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muzik wrote:Is this identical to a sphere?

Yes.

Macbi

Posts: 659
Joined: March 29th, 2009, 4:58 am

...and is this identical to anything? Is there anywhere I can find a list of such 2-manifolds?

`x = 119, y = 113, rule = B3/S2349bo\$50b2o\$52bo\$53b2o\$55bo\$56b2o\$58b2o\$59b2o\$60b2o\$8b97o\$8bo53bo41bo\$8bo52bo42bo\$8bo51bo43bo\$8bo49b2o44bo\$8bo48bo46bo\$8bo47bo47bo\$8bo45b2o48bo\$8bo43b2o50bo\$8bo42bo13b3o36bo\$8bo41bo14bobo36bo\$8bo40bo15b3o36bo\$8bo40bo15bobo36bo\$8bo56bobo36bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo11b2o82bo13bo\$8b3o9bobo81bo12bo\$8b3o9b2o82bo12bo\$7b2o2bo8bobo69bo11bo11bo\$6bobo2bo8b2o71bo10bo11bo\$5bo2bo3bo80bo10bo10bo\$4bo3bo3bo81b2o8bo9bo\$3bo4bo4bo82bo7bo8bo\$2bo5bo5b2o80bo7bo7bo\$bo6bo7bo70b3o7bo6bo6bo\$o7bo78bobo8bo5bo5bo\$8bo78b3o9bo4bo4bo\$8b4o75bobo10bo3bo4bo\$7b2o2bo75bobo11bo2bo3bo\$6bobo3bo89b3o2bo\$5bo2bo4b2o88b2obo\$4bo3bo6bo88b2o\$4bo3bo95bo\$3bo4bo95bo\$3bo4bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo45b2o48bo\$8bo45bobo47bo\$8bo45b2o48bo\$8bo45bobo47bo\$8bo45b2o48bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo95bo\$8bo47bo47bo\$8bo45b2o48bo\$8bo45bo49bo\$8bo44bo9bo40bo\$8bo43bo7b4o40bo\$8bo41b2o7b2o43bo\$8bo40bo8bo45bo\$8bo40bo6b2o46bo\$8bo38b2o7bo47bo\$8b97o\$48bo6b2o\$49bo6bo\$50b2o5bo\$52bo5b2o\$53bo6bo\$54b2o\$56b2o!`
Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!
muzik

