Exact(6)
Case 2. Suppose that k = n − m is even, where k ≥ 1.
when m is even, where r = u 1 + 2 u 2 + ⋯ + ( m − 1 ) u m − 1. Proof Let m be odd.
Most of the progress for the existence of {4}-GDDs of type gum1 is on the case when gu is even, where the existence for small g has played a key role.
Then a i = 0 if i is odd; and a i = ∑ H ( − 1 ) c + ( H ) 2 c ( H ) if i is even, where the summation is over all basic oriented graphs ℋ, of G σ, having i vertices, and c + ( H ) and c ( H ) are respectively the number of evenly oriented even cycles and even cycles contained in ℋ.
Obviously, the mapping criterion is x ( n ) = s 1 ( n + 1 2 ), if n is odd s 2 ( n 2 ), if n is even, where s i (l) denotes the l-th (1≤l≤L) codeword of S i in one cooperative frame.
where c ¯ i are suitable constants and K 2 is a continuous function on [ 0, ∞ ) and lim t → ∞ K 2 ( t ) = 0. Choosing ν = ( n − 2 ) / 2 in (39), from (42) the assertion follows. □. Proof of Theorem 2 By Lemma 5, (2) has a solution u ′ ( t ) + ε ( t ) if n is odd, and u ( t ) + ε ( t ) if n is even, where lim t → ∞ ε ( t ) = 0.
Similar(53)
"Everything, down to the last comma, had to be left as it was, even where — an admitted rarity — improvement was possible".
And there is even more where that came from.
The contrast is even starker where the crisis has been most acute.
As a simple example, consider EFSM M 1 in Fig. 1 and the sets C′ = {(s, z) : s ∈ S and z is odd} and C′′ = {(s, z) : z is even} where these two sets are infinite.
It is even harder internationally, where different electorates are involved.
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Since I tried Ludwig back in 2017, I have been constantly using it in both editing and translation. Ever since, I suggest it to my translators at ProSciEditing.

Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com