OK the part inside the [] brackets should be interpreted as (let me
solve that equation): f(p5)=(1-p5)/(p5-1). So except for p5=1 f(p5)=-1
and can be anything if p5=1. Hmmm... Does that make sense? f(1) arbitrary?
I think the guessing package must improve on the output. Have you seen
that there is a minus sign in front of the last \prod symbol. That was
what confused me most. But probably that minus is just
(-1)*\prod [...]
. And, of course, the + 2 + 2 looks pretty strange.
Unfortunately, I cannot find where the display of %defprod is defined.
In fact, I find it pretty hard to work with the result. How to take that
apart to use in further computation if needed? Needs a tutorial.
Ralf
On 23.08.22 17:22, Waldek Hebisch wrote:
On Tue, Aug 23, 2022 at 04:26:02PM +0200, Ralf Hemmecke wrote:
Martin, Waldek,
can you tell me how I am supposed to interpret the result of this:
(153) -> guess([2,3,5,7,11,13,17,19])
(153)
[
s - 1 p - 1 s - 1
n - 1 8 7 6
--+ ++-++ --+ ++-++
> | | > - | | [f(p ): (p - 1)f(p ) + p - 1 = 0]
+ 2
--+ | | --+ | | 5 5 5 5
s = 0 p = 0 s = 0 p = 0
8 7 6 5
+
2
]
Martin can probably give better explanation, but I will try. 'guess'
tries to represent seqence as sums or product of simpler sequence
and is doing this recursivly. So we have sum of products of sums
of producs where inner term is f(p_5) and f satifies given
equation. There is display problem: first 2 is added to inner sum
(so we really should have parentheses around inner sum + 2).
Scalar term should be put before sum (but unfortunately current
diplay logic puts them last creating extra confusion).
Also, equation defines rational function. Looking at InputForm
I see that this is reccurence equation of order 0. 'guess'
probably should simplify this to rational function (in this case
constant 1).
Of course, as ususal there is trouble if we have enough data
to support out guess. Funnily, if I add next prime I get
the same formula, so it passes normal sanity check.
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