Posts: 3301
Joined: January 28th, 2016, 2:47 pm
Location: Scotland

That's a Klein bottle. You can see by cutting it and rearranging it like this:
`x = 325, y = 120, rule = B3/S2349bo219bo\$50b2o218b2o\$52bo219bo\$53b2o218b2o\$55bo219bo\$56b2o218b2o\$58b2o218b2o\$59b2o218b2o\$60b2o218b2o\$8b97o123b97o\$8bo53bo40b2o123bo53bo40b2o\$8bo52bo40b3o122b2o52bo40bo\$8bo51bo40b2obo121b3o51bo40b2o\$8bo49b2o41bo2bo121bobo49b2o41bo\$8bo48bo42bo3bo120b2obo48bo42bo\$8bo47bo43bo3bo119b2o2bo47bo43bo\$8bo45b2o43bo4bo119bo3bo45b2o43bo\$8bo43b2o44bo5bo118b2o3bo43b2o44bo\$8bo42bo13b3o28b2o6bo118bo4bo42bo13b3o28b2o\$8bo41bo14bobo27b2o7bo116b2o5bo41bo14bobo27b2o\$8bo40bo15b3o27bo8bo115b2o6bo40bo15b3o27bo\$8bo40bo15bobo26bo9bo115bo7bo40bo15bobo26bo\$8bo56bobo25bo10bo115bo7bo56bobo25bo\$8bo82b2o11bo114bo8bo82b2o\$8bo80b2o13bo114bo8bo80b2o\$8bo80bo14bo113bo9bo80bo\$8bo79bo15bo112bo10bo79bo\$8bo78b2o15bo111bo11bo78b2o\$8bo77bo17bo110bo12bo77bo\$8bo76bo18bo108b2o13bo76bo\$8bo75b2o18bo107b2o14bo75b2o\$8bo74bo20bo107bo15bo74bo\$8bo73bo21bo106bo16bo73bo\$8bo73bo21bo104b2o17bo73bo\$8bo72bo22bo103b2o18bo72bo\$8bo72bo22bo101b2o20bo72bo\$8bo70b2o23bo99b2o22bo70b2o\$8bo69bo25bo98bo24bo69bo\$8bo68bo26bo96b2o25bo68bo\$8bo67bo27bo95b2o26bo67bo\$8bo65b2o28bo93b2o28bo65b2o\$8bo63b2o30bo92b2o29bo63b2o\$8bo11b2o30b3o16b2o31bo13bo78bo30bo11b2o30b3o16b2o\$8b3o9bobo29bo18bo32bo12bo78bo31b3o9bobo29bo18bo\$8b3o9b2o29bo17b2o33bo12bo78bo31b3o9b2o29bo17b2o\$7b2o2bo8bobo27bo17bo23bo11bo11bo78bo31b2o2bo8bobo27bo17bo\$6bobo2bo8b2o28bo16b2o24bo10bo11bo77bo31bobo2bo8b2o28bo16b2o\$5bo2bo3bo37bo15b2o25bo10bo10bo77bo31bo2bo3bo37bo15b2o\$4bo3bo3bo37b5o10b2o27b2o8bo9bo76b2o31bo2b2o3bo37b5o10b2o\$3bo4bo4bo50b2o30bo7bo8bo72bo3bo32bo3b3o3bo50b2o\$2bo5bo5b2o48bo31bo7bo7bo73bo2bo32bo2b2obobo3b2o48bo\$bo6bo7bo46bo23b3o7bo6bo6bo74bobo32bo2bo3bo2bo4bo46bo\$o7bo45b11o22bobo8bo5bo5bo69bo5b2o32bo3bo3bo2bo42b11o\$8bo51bo3bo22b3o9bo4bo4bo70bo5b2o35bo4bo3bo47bo3bo\$8b4o47bo4bo22bobo10bo3bo4bo70bo5b2o34bo5b4obo45bo4bo\$7b2o2bo47bo4bo22bobo11bo2bo3bo66bo4bo3b2o2b5o27b2o5b2o2bo2bo44bo4bo\$6bobo3bo43bobo5bo37b3o2bo67bo4bob2o36bo5bob2o2bobo41bobo5bo\$5bo2bo4b2o36b7o6bo38b2obo68bo3b3o37bo5b6o2b3o35b7o6bo\$4bo3bo6bo40b3o5b2o38b2o69bo4b2o3b6o33bobobo2bo3bo40b3o5b2o\$4bo3bo46bo2bo6bo38bo70bo3bob4o39b2o2bo3bo42bo2bo6bo\$3bo4bo46bo2bo6bo38bo70bob2o44b2o3bo3bo42bo2bo6bo\$3bo4bo45bo3bo5bo39bo70b2o46b2o3bo4bo40bo3bo5bo\$8bo45bo2bo46bo70bo47bo4bo45bo2bo\$8bo39b6o3bo46bo70bo47bo4bo39b6o3bo\$8bo42b3o3bo46bo69b2o46b2o4bo42b3o3bo\$8bo41bo3bo2bo46bo67b2ob7o2bo5b2o36bo41bo3bo2bo\$8bo40bo4bo2bo46bo66b2o8b3o5bo38bo40bo4bo2bo\$8bo39bo5bo2bo46bo65b2o17bo38bo39bo5bo2bo\$8bo38bo6bo49bo64b2o18bo38bo38bo6bo\$8bo38bo6bo49bo63b2o20b2o36bo38bo6bo\$8bo37bo57bo62bo60bo37bo\$8bo36bo58bo62bo60bo36bo\$8bo35bo59bo60b2o61bo35bo\$8bo33b2o60bo59b2o62bo33b2o\$8bo31b2o62bo59bo63bo31b2o\$8bo30b2o63bo58bo64bo30b2o\$8bo30bo64bo58bo64bo30bo\$8bo29bo65bo57bo65bo29bo\$8bo28b2o65bo56bo66bo28b2o\$8bo28bo66bo55bo67bo28bo\$8bo27bo67bo54bo68bo27bo\$8bo26bo68bo54bo68bo26bo\$8bo26bo68bo52b2o69bo26bo\$8bo25bo69bo50b2o71bo25bo\$8bo25bo69bo49bo73bo25bo\$8bo24bo70bo48bo74bo24bo\$8bo23b2o20b2o48bo47b2o74bo23b2o\$8bo23bo21bobo47bo46bo76bo23bo\$8bo22bo22b2o48bo45bo23b3o51bo22bo\$8bo20b2o23bobo47bo44b2o23bobo51bo20b2o\$8bo19b2o24b2o48bo42b2o25b3o51bo19b2o\$8bo19bo75bo42bo26bobo51bo19bo\$8bo18bo76bo41bo27bobo51bo18bo\$8bo17bo77bo41bo21bo59bo17bo\$8bo16bo78bo40bo23b2o57bo16bo\$8bo15bo79bo39bo26bo56bo15bo\$8bo13b2o32bo47bo37b2o27bo56bo13b2o\$8bo10b3o32b2o48bo36b2o29b2o54bo10b3o\$8bo9b2o34bo49bo36bo32bo53bo9b2o\$8bo9bo34bo9bo40bo35bo34bo52bo9bo\$8bo7b2o34bo7b4o40bo34bo36bo51bo7b2o\$8bo5b3o33b2o7b2o43bo32b2o38bo50bo5b3o\$8bo4b2o34bo8bo45bo30b2o41bo49bo4b2o\$8bo2b3o35bo6b2o46bo29b2o43bo48bo2b3o\$8bob2o35b2o7bo47bo29bo44b2o47bob2o\$8b97o27b100o\$48bo6b2o124bo\$49bo6bo123bo\$50b2o5bo121bo\$52bo5b2o118bo\$53bo6bo115b2o\$54b2o119bo\$56b2o116bo\$173bo\$172bo\$171bo\$170bo\$168b2o\$166b2o\$165bo!`
For a classification of surfaces, see here: https://en.wikipedia.org/wiki/Surface_(topology)#Classification_of_closed_surfaces

