@LarrySchweikart My take:
SCOTUS must not be seen to play favourites - yet has an obligation to ensure that corruption is voided.

So wherever widespread and continuous failure to observe due process (as per State law) is seen, void all ballots from that county / counting centre.

1/
@LarrySchweikart If the number of voided counties exceeds more than 1 or 2, then void the entire election for the State.

In the case of House and Senate representatives, require special elections.

2/
@LarrySchweikart For Presidential Electors, legislation now in effect gives the State legislature responsibility and authority to select the State's slate of Electors - as per the original Constitutional design.

3/
@LarrySchweikart If this is done quickly enough, the State Legislature could try to run another election - again all per the legislation now already in effect.

If not, they get to make the choice as they can.

4/
@LarrySchweikart This has the effect of,
as per the original constitutional design,
removing capability for selecting the Electors from the State/County Executives and restoring it to where the Founders intended: the State Legislators

/5
@LarrySchweikart Depending on how Legislatures do their business, this may or may not thrill either party.

But it:
- is consistent with existing legislation;
- avoids partisan decisions being forced on SCOTUS; and
- removes from State/County Executives an authority/capability they shouldn't have
@LarrySchweikart In the case of Senate and House representatives - any State that continues to run elections so corrupt that they are thrown out over and over - then they sit without representation in Congress until they get their act together.

/7
@LarrySchweikart Once again SCOTUS is not forced to make partisan selections, while still as unbiased referee enabled to punish corrupt practices.

/8

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More from @pgeerkens

20 Mar
@GWOMaths Define
y ≡ x - 16
to allow factoring the term 2¹⁶ completely.

Then 2¹⁶ + 2¹⁹ + 2ˣ
= 2¹⁶ + 2¹⁹ + 2ʸ⁺¹⁶
= 2¹⁶ . (1 + 2³ + 2ʸ)
= 2¹⁶ . (9 + 2ʸ)

Now a solution, viz y=4, is readily obvious as yielding
= 2¹⁶ . (9+16)
= 2¹⁶ . 25
= (2⁸ . 5)²

∴ x = y + 16 = 20.
@GWOMaths However - is our solution unique?

Suppose
∃ a ∈ Z
such that
9 + 2ʸ ≡ a²

Then
2ʸ = a² - 9 = (a-3) . (a+3)
and both (a-3) and (a+3) must be powers of 2.

This only occurs for
a = 5 => 2ʸ = 16 => y = 4 => x = 20.

Our solution is unique.
@GWOMaths Finally - as a fun calculating observation,for those who have memorized powers of two at least to 2¹⁶:

(2⁸ . 5)²
= (2⁷ . 2 . 5)²
= (2⁷ . 10)²
= (128 . 10)²
= 1280²

= 2¹⁶ . 25
= 2¹⁴ . 4 . 25
= 2¹⁴ . 100
= 16,384 . 100
= 1,638,400.
Read 5 tweets
7 Mar
@nklym143 @GWOMaths Then I'd note:

- Generating Function f(x) for a single standard die is
x.(1+x+x²+x³+x⁴+x⁵)
= x+x²+x³+x⁴+x⁵+x⁶
= Σ xⁿ . C(n,1)

1/
@nklym143 @GWOMaths - For two Std. die f(x)² is
= 1.x²+1.x³+1.x⁴+ 1.x⁵+ 1.x⁶+ 1.x⁷
+ 1.x³+1.x⁴+1.x⁵+ 1.x⁶+ 1.x⁷+ 1.x⁸
+ 1.x⁴+1.x⁵+1.x⁶+ 1.x⁷+ 1.x⁸+ 1.x⁹
+ 1.x⁵+1.x⁶+1.x⁷+ 1.x⁸+ 1.x⁹+1.x¹⁰
+ 1.x⁶+1.x⁷+1.x⁸+ 1.x⁹+1.x¹⁰+1.x¹¹
+ 1.x⁷+1.x⁸+1.x⁹+1.x¹⁰+1.x¹¹+1.x¹²

