Hypothesis

[Rainer Rosenthal]

Artur wrote:
> sum of squares of three 
> consucutive primes A133529 
> is divisable by 3 (with exception 38 and 83)

Each prime p > 3 is not divisible by 3, so it has
remainer 1 or 2 modulo 3. Thus its square is 1 mod 3.
Adding three numbers, which are 1 mod 3, gives a
number divisible by 3: 1 + 1 + 1 = 0 (mod 3).

Cheers,
Rainer Rosenthal
r.rosenthal at web.de

P.S. Wouldn't it be a good idea to ask such questions
     in newsgroup sci.math? A hint in SeqFan will be
     enough then, and leaves discussion where it belongs
     to. In my opinion SeqFan is related to discussing
     OEIS organisation and internal references etc.



Apparently there have been problems with our
Internet connection to the outside world today.

They say they are working on the problem.

Neil

> Does anyone know if the web page of EIS is working  now ?
>  I have sent a comment  twice but got no confirmation by email .   Karol 
> A. Penson



Rainer said:

> P.S. Wouldn't it be a good idea to ask such questions
>      in newsgroup sci.math? A hint in SeqFan will be
>      enough then, and leaves discussion where it belongs
>      to. In my opinion SeqFan is related to discussing
>      OEIS organisation and internal references etc.

are perfectly valid topics for the seqfan list.
And I certainly don't have time to read any newsgroups!
Neil



Dear Roland,
I have just returned from a vacation.
Thanks for spotting my stupid mistake.
The correct comment should be:

Number of ordered trees with n edges and having nonleaf
nodes of outdegree 1
or 3

as corrected on Sep 25 by Christian Bower; THANKS Christian.

Sequence is a Motzkin-like sequence. Motzkin sequence counts
ordered trees
with n
edges