Macbi

Posts: 659
Joined: March 29th, 2009, 4:58 am

Probably getting a bit ridiculous here. This would result in something like a cone, which would end up being a hemisphere, which can be flattened out into a circle, which is a plane, right?

`x = 111, y = 106, rule = B3/S2341bo\$42b2o\$44bo\$45b2o\$47bo\$48b2o\$50b2o\$51b2o\$52b2o\$97o\$o53bo41bo\$o52bo42bo\$o51bo43bo\$o49b2o44bo\$o48bo46bo\$o47bo47bo\$o45b2o48bo\$o43b2o50bo\$o42bo52bo\$o41bo53bo\$o40bo54bo\$o40bo54bo\$o95bo\$o48b3o44bo\$o48bobo44bo\$o48b3o44bo\$o48bobo44bo\$o48bobo44bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95b2o\$o95b3o\$o94b2obo\$o93b3o2bo\$o93bobo3bo\$o92b2obo3bo\$o92bo2bo4bo\$o91bo3bo4b2o\$o91bo3bo5bo\$o90bo4bo5b2o\$o90bo4bo6bo\$o89bo5bo7bo\$o79b3o7bo5bo7bo\$o79bobo6bo6bo7b2o\$o79b3o6bo6bo8bo\$o79bobo5bo7bo8bo\$o79bobo4bo8bo9bo\$o86bo8bo10bo\$o86bo8bo10bo\$o85bo9bo10b2o\$o85bo9bo11bo\$o84bo10bo11bo\$o95bo11b2o\$o95bo12b2o\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$o95bo\$97o!`
Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!
muzik

Posts: 3301
Joined: January 28th, 2016, 2:47 pm
Location: Scotland

Yeah.

