The 1335 Days - The Power Of The Arm

We may pick up the end of the passage;

Dan 12:12 Blessed is he that waiteth, and cometh to the thousand three hundred and five and thirty days. (KJV)

There is very little in the way of a clue here - except for we can state rather haphazardly that 1335 = (2*666) + 3.

In our calculation of 666 we held fixed a right handed octal and counted the number of possible antichrist "bows" outside of the seven right handed K4 subgroups, excluding the one left handed bow that corresponded to the static right handed K4 group. Then, we counted the six octals under each such bow and we calculated in effect, 666 = (((35-7)*4) - 1)*6.

However, what if we were to have symmetry with one octal on the left as well? By excluding the right handed K4 subgroup as it were also the one "bow" we subtracted, then we have a possible "right handed" antichrist also numbering 666, from which we have excluded the seven subgroups of our left handed octal.

So, plausibly we recover 2*666=1332

How do we recover the extra three? We may state that each static triple is related by the principle in revelation of mingling "hail and fire with blood" that we have the asociations of groups of K4 subgroups to there intersection, each of those three groups corrsponding to the singletons in the alternate hands static subgroup.

So for our familiar reference or sun octal given a=1 we recover

a = [b,d,f] = [c,e,g] = 1 = {[a,f,g], [a,b,c],[a,d,e]} = {[a,e,f],[a,b,g],[a,c,d]}
b = [c,d,g] = [a,e,f]
d = [c,e,f] = [a,b,g]
c = [a,f,g] = [b,d,e]
f = [b,e,g] = [a,c,d]
g = [a,d,e] = [b,c,f]
e = [a,b,c] = [d,f,g]

So without loss of generality assuming a = 1 we may state that on excluding one subgroup/bow pair we find corresponding to them three groups of our octal from the same hand that we had excluded in our calculation. (I.e. excluded from the 35 possible triples in seven symbols - reduced by 7 within our reference octal to 28 without.)

In our calculation these three groups or triples sit "in the centre" between the two sets of 666 antichrist bows on the left or the right. There is the verse near the end of the sequence that reveals this spiritual construction of Christ between antichrist.

Dan 11:45 And he shall plant the tabernacles of his palace between the seas in the glorious holy mountain; yet he shall come to his end, and none shall help him. (KJV)

There are also the verses in Zechariah that show the two systems of 666 carrying the vessel of antichrist into captivity in the land of spiritual Babylon:

Now, if Daniel's vision concerned the mystery of the seven lampstands, then by excluding one octal to calculate our left and right handed "666 sets" of antichrist bows, we may safely assume that these sets agree on the same central seven cycle: We then place that seven cycle (which is only valid on those two octals) as our "angel's circuit" which traverses the seven spirits of God (as in the churches of the letters).

By excluding that one octal as "rest" or "sabbath" the remaining seven cycles agree on one octal between them, and there are also seven others that are paired with each of them in turn. (Each seven cycle is valid on two octals only, but each octal has eight seven cycles upon it.) The triples of the one common octal are excluded in the calculation of 666 upon the bows opposite to that hand, though the calculation of 666 depends only on the combination of unity with a static subgroup. (The left hand is free to float if the right hand is totally excluded.)

At this early stage we must note that to the left and right we have systems numbering 666, and the action "takes place in the centre". What is relevant is that although the left and right hands are excluded, the unity element is not - it is common to both the left and right handed fields, and to state that a = 1 is to state that:

a = [b,d,f] = [c,e,g] = 1 = {[a,f,g], [a,b,c],[a,d,e]} = {[a,e,f],[a,b,g],[a,c,d]}

Our calculations of 666 excluded one "bow" corresponding to [c,e,g] or [b,d,f] (with right hand and left hand excluded respectively) What remains inside the system of 666 is the three remaining choices of unity when one group/bow remains static.

for [b,d,f] static, one of {a,c,e,g} = 1
for [c,e,g] static, one of {a,b,d,f} = 1

Given [b,d,f] static, there are three further choices for unity and these are in collision with the selection of group/bow in the opposing hand. For both hands to remain in a state of "synthetic rest" the unity element must remain fixed: Yet if the right hand is changing it's unity then the left hand is totally "uprooted" or "swept away". (c.f. "overflown" in the text.)

