[seqfan] Using floretions for primality proofs
Creighton Kenneth Dement
creighton.k.dement at mail.uni-oldenburg.de
Wed Dec 2 00:25:05 CET 2009
I would like to give a quick demonstration of how floretions may be used
to (hopefully elegantly) prove some prime number properties of sequences.
Previously, they have been used to generate sequences with interesting (in
my opinion) prime number properties. For example, A113166 has the property
A113166(p) = Fib(p-1) for all primes. However, I've never really thought
about using floretions to provide proofs of primality for sequences until
today. As a hobby mathematician, I certainly do not wish to claim that
other methods can't be used to do to the same or something similar.
A reminder: tes(X) is the coefficient of the unit vector (written 'ee' or
sometimes simply as 1) of the floretion X. See Robert Munafo's page
http://mrob.com/pub/math/seq-floretion.html for an overview.
If tes(X) = q and tesseq(X) = (tes(X), tes(X^2), tes(X^3), ...) is a
sequence of integers, then
p divides tes(X^p) - q^p for all primes > 2.
If I haven't made a mistake, this has now been proven for all q in the
cases of p = 3 and p = 5 and I'm edging closer to a proof of the general
case. To see how this plays out on a concrete example, suppose we wish to
For prime p, A001333(p) congruent to 1 mod p
Define X = 'ii' + 'ik' + 'ee'
Then tes(X) = 1 and we see that
tesseq(X): [1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363, 8119, 19601,]
resembles A0001333, apart from the initial term.
Theorem. tesseq(X) is always a 4th order linear recurrence or less for any X.
From this it follows that tesseq(X) does, in fact, return A0001333. Thus,
the requirements of the conjecture are met and we conclude that, if the
conjecture is correct, p divides tes(X^p) - 1^p = tes(X^p) - 1 =
A0001333(p) - 1 for all primes.
Note also that (1, tes(X+ee), tes((X+'ee')^2), ...) is the binomial
transform of (1, tes(X), tes(X^2), ...) and that if tes(X) = 0, then of
course tes(X+'ee') = 1.
Let's now pick a floretion "at random" and see how the conjecture works in
Choose X = 2'i + 5i' - k' + 'ii' + 'kk' + 'ik' + 3'jk' - 2'kj' + 3'ee'
We have tesseq(X) = (3, -5, -15, 1505, 17683, 82411, 121201, 1724769,
44697027, 451106875, 2430789969, ...)
which has the generating function (-3 + 41*z - 291*z^2 - 111*z^3)/(-1 +
12*z - 82*z^2 + 388*z^3 + 111*z^4)
tes(X) = 3. Therefore
3 must divide tes(X^3) - 3^3 = -15 + 27 = 12
5 must divide tes(X^5) - 3^5 = 17683 = 17440
7 must divide tes(X^7) - 3^7 = 119014 = 2*7*8501
11 must divide tes(X^11) - 3^11 = 2430612822 = 2*3*11*36827467
I'm looking forward to proving the conjecture in its entirety.
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