Macbi

Posts: 659
Joined: March 29th, 2009, 4:58 am

I noticed this thread and I hope that at least through affinity, somebody here might be able to help me. Is there a name for the kind of oriented rhombus tiling shown in the picture below?
49726547_2255109781479246_6600225566891180032_o (2).jpg (30.83 KiB) Viewed 3860 times

To start with, it's just a rhombus with 60° acute corners, composed of two equilateral triangles. This was known to the ancient Romans among others who used it in mosaics that give the illusion of 3D cubes, and it was used in the same way much later in the video game Q*bert. Irregular rhombus tilings are analogous to domino packings on a rectangular grid and, like them, reducible to bipartite matching (each covers one upsidedown and one rightsideup triangle). So this is not my question. (I am also not asking about Penrose tilings; these rhombuses are the 60° kind.)

But what interested me was to replace the flat sides with alternating convex and concave arcs that match up. In this case, they are circular arcs with a radius of curvature equal to the triangle side, but they could be any shape as long as they match up, such as jigsaw nubs and voids.

Adding curvature, we get a "z" orientation, shown here in a dark color, and an "s" orientation, shown here in a light color. You can make one of these out of two-sided material, so that you flip it over to change the color. In fact, I made some with paper and cardboard sides:
49899078_2256694581320766_2710552348677111808_o.jpg (25.07 KiB) Viewed 3860 times

These kinds of tilings are pretty easy to analyze, but it is fun, at least for small tilings to see how the orientations impose a coloring. Colors need to invert back and forth to go in a straight line but stay the same color around a turn.

(In fact, this reminds me of something I found very confusing (years back) about Madachy's method for constructing flexagons, which have a similar alternating rule when making the paper strips of triangles. Is there a name for this property of flipping sign/color going in a straight line?)

The shapes themselves lend themselves to decorative tiling. Surely I cannot be the first to have played around with this kind of rhombus tiling. Any leads?
pcallahan

Posts: 298
Joined: April 26th, 2013, 1:04 pm

Which kind of curves(not functions) has the following property:
All tangent lines intersects with the curves only once?

An example is circle.

Is it possible to have one or multiple(but finite number of) formulas that can describe all of them?
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Hunting

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Hunting wrote:Which kind of curves(not functions) has the following property:
All tangent lines intersects with the curves only once?

An example is circle.

Is it possible to have one or multiple(but finite number of) formulas that can describe all of them?

My gut feeling is that those curves are precisely the boundaries of convex sets.

My gut feeling is also that you can't describe them all using a finite set of formulae (even allowing for parametrization[1]), for you can take any concave, monotonically increasing function on an interval [a, b], and mirror it four-fold (think D4 symmetry) to yield such a curve (and set). Since there's more than 2^{\aleph_0} concave, monotonically increasing functions (proof left as an exercise for the reader), these cannot be described by a finite number of formulae, and as such neither can the curves you're interested in.

But I'm no mathematician, and no doubt what I wrote above is all wrong, for reasons that will be pointed out by the more competent denizens of this forum at most two minutes after I post this.
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Apple Bottom

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Joined: July 27th, 2015, 2:06 pm

I also thought that the answer would be the boundaries of convex sets, but I think the question is impossible to answer without a more precise definition of "tangent line". If the definition was "line with the same derivative at a point where the curve is differentiable" then something like the Weierstrass function would work. It's not differentiable anywhere, so it has no tangents, so all of its tangents intersect it at one point!

