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.