User talk:J. Lowell

Numbers that -k is never a square mod

As mentioned in A240920 I used the law of quadratic reciprocity. You can Google it, or guess the pattern by computing the primes modulo k, 2k or 4k. Jens Kruse Andersen 10:36, 11 August 2014 (UTC)

Re: A171604

Definition

A171604 Take the standard 2-D lattice packing of pennies; a(n) = number of ways to pick n pennies (modulo rotations and reflections) such that if we form a linkage with centers of pennies as hinges and with struts between centers of two touching pennies, the linkage is rigid.

Clarifications:

• The pennies are laid flat on a horizontal plane.
• The struts have length (1) equal to the diameter (1) of pennies, since pennies must touch but not overlap.

It seems that to get a rigid packing, the pennies must positioned on an hexagonal lattice.

Examples

Example of a rigid packing (of 38 pennies) with a hole, through which we have a taut chain of pennies. (The chain is rigid, since the struts are infinitely rigid and a straight line has the shortest length.)

...o.o.o.o.o
..o.o.o.o.o.o
.o.o.......o.o
o.o.o.o.o.o.o.o
.o.o.......o.o
..o.o.o.o.o.o
...o.o.o.o.o

— Daniel Forgues 04:32, 11 October 2016 (UTC)

We can form a rigid hexagonal packing (of 34 pennies, which happens to be the 9th Fibonacci number...) if the sides have a thickness of at least 2 pennies.

...o.o.o.o.o
..o.o.o.o.o.o
.o.o.......o.o
o.o.........o.o
.o.o.......o.o
..o.o.o.o.o.o
...o.o.o.o.o

— Daniel Forgues 04:46, 11 October 2016 (UTC)

Is such a loose packing admissible as a packing? - — Daniel Forgues 04:48, 11 October 2016 (UTC)

A rigid triangle of 39 pennies:

........o
.......o.o
......o.o.o
.....o.o.o.o
....o.o...o.o
...o.o.....o.o
..o.o.......o.o
.o.o.o.o.o.o.o.o
o.o.o.o.o.o.o.o.o

We cannot place a penny at x, because the distance from x to either the penny above or the penny below is 

2√  3 2

, which is less than the length (1) of the struts:

........o
.......o.o
......o.o.o
.....o.o.o.o
....o.o.o.o.o
...o.o..x..o.o
..o.o...o...o.o
.o.o.o.o.o.o.o.o
o.o.o.o.o.o.o.o.o

— Daniel Forgues 05:19, 11 October 2016 (UTC)

I just realized: since the struts all have the same length (1), we cannot have those vertical taut chains on an hexagonal lattice since the pennies centers would not be distant by 1 unit. I think that to get a rigid packing, all the pennies must necessarily be on some locations belonging to an hexagonal lattice on the plane. This one with 42 pennies we could have:

.........o
........o.o
.......o.o.o
......o.o.o.o
.....o.o...o.o
....o.o.....o.o
...o.o.o.o.o.o.o
..o.o.........o.o
.o.o.o.o.o.o.o.o.o
o.o.o.o.o.o.o.o.o.o

We could also have diagonal chains (maybe a whole spider web of chains and subchains and subsubchains and... along the any three directions of the hexagonal lattice). E.g., with 47 pennies:

...........o
..........o.o
.........o.o.o
........o.o.o.o
.......o.o...o.o
......o.o...o.o.o
.....o.o...o...o.o
....o.o...o.....o.o
...o.o...o.o.o.o.o.o
..o.o...o.........o.o
.o.o.o.o.o.o.o.o.o.o.o
o.o.o.o.o.o.o.o.o.o.o.o

— Daniel Forgues 05:49, 11 October 2016 (UTC)

A rigid triangular spider web of pennies (all the chains are taut):

......................o
.....................o.o
....................o.o.o
...................o.o.o.o
..................o.o...o.o
.................o.o.....o.o
................o.o.o.....o.o
...............o.o...o.....o.o
..............o.o.....o.....o.o
.............o.o.....o.o.....o.o
............o.o.....o...o.....o.o
...........o.o.....o.....o.....o.o
..........o.o.....o.o...o.o.....o.o
.........o.o.....o...o.o...o.....o.o
........o.o.....o.....o.....o.....o.o
.......o.o.....o.....o.....o.o.....o.o
......o.o.....o.o.o.o.o.o.o...o.....o.o
.....o.o.....o...........o.....o.....o.o
....o.o.....o.o.o.o.o.o.o.o.o.o.o.....o.o
...o.o.....o.....................o.....o.o
..o.o.....o.......................o.....o.o
.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o
o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o

— Daniel Forgues 06:18, 11 October 2016 (UTC)

A rigid spider web of pennies (all the chains are taut):