My gut feeling is also that you can't describe them all using a finite set of formulae (even allowing for parametrization[1]), for you can take any concave, monotonically increasing function on an interval [a, b], and mirror it four-fold (think D4 symmetry) to yield such a curve (and set). Since there's more than 2^{\aleph_0} concave, monotonically increasing functions (proof left as an exercise for the reader), these cannot be described by a finite number of formulae, and as such neither can the curves you're interested in.
This part is definitely correct. Even if we limit ourselves to D8 symmetric infinitely differentiable convex curves there will still be 2^ℕ of them, which is Too Many.

Macbi

Posts: 659
Joined: March 29th, 2009, 4:58 am

Apple Bottom wrote:
Hunting wrote:Which kind of curves(not functions) has the following property:
All tangent lines intersects with the curves only once?

An example is circle.

Is it possible to have one or multiple(but finite number of) formulas that can describe all of them?

My gut feeling is that those curves are precisely the boundaries of convex sets.

My gut feeling is also that you can't describe them all using a finite set of formulae (even allowing for parametrization[1]), for you can take any concave, monotonically increasing function on an interval [a, b], and mirror it four-fold (think D4 symmetry) to yield such a curve (and set). Since there's more than 2^{\aleph_0} concave, monotonically increasing functions (proof left as an exercise for the reader), these cannot be described by a finite number of formulae, and as such neither can the curves you're interested in.

Thanks!
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Favorite gun ever:
`#C Favorite Gun. Found by me.x = 4, y = 6, rule = B2e3i4at/S1c23cijn4ao2bo\$4o3\$4o\$o2bo!`
Hunting

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Location: Ponyville, Equestria

Hunting wrote:Which kind of curves(not functions) has the following property:
All tangent lines intersects with the curves only once?

An example is circle.

Is it possible to have one or multiple(but finite number of) formulas that can describe all of them?

This set contains the set of closed convex curves as a subset.
If we require the curves to be closed, this is the entire set, since the tangent line will sweep out the entire plane during a circumnavigation of the curve.
Otherwise, we can also use sets of open segments of convex curves, arranged such that no piece touches any sector swept out by the tangent of any other (for example, two halves of a circle separated by one unit perpendicular to the slice).

P.S. I'm fairly sure the set of convex closed curves has cardinality on the order of \aleph_1^{\aleph_1}, since unrolling the curve gives a segment which can be bijected onto the positive reals and each point on the curve can have any(?) positive real curvature.
P.P.S. This doesn't count curves with cusps, but I don't think adding the possibility of jump discontinuities in the first derivative changes the final answer.
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Hunting wrote:Which kind of curves(not functions) has the following property:
All tangent lines intersects with the curves only once?

An example is circle.

Is it possible to have one or multiple(but finite number of) formulas that can describe all of them?

This set contains the set of closed convex curves as a subset.
If we require the curves to be closed, this is the entire set, since the tangent line will sweep out the entire plane during a circumnavigation of the curve.
Otherwise, we can also use sets of open segments of convex curves, arranged such that no piece touches any sector swept out by the tangent of any other (for example, two halves of a circle separated by one unit perpendicular to the slice).

P.S. I'm fairly sure the set of convex closed curves has cardinality on the order of \aleph_1^{\aleph_1}, since unrolling the curve gives a segment which can be bijected onto the positive reals and each point on the curve can have any(?) positive real curvature.
P.P.S. This doesn't count curves with cusps, but I don't think adding the possibility of jump discontinuities in the first derivative changes the final answer.

Alright. Another example(I forgot to mention) is any exponential curve which is not closed.

Is there a proof for the "closed convex curves as a subset"?
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Working on:

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Favorite gun ever:
`#C Favorite Gun. Found by me.x = 4, y = 6, rule = B2e3i4at/S1c23cijn4ao2bo\$4o3\$4o\$o2bo!`
Hunting

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BlinkerSpawn wrote:P.S. I'm fairly sure the set of convex closed curves has cardinality on the order of \aleph_1^{\aleph_1}, since unrolling the curve gives a segment which can be bijected onto the positive reals and each point on the curve can have any(?) positive real curvature.
P.P.S. This doesn't count curves with cusps, but I don't think adding the possibility of jump discontinuities in the first derivative changes the final answer.