2/
@nklym143 @GWOMaths = 1.x²+2.x³+3.x⁴+4.x⁵+5.x⁶+6.x⁷+5.x⁸+4.x⁹+3.x¹⁰+2.x¹¹+1.x¹²
Read 11 tweets
6 Mar
⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ - supers
Α α - Alpha
Β β - Beta
Γ γ - Gamma
Δ δ - Delta
Ε ε - Epsilon
Ζ ζ - Zeta
Θ θ - Theta
Λ λ -Lambda
Μ μ - Mu
Ν ν - Nu
Π π - Pi
Ρ ρ - Rho
Σ σ - Sigma
Τ τ - Tau
Υ υ - Upsilon
Φ φ - Phi
Χ χ - Chi
Ψ ψ - Psi
Ω ω - Omega

≅ - congruence
∴ - therefore
∈ : Element
Δ : Triangle
± : Plus/minus
× ÷ : Times & Division
≤ ≠ ≥ : Inequality
∠ : Angle
° : Degree
⊥ ∥ : Perpendicular & Parallel
~ : Similarity
≡ : Equivalence
∝ : Proportional to
∞ : Infinity
≪ ≫ : Mush less/greater than
∘ : Function composition
† * : Matrix
More superscripts:
⁺ ⁻ ⁽ ⁾
ᵃ ᵇ ᶜ ᵈ ᵉ ᶠ ᵍ ʰ ⁱ ʲ ᵏ ˡ ᵐ ⁿ ᵒ ᵖ ʳ ˢ ᵗ ᵘ ᵛ ʷ ˣ ʸ ᶻ
ᴬ ᴮ ᴰ ᴱ ᴳ ᴴ ᴵ ᴶ ᴷ ᴸ ᴹ ᴺ ᴼ ᴾ ᴿ ᵀ ᵁ ⱽ ᵂ
ᵝ ᵞ ᵟ ᶿ ᵠ ᵡ
Read 4 tweets
22 Feb
@GWOMaths The game Nim is played by two players alternately removing items from several piles of items. Absent physical materials, with pen and paper a game can be setup as rows of vertical lines, players stroking them out to "remove" them.
@GWOMaths Each turn a player removes 1 or more items from precisely one pile, or row.

The winner is the player who does not (or, as a variant, does) remove the last item.

Consider the setup of two piles of 2 items:

xx
xx

The player to move can be forced to lose by his opponent.
@GWOMaths Note that this is true in both variants.

a) Last item loses:
-Remove one item and your opponent takes both items from the other pile;
- Take both items from one pile and your opponent removes just one from the other.

b) Last item wins:
Your opponent wins by mirroring your move.
Read 16 tweets
17 Feb
@GWOMaths @MathsTim Hint and then solution follows
@GWOMaths @MathsTim Use modulo arithmetic to narrow the possible values of x and y.
@GWOMaths @MathsTim From inspection, x>=5.
As difference of even numbers is always even, x ≅ 1 mod 2.

Evaluating mod 3, noting 18 ≅ 0:
0² ≅ 0; 0².0 ≅ 0
1² ≅ 1; 1².0 ≅ 0
2² ≅ 1; 2².0 ≅ 0
∴ x x ≅ 1 or 2
Read 10 tweets
15 Feb
@GWOMaths @MathsTim Recall from Feb. 1 this about generating functions for the odd and distinct partitions of n.


Then the generating function for unrestricted partitions of n, P(n), is given by

SUM(P(n) . q^n) = PROD(1 / (1 - q^i) )
where n ∈ {0, ... } and i ∈ {1, ...}.
@GWOMaths @MathsTim Then P(7) will be the coefficient of q^7 from the expansion of the RHS, noting that
1 / (1 - q^i)

is just
SUM(1 + q^i + q^2.i + q^3.i + q^4.i + ...)

and that we only require in each expansion powers of q ≤7.
@GWOMaths @MathsTim Se we must find q^7 term in

G
= (1+q+q^2+q^3+q^4+q^5+q^6+q^7)
.(1+q^2+q^4+q^6).(1+q^3+q^6)
.(1+q^4).(1+q^5)(1+q^6).(1+q^7)

Evaluating from right
G
= (1+q+q^2+q^3+q^4+q^5+q^6+q^7)
.(1+q^2+q^4+q^6).(1+q^3+q^6)
.(1+q^4).(1+q^5)(1+q^6+q^7)

...
Read 13 tweets

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