Likewise the "arm" or octal underneath is not retained in the other hand when the unity element is changed. We may without loss of generality assume the "arm" to be the static triple, (whether subgroup or bow) that maintains whichever happens to be the "current octal".

The kings of the north and south are in collision in the text, that much remains clear - yet is there anything else we need to ascertain? Clearly we need to account for three days and two sets of 666. Each day in those sets of 666 is equal to a static triple and choice of unity with it.

In each set of 666 there are two distinct and disjoint sets of triples in the reference hand or octal. That would mean that out of our sets of bows, we have 35 - 7 bows not in each hand, but then 35 - 14 = 21 bows common to both left and right systems of 666.

There are then (4*21) = 84 bows paired with unity, from which we must exclude only one pair in the centre common in both left and right hands, giving us 83 bows of our conjoined or paired 666 system. Now, as before each is valid in six octals, so 83*6 = 498 "days".

Essentially then, there are 498 days in the intersection of the two systems of 666. We have 666 - 468 = 168 subgroup/bow pairs not common to both. Now, 168 = 24*7, which corresponds nicely to the 144 octal reversals and the 24 elders in the seven spirits of God - they are the sealed or elect.

Then clearly in the center we see that between the two opposing identities of the north and south are the people of God. Clearly the text is truly referring to a vision of what will befall Daniel's people in the latter days.

We stated that the unity element was now excluded - we see that the 144 floating unity states in eight C7 groups with three automorphisms correspond to the sealed. Yet these 144 are not in the "intersection" but are outside - in the symmetric difference of the two "666 sets".

The three days in the middle added into the two sets of 666 are therefore the elements not common to either set of 666. From where do these come? We may state they are the "Glorious holy mountain between the seas". We must reexamine our equivalent to the "angel's circuit".

Now, common to both left and right hands is one element - unity. Unity in each hand corresponds to the intersection of three K4 subgroups. Yet then, we were counting pairs of subgroups and bows and the floating unity was completely determined from the pairing of left and right hands.

However, common to both systems under the angel's circuit is only one octal: the reference octal for the right hand common to the seven churches is the alternate octal of the angels circuit in the other hand.

I.e.

a b d c f g e
c f g e a b d
d c f g e a b
b d c f g e a
g e a b d c f
f g e a b d c
e a b d c f g

As our familiar walk on the right hand has every column and row on our familiar right handed octal: yet if we use the same angel (from walking each row) on the opposing octal we would then have:

a e g f c d b
e g f c d b a
c d b a e g f
f c d b a e g
g f c d b a e
b a e g f c d
d b a e g f c

(Where the two sets of rows are acted on by inverses of the same cycle to preserve the automorphisms.)

Now no two columns in either set share the same octal.

So, back to Daniel's opening of the 10th chapter, There were three weeks or 21 days remaining of the 24 possible automorphisms: We should therefore add back in the three automorphisms excluded - where are they?

A couple of verses speak to mind:

Psa 23:4 Yea, though I walk through the valley of the shadow of death, I will fear no evil: for thou art with me; thy rod and thy staff they comfort me. (KJV)

As well as;

2Th 2:3 Let no man deceive you by any means: for that day shall not come, except there come a falling away first, and that man of sin be revealed, the son of perdition;
2Th 2:4 Who opposeth and exalteth himself above all that is called God, or that is worshipped; so that he as God sitteth in the temple of God, shewing himself that he is God. (KJV)

In the former, walking the path of the sealed believer between the two systems of 666 (aka the fourth pale horse, whose rider is named "Death") is compared with the latter, where the antichrist sets himself up in the multitude acting as if he were sat in the place of God between "the seas" judging as God.