It's not quite that large. Since it's continuous, you can specify a curve (a map ℝ ⟶ ℝ^2, or from some interval in ℝ) by specifying where it sends the rationals. So it's only size \aleph_1^{\aleph_0}, which is the same as 2^{\aleph_0}.

Macbi

Posts: 659
Joined: March 29th, 2009, 4:58 am

Hunting wrote:
BlinkerSpawn wrote:This set contains the set of closed convex curves as a subset.
If we require the curves to be closed, this is the entire set, since the tangent line will sweep out the entire plane during a circumnavigation of the curve.
Otherwise, we can also use sets of open segments of convex curves, arranged such that no piece touches any sector swept out by the tangent of any other (for example, two halves of a circle separated by one unit perpendicular to the slice).
[...]

Alright. Another example(I forgot to mention) is any exponential curve which is not closed.

Is there a proof for the "closed convex curves as a subset"?

Follows from the definition of "convex".
And the exponential is another interesting case; an open convex curve of any (finite or infinite) length works.
It should be straightforward to show that infinite open curves can only exist in isolation.
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X-post from random posts:
Moosey wrote:1
11
21
1211
111221
312211
13112221
What rate does the sequence grow at?
It seems erratic but around exponential.
(The code is n X’s in a row is nX. Thus 333 becomes 33. 22 becomes 22.)

If you haven’t heard of it before, I definitely did not make up the sequence.

EDIT:
http://www.conwaylife.com/forums/viewtopic.php?f=12&t=2703&start=1925#p71827
My rules:
They can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Moosey

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Can E6, E7 and E8 polytopes be stellated, and, if so, have there been any attempts so far to enumerate these?
Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!
muzik

Posts: 3301
Joined: January 28th, 2016, 2:47 pm
Location: Scotland

Can every number besides powers of 2 be expressed as a sum of at least two consecutive integers? I know all odds besides 1 can be, but here's a Lua program that checks the evens
she/they // Please stop using my full name. Refer to me as dani.

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danny

Posts: 944
Joined: October 27th, 2017, 3:43 pm
Location: New Jersey, USA

danny wrote:Can every number besides powers of 2 be expressed as a sum of at least two consecutive integers? I know all odds besides 1 can be, but here's a Lua program that checks the evens

0 is an integer, so one is also an odd that can be expressed as the sum of two consecutive integers.
My rules:
They can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Moosey

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danny wrote:Can every number besides powers of 2 be expressed as a sum of at least two consecutive integers? I know all odds besides 1 can be, but here's a Lua program that checks the evens

If negative numbers are allowed, any positive integer n can be written as (1-n)+(2-n)+...0+1+2+...(n-1)+n.
`x = 11, y = 5, rule = B2ck3-ij5n78/S01e2ei3-k5ai8b2o\$2o3b2o\$bo4bo3bo\$2bo2b2o\$o7bo!`

(Check Gen 2)

toroidalet

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danny wrote:Can every number besides powers of 2 be expressed as a sum of at least two consecutive integers? I know all odds besides 1 can be, but here's a Lua program that checks the evens
Yep.

Suppose k = n+1 + n+2 + ... + m. Then since n+1 + n+2 + ... + m = m(m+1)/2 - n(n+1)/2 = (m^2 - n^2 +m - n)/2 = (m - n)(m + n + 1)/2 we have 2k = (m - n)(m + n + 1). So if we pick a factorisation PQ of 2k then we can find m and n by solving the simultaneous equations m - n = P and m + n + 1 = Q. These have integer solutions so long as P and Q have different parities. So we need to write 2k as a product of an odd number and an even number. We can do this so long as k has an odd factor, so we'll succeed so long as k isn't a power of 2. (Of course any k divides by 1 and -1, but if you substitute these in you find the solution where k is the sum of one consecutive number, namely itself, and the solution where k = -(k-1) + ... + (k-1) + k.)

