Number factorization and a highly suspicous claim

[Creighton Dement]

Dear Seqfans, 

In reference to sequence 

%S A111108
0,5,3,25,25,131,175,705,1137,3875,7095,21649,43225,122435,259423,
%T A111108
698625,1541985,4011971,9107175,23143825,53559817,133933475,314086735,
%U A111108
776787009,1838300625,4512108515,10745077143,26237143825,62749602745
%N A111108 a(n) = A001333(n) - (-2)^(n-1), n > 0.
%C A111108 Conjecture: for prime p, p divides a(p). Note that (a(n)) and
A001333 have different offsets.
%D A111108 C. Dement, Floretion Integer Sequences (work in progress)
%F A111108 G.f. -x*(3*x+5)/((2*x+1)*(x^2+2*x-1))
%Y A111108 Cf. A001333, A000204.
%K A111108 easy,nonn,new
%O A111108 1,2
%A A111108 Creighton Dement (crowdog(AT)crowdog.de), Oct 14 2005


(question reposted with permission)  

Date: Sun, 16 Oct 2005 11:18:54 -0700 (PDT)
From: jonathan post <jvospost2 at yahoo.com>
Subject: Correction to Conjecture in A111108
To: njas at research.att.com, Creighton Dement <crowdog at crowdog.de>
Cc: jvospost2 at yahoo.com

Dear Dr. Sloane and Dr. Creighton Dement,

Should not the fascinating conjecture be corrected to:
"Conjecture: for ODD prime p, p divides a(p)"?

After all, a(2) = 5 and 2 does not divide 5.  I'm sure
that this statement was intended, and accidently
truncated.

I suspect that it is worth having someone with access
to significant computational resources test the
conjecture empirically (i.e. "Experimental
Mathematics") concurrently with attacking this on
theoretical grounds.

Thank you for your attention and consideration,

Sincerely,

Jonathan Vos Post


My reply:

> Dear Dr. Jonathan Vos Post,
>
> Thanks for that comment!  I overlooked p = 2.
> (I will wait for your permission to post anything to
> the list.)
>
> Sincerely,
> Creighton
> 
 
If part 2. of the claim, below, holds (which I currently regard as
highly suspicious), there is probably a good basis for a proof of the
conjecture (for odd prime).  By the way, I would be most grateful to
anyone who can find the flaw in my proof of proposition 2.0.3 (Pure
Quaternions and 2nd Order Recurrence Relations) of
http://www.crowdog.de/Flointseq.html I checked all the relations from he
previous lemma several times over and must have blinded myself to
something (see end of proof for comment on what I was expecting). 
 
 I hope the claim, below, makes sense the way it is stated (this is
partly a recount of what has already been written).  

Note that the floretion algebra is merely providing an example in the
case below:
 
If F is the floretion algebra over the reals and x = x_0'i + x_1'j +
x_2'k is any "pure quaternion", then a = Fx and b = Gx have the 
properties a^2 = 0 and ab != ba where F = .25('ii' + 'jj' + 'kk' + e)
and G = .25('i + i' + 'jk' + 'kj'). Let e be the unit in F. For any y =
y_0'i + y_1'j + ... + y_(15)e in F, define tes : F -> reals, tes[y] =
y_(15). Observe that tes is a linear function. 

It can be shown that for all x, y in F: tes[xy] = tes[yx]. Despite the
previous fact, it came as a small surprise (to me) that the binomial
theorem does not hold (with the goal of breaking down tes[(a+b)^n])
because tes[a^2b^2] = tes[ba^2b] = tes[b^2a^2] cannot be rearranged into
tes[abab] = tes[(ab)^2] for general x = x_0'i + x_1'j + x_2'k.   

Lastly, assume that an x exists such that 4tes((Fx + Gx)^n) is an
integer for all n.   


For each n, the integer tes[(a+b)^n]) can partitioned into a sum of
integers as 
4tes[(a+b)^n]) = 4(tes[b^n] + s_1tes[y_1] + s_2tes[y_2] + s_3tes[y_3] +
... s_4tes[y_m])

where 

1. No two of the above (m+1)y's (say, y_u and y_v) can be rearranged to
equal each other by writing one as a product of elements  and bringing
the last element in the product "up to the front", i.e. if 
y_u = b^qa^r...a^s  then  b^qa^r...a^s != y_v ; a^sb^qa^r... != y_v,
etc.

2. y_u is not written in the form of a sum of two or more elements.  

3. y_u cannot be written as y_u = t*b^qa^r...a^s with integer t > 1


As noted before: 

tes[(a + b)^8] = tes[b^8] + 8tes[b^7a] + 8tes[b^5aba] + 8tes[b^4ab^2a] +
4tes[b^3ab^3a] + 8tes[b^3ababa] + 8tes[b^2ab^2aba] + 2tes[abababab]

tes[(a + b)^7] = tes[b^7] + 7tes[b^6a] + 7tes{b^4aba] + 7tes[b^3ab^2a] +
7tes[b^2ababa]

tes[(a + b)^6] = tes[b^6] + 6tes[b^5a] + 6tes[b^3aba] + 3tes[b^2ab^2a] +
2tes[ababab]

tes[(a + b)^5] = tes[b^5] + 5tes[b^4a] + 5tes[b^2aba]

tes[(a + b)^4] = tes[b^4] + 4tes[b^3a] + 2tes[baba]

tes[(a + b)^3)] = tes[b^3] + 3tes[b^2a]


Claim: 
1. if tes[y_m] = tes[b^(n-1)a] then s_m = n 
2. for all s_n, s_n divides n  
3. 1 + s_1 + ... + s_m = Lucas(m)


Of course, point 2 of the claim makes up for most of the "juice"- should
it indeed hold. Empirical evidence that it does hold is highly limited
and currently only based on a few properties. For example:

For x = -2'i -'j -'k,  the sequence 
(tes[(Fx+Gx)], tes[(Fx+Gx)^2], tes[(Fx+Gx)^3], tes[(Fx+Gx)^4], ) "looks
a lot like" two times the sequence
http://www.research.att.com/projects/OEIS?Anum=A001333
Numerators of continued fraction convergents to sqrt(2)

and by proposition 2.0.3 (Pure Quaternions and 2nd Order Recurrence
Relations) 
of http://www.crowdog.de/Flointseq.html 

********
tes[(Fx+Gx)^n] = -x_0tes[(Fx+Gx)^(n-1)] - x_1x_2tes[(Fx+Gx)^(n-2)]
for x = x_0'i + x_1'j + x_k'k
********
the above sequence "really is" two times the OEIS sequence A001333
(though a proof of the last proposition currently incomplete - see my
SOS at the top of the page). 

We know from above that 4tes[(Fx+Gx)^n] can be partioned as:

4tes[(Fx+Gx)^n}] =  4(tes[(Gx)^n] + ntes[(Gx)^(n-1)Fx] + ...)

 From the power sequence proposition 1.2.2 (same link as above), it
follows that 
4tes[(Gx)^n] = (-x_0)^n = 2^n. It follows that 

2*A001333(n) = 4tes[(Fx+Gx)^n}] =  4(2^n + ntes[(Gx)^(n-1)Fx] + ...)

If p is a prime, it follows directly from point 2. of the (highly
suspicious!) above claim 
that  p divides 2*A001333(p) - 4(2^p)