a[0],
a[1], ...,
a[s] be an increasing arithmetic progression
with successive difference d.
Assume that a[0] and d are coprime and greater than 1.
Considering S = {a[0],...,a[s]} as
a position in Sylver Coinage,
we can also assume that s<a[0],
since for any k,
a[k+a[0]] =
a[k] + da[0],
so that a[0] and a[k]
together eliminate a[k+a[0]].
Let t = t(S).
The value of t is given by
Roberts's Formula.
This is the Progression Hypothesis:
s=1 then S is an
ender as shown by the proof of
Hutchings's Theorem.
In fact any position satisfying the conditions of
Hutchings's Theorem is a quiet ender.
a[0]=3 and s=2 then S is
a non-quiet ender.
a[0]>3 and s=a[0]-1,
then t=sd, and the moves that fail to eliminate t
are the values of kd for all k less
than or equal to s
that do not divide s.
Since in particular s-1 does not divide s,
S is not an ender.
a[0]>3 and s<a[0]-1,
then the moves that fail to eliminate t
are the values of t-kd
for k from 1 to (a[0]-2) mod s inclusive.
If s divides (a[0]-2), then
S is a quiet ender; otherwise it is not an ender.
S={2,9}.
Since a[0]=2, S is a quiet ender.
Let S={3,7,11}.
Since a[0]=3 and s=2, S is a non-quiet ender.
Let S={5,6,7,8,9}.
Since a[0]>3 and s=a[0]-1,
S is not an ender.
Here t=4, s=4, and d=1;
and 4 fails to be eliminated
by the one multiple of 1 that does not divide s: 3.
Let S={7,11,15,19,23,27,31}.
Again a[0]>3 and s=a[0]-1,
so S is not an ender.
Here t=24, s=6, and d=4;
since neither 4 nor 5 divides 6, 24 fails to be eliminated by 16 and 20.
Let S={6,13,20,27}.
In this case a[0]>3 and s<a[0]-1.
We find that t=41, s=3, and d=7.
Since s leaves a remainder of 1 when divided into
a[0]-2,
41 fails to be eliminated by (41-1*7)=34.
Let S={8,11,14,17,20,23,26}.
Again a[0]>3 and s<a[0]-1.
Here t=29, s=6, and d=3.
Since s | a[0]-2, S is a quiet ender.