Macbi

Posts: 659
Joined: March 29th, 2009, 4:58 am

How many numbers can be expressed by adding/subtracting/multiplying (but no dividing) consecutive positive integers starting with one and ending at most at 10? (They must be in order.)
Is there a function, f(n), for quickly finding it starting with one and ending with most n?

Just as an example of a way a number can be expressed this way:
99 = (1+2)(3)(4+5-6+7-8+9)

EDIT:
I’ll make a list (going up to ten, with the up to ten rule.)
Format:
n [example, not necessarily minimal way to make n]

1 (1)
2 (1)(2)
3 (1+2)
4 (1+2-3+4)
5 (1+2+3+4-5)
6 (1+2+3)
7 (1+2+3-4+5)
8 (1+2-3+4)(-5+6-7+8)
9 (1+2)(3)
10 (1+2+3+4)

What are the positive integers who can be expressed this way so that the largest number in the minimalist (i.e. legal expression whose largest number is as small as possible. (1)(2) is a more minimal expression for 2 than (1+2+3-4)) expression of the integer <= the integer in question?

Again, because that was confusing:
What are the positive integers who can be expressed this way so that the largest number in the minimalist expression of the integer <= the integer in question?

It is easy to show that all triangular numbers would be in the sequence.
It seems probable that EVERY positive integer is in the sequence.

Can anyone find a positive integer not in the sequence? Can anyone prove one/none exist? Can anyone prove no proofs of either can exist, or that one can be proven (a meta proof)? Can anyone prove that there is/is not a meta proof in one direction? (A metametaproof)
My rules:
They can be found here

Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

Moosey

Posts: 1412
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board.

(The following proof builds on Macbi's proof that every number (except powers of 2) can be represented as a sum of 2+ consecutive integers)
Assume k=n+n+1+n+2+...+m. If n is odd, then (-1+2)(-3+4)...(-(n-2)+(n-1))(n+n+1+...+m) works; if n is even then (1)(-2+3)(-4+5)...(-(n-2)+(n-1))(n+...m) works. Thus any k can be represented in this way with the largest term ≤k, and strictly less if k is not a power of 2 (for a power of 2 n=m=k).

If k 2^n, then the following sequence works (assuming n≥2 (the first term is a factor of 4), but the other cases are easily proven empirically):
(1+2-3+4)(-5+6-7+8)(-9+10-11+12)...(-(4n-7)+4n-6-(4n-5)+4n+4)
EDIT: The second proof is unnecessary; we can already represent powers of 2 with the general case.
`x = 11, y = 5, rule = B2ck3-ij5n78/S01e2ei3-k5ai8b2o\$2o3b2o\$bo4bo3bo\$2bo2b2o\$o7bo!`

(Check Gen 2)

toroidalet

Posts: 946
Joined: August 7th, 2016, 1:48 pm
Location: my computer

Well clearly there must be SOME end, since there are a finite number of operations.
I'm writing a python program and the results are coming out soon.

EDIT: attached are the results. Smallest un-doable is 59
Attachments
moosey-problem.txt

testitemqlstudop

Posts: 696
Joined: July 21st, 2016, 11:45 am
Location: very very very very boats

1-2+(3)(4)(5)
Sarp

Posts: 189
Joined: March 1st, 2015, 1:28 pm

10,958 is the smallest number that can't be written this way:
https://arxiv.org/pdf/1302.1479.pdf
Life is hard. Deal with it.
My favorite oscillator of all time:
`x = 7, y = 5, rule = B3/S2-i3-y4i4b3o\$6bo\$o3b3o\$2o\$bo!`

Hdjensofjfnen

Posts: 1079
Joined: March 15th, 2016, 6:41 pm
Location: r cis θ

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