The latter we can equate to the system of the image facilitating between two opposing groups, as conjugation on "the crowd" with gHg^-1 with H as the group A5 - over the multitude, having "mapped the room". This is a subject best left until the end of the vision, but the three elements correspond to the unity element common to both in each case, and there are three days in each the left and right that contain the unity element.

We saw that to preserve the automorphisms of the second, fouth and eighth powers of the seven cycle (used in the angel's circuits above), we simply used the inverse from the same cycle. Now, there are three powers of the element for holding static the right handed octal and three of the same for the left handed. Could it be possible that we counted all three powers of the same circuit twice? (actually, no we didn't) We would have to remove one half of those six, and add three back in! (Not necessary.)

We know if we keep the unity fixed and vary the static triple we also number the same set of 666 (as in the trumpets rather than the seals.) Knowing that the right handed octal is also excluded from the right handed system of 666 when it is reduced to the 21 days of the seven "churches" (the angels circuit is the only group with the right handed octal) We must add back in three triples, and these may only be the triples corresponding to unity.

In the K4 form which we are principally interested in,

a = 1 = [a,b,c]
b = [a,d,e]
c = [a,f,g]

Whereas the octal form of the same three groups is:

e = 1 = [a,b,c]
g = [a,d,e]
c = [a,f,g]

In the K4 form we reduce the three cycles of the static subgroup and bow in the octal [b,d,f] and [c,e,g] to cycle the three subgroups of the right hand that are preserved under frobenius in a three cycle. We should also acknowledge the systems:

a = 1 = [a,d,e]
d = [a,f,g]
e = [a,b,c]

a = 1 = [a,f,g]
f = [a,b,c]
g = [a,d,e]

Which are preserved in the octal under frobenius.

Now each of these K4 forms of the ultrafilter are valid in the trinity, and each correspond to the same choice of unity and the same two static subgroup/bow triples. Can there be any answer in the octal other than these three? Clearly in the psalm the author was comforted by God in the face of "Death", (There was a little help given) and in the latter passage antichrist sat as Christ in the body of Christ judging as if he were Christ. The triples of Christ as sat in the right hand groups of [a,b,c],[a,d,e],[a,f,g] are outside the system of the octal yet are valid for those particular bows and groups paired in the two systems of 666.

Christ then is a perfect solution to the conundrum of where the three extra days appear from. We can simply exchange from right hand to left in a similar manner:

a = 1 = [a,e,f]
e = [a,b,g]
f = [a,c,d]

So, each hand has a similar access to the K4 form of the filter. Is there a real difference? If the distinction is that [b,d,f] and [c,e,g] are static and a=1 then no, there is no difference: We could posit that for every subgroup/bow pair of the 666 we would have to add in a factor of three and entail 3*666 = ... However, there is only one octal group shared in both left and right handed 666 sets - underneath the angel's circuit itself. Therefore, we would only have to add in an extra three days. (and not merely two - as we had excluded both left and right triples in each case of 666.)

The extra three days are then the three states of the K4 filter's multiplicative group as induced by frobenius acting on the octal. Irrespective of whatever octal is underlying them, there is no factor of three, just a constant to add that depends on every particular choice of group/bow held static.

The action that is in the centre is then the king of the north and south fighting over whether the central K4 form is "on their side" or not. I.e. whether as above a = 1 = [a,b,c] or whether a = 1 = [a,e,f] etc. The preeminence is to own the centre - and they are in collision for the purpose of undermining the kingdom of God. When one owns the centre, one is in control, yet only God is in control. The collision over which element corresponds to unity is the battle in the text: and the kingdom of God is not decided by the warring of men.

This seems quite esoteric, that history could be reduced to a spiritual tug of war in an abstraction, but in order to discern the nature of Daniel's vision we must continue to examine the time periods given at the passage's end.

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