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User talk:J. Lowell
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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 2D 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.oWe cannot place a penny at x, because the distance from x to either the penny above or the penny below is

........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.oThis is getting awfully complicated! How can one hope to find an algorithm giving all the rigid packings of
n 
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
...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
 — 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)