Re: factorial sequence proposal

wugi wrote:
"I.N. Galidakis" :
wugi wrote:

The basic idea is, implementing factorials of factorials, or
"powered" factorials.
So I came to use the notation !, !!, !!!, .... !^k, although I found
out that !! is being used for less exciting (in my view)
definitions.

A sequence needs a starting number, or seed n : hence,
Seq = Seq(n), with terms a(k)=Seq(n,k) as follows:
a(0) = n
a(1) = n!
a(2) = a(1)! = (n!)! == n!!
.....
a(k) = a(k-1)! = n(!!...!!) == n(!^k)

The idea is not really new. People have worked with iterated
factorials before.

For example:

http://ioannis.virtualcomposer2000.com/math/hfseries.html

There is a reference given to me by Dave L. Renfro in AMM Problems
(which I can't find right now) which gives bounds for it. Using your
notation (and if memory serves right):

n^^k < n(!^k) < n^^(k+1),

where ^^ is the tetration operator. Perhaps that's why your sequence
didn't make it to OEIS.

Thank you.

You are welcome.

But didn't OEIS aspire comprehensiveness?

Most certainly.

So if one signals not
having found such or so a series in their site, one likes at least to
be pointed to it, or any other reply.

I really doubt that Mr. Sloane will bother emailing back to you a negative
response, or even a positive one for that matter. I don't think he has the time,
anyway. The way I check for my sequences, is to wait one-two months, until they
finally show up in the list of new sequences. I received no notification from
Mr. Sloane for my submissions.

The application procedure also
does (apparently) only take specific number examples; parametered
formulas seem out of question, ruling out god knows how many cases.

I don't think the OEIS is as much paranoid as you state. Did you use the
standard submission form that Mr. Sloane has available for upload?

Another reason for failure might be that a(n) grows very rapidly and Sloane
needs at least 5-6 terms from any new sequence. I don't think your a(n) is very
friendly in that respect.

And yet another reason your sequence didn't make it, might be that your sequence
is really infinitely many sequences a(k,n), k,n\in N. Depending on the arguments
k and n, you get many different sequences. For example:

a(k,1) = {1, 1, 1, ....}
a(k,2) = {2, 2, 2, ....}
a(k,3} = {3, 720, .2601218944e1747, ...}
a(k,4) = {4, 620448401733239439360000, ....}
.....

A by-observation for the series above:
2 is apparently a unit seed (besides 1), as we get
2=2!=2!!=2!!!=....

Yes. The exact terminology you have in mind, is probably that of "fixed point".

2 is a fixed point of your a(n). So is 1 in fact, since:

a(k,2) = 2, for all k\in N and a(k,1) = 1, again for all k\in N.

guido
http://home.scarlet.be/~pin12499
--
I.N. Galidakis

.

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