sequence is: (I have not confirmed the n = 15 term is 0, but it seems that it is 0, i.e. it seems that there is no triangular number of the form p^3*q^2 with p, q both primes)

1, 3, 0, 6, 0, 28, 0, 136, 66, 0, 36, 496, 276, 0, 0, 118341, 120, 0, 1631432881, 300, 8128, 210, 0, 528, 0, 29403, 1176, 32896, 630

Is it possible to extend this sequence to n = 100? 1.165.194.85 (talk) 00:51, 16 March 2024 (UTC)

This paper by Cohn implies that there are no prime solutions $(p,q)$ to $p^{3}+1=2q^{2}$ and $2p^{2}+1=q^{3}$. If anyone here can prove that the elliptic curves $y^{2}=x^{3}\pm 4$ have no nontrivial integral points (for $-4$ that would be $(2,\pm 2)$ and $(5,\pm 11)$, while for $4$ that would be $(0,\pm 2)$), then that implies that there are no triangular numbers of the form $p^{3}q^{2}$ at all. GalacticShoe (talk) 06:22, 16 March 2024 (UTC)

I'm not sure what you mean by "at all". If you drop the requirement that q is prime then there are many solutions, for example 242⋅243/2 = 3³33², 12167⋅12168/2 = 23³78². --RDBury (talk) 08:42, 16 March 2024 (UTC)

Sorry, shoulda clarified; no triangular numbers of the form $p^{3}q^{2}$ at all where $p,q$ are prime. Naturally, the two cases mentioned earlier (the latter example of which I hastily posted before remembering that I completely forgot about the prime signature part) are either not covered by Cohn's paper, with the $242$ case being invalid since we spread powers of $3$ among the cube and the square, or they are one of Cohn's special case, as with $12167$. GalacticShoe (talk) 09:51, 16 March 2024 (UTC)

Thanks for the clarification. I don't have access to the Cohn paper, but it seems to me that the p³+1=2q² is trivial to exclude, and similarly for p³-1=2q². That leaves 2p³±1=q². Using brute force I found 2⋅23³+2=156² but none, even when p is only required to be odd and q can be anything, where the difference is one. Note that this generates the 12167 example, but that's not the way I found it. I'm not sure how y²=x³±4 is related, but I'm no expert on elliptic curves so I may have to take your word on that. In any case, given that the p³q² is proving so difficult, it seems unlikely that getting to n=100 is feasible. Finding entries where there is a solution should only require a computer search, but if none are found then proving that there are none to be found can be difficult. It seems ironic that while there is apparently no p³q² solution, there is a p⁵q², though this seems less likely at first glance. RDBury (talk) 18:24, 16 March 2024 (UTC)

The $y^{2}=x^{3}\pm 4$ condition arises because, if we write $x=2p$ and $y=2q$, then an integer solution to $y^{2}=x^{3}\pm 4$ implies $(2q)^{2}=(2p)^{3}\pm 4\Rightarrow 4q^{2}=8p^{3}\pm 4\Rightarrow q^{2}=2p^{3}\pm 1$ is a rational, possibly integer solution to the equalities we are looking at. All integer solutions to $q^{2}=2p^{3}\pm 1$ can be generated this way, so the lack of a nontrivial integer solution to $y^{2}=x^{3}\pm 4$ (or the existence only of solutions where $x,y$ are odd) would rule out any further triangular numbers of the form $p^{2}q^{3}$ with $p,q$ prime. GalacticShoe (talk) 21:33, 16 March 2024 (UTC)

Right, I should have seen that. The 2p³+1=q² case may be easier than I thought; it reduces to 2p³=(q+1)(q-1). You can rule out p=2, q=2, giving q odd, but then by unique factorization q±1 divides p, which is impossible. I thought of applying a similar trick to 2p³-1=q², or 2p³=(q+i)(q-i), and using unique factorization over Gaussian integers. But I quickly got bogged down with this. --RDBury (talk) 02:10, 17 March 2024 (UTC)