..................o.o.o.o.o.o.o.o.o.o
.................o.o.o.o.o.o.o.o.o.o.o
................o.o.o.............o.o.o
...............o.o...o...........o...o.o
..............o.o...o.o.........o.....o.o
.............o.o...o...o.o.o.o.o.......o.o
............o.o...o.....o.....o.o.......o.o
...........o.o...o.....o.o...o...o.......o.o
..........o.o...o.....o...o.o.....o.......o.o
.........o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o.o
..........o.o.....o.......o.o.......o.....o.o
...........o.o.....o.....o...o.....o.....o.o
............o.o.....o...o.o.o.o...o.....o.o
.............o.o.....o.o.......o.o.....o.o
..............o.o.....o.........o.....o.o
...............o.o...o.o.o.o.o.o.o...o.o
................o.o.o.............o.o.o
.................o.o.o.o.o.o.o.o.o.o.o
..................o.o.o.o.o.o.o.o.o.o

This is getting awfully complicated! How can one hope to find an algorithm giving all the rigid packings of 

n

pennies?! — Daniel Forgues 06:41, 11 October 2016 (UTC)

Ordering

Examples for n = 2, 3, 4, 5, 6, 7: ordered by increasing number of rows, pennies on lowest rows first.

n = 2: (2)

.o.o

n = 3: (2, 1)

..o
.o.o

n = 4: (2, 2)

..o.o
.o.o

n = 5: (3, 2)

..o.o
.o.o.o

n = 6: (3, 3), (3, 2, 1), (2, 3, 1)

..o.o.o
.o.o.o

...o
..o.o
.o.o.o

...o
..o.o.o
...o.o

n = 7: (4, 3), (3, 3, 1), (3, 2, 2), (2, 3, 2), (2, 3, 2)

..o.o.o
.o.o.o.o

...o
..o.o.o
.o.o.o

...o.o
..o.o
.o.o.o

(invalid because it is a reflection of the immediately above packaging) (I see: reflection about the 30° upwards diagonal. - D.F.)

...o.o
..o.o.o
...o.o

...o.o
..o.o.o
.o.o 

Is there a fifth one for n = 7? — Daniel Forgues 04:08, 12 October 2016 (UTC)

n = 8: (4, 4), (4, 3, 1), (3, 3, 2), (3, 3, 2), ...?

..o.o.o.o
.o.o.o.o

...o
..o.o.o
.o.o.o.o

...o.o
..o.o.o
.o.o.o

...o.o
..o.o.o
...o.o.o

(invalid because it is a reflection of the immediately above one) (I see: horizontal reflection. - D.F.)

— Daniel Forgues 04:25, 12 October 2016 (UTC)

Did you find more patterns for n = 8? (There should be more than for n = 7, since I think that the sequence should be increasing...) — Daniel Forgues 05:17, 14 October 2016 (UTC)

.o.o
..o.o.o
.o.o.o

— Daniel Forgues 05:17, 14 October 2016 (UTC)

Chains are infinitely taut

Since all the struts (of length equal to pennies diameters) between the pennies centers are infinitely rigid, all the chains are infinitely stiff/taut since the chains are shortest paths: moving a penny sideways by an amount epsilon would have to increase the length of the chain by a tiny amount, which you cannot do since you cannot stretch the struts by even an epsilon amount! — Daniel Forgues 03:51, 22 October 2016 (UTC)

4366

Using the prime 734550668324980890902704447022133535769 you can extend your sequence starting from 4366 from a C85 to a C332:

95204106181768419529010752287946033126591729749489177872915361782936572326748820081677688881479443392043561574325305173168828244377762100893639072247908086626163887914781715187902615087140713961250872014078602223277620224147796991275538568406831736957888778388017386746799201810685266791096241995365865748643751986914049638315718411

The next prime is almost surely 28664812261310161628639 but I don't have a good way to prove it (though it's not hard to show that it holds with high probability, 99.9% or better, with ECM).

Charles R Greathouse IV (talk) 17:32, 20 February 2018 (EST)

To get an automatic scroll bar, you may use

<pre style="overflow: auto;">
95204106181768419529010752287946033126591729749489177872915361782936572326748820081677688881479443392043561574325305173168828244377762100893639072247908086626163887914781715187902615087140713961250872014078602223277620224147796991275538568406831736957888778388017386746799201810685266791096241995365865748643751986914049638315718411
</pre>

which yields

95204106181768419529010752287946033126591729749489177872915361782936572326748820081677688881479443392043561574325305173168828244377762100893639072247908086626163887914781715187902615087140713961250872014078602223277620224147796991275538568406831736957888778388017386746799201810685266791096241995365865748643751986914049638315718411

— Daniel Forgues 20:14, 20 February 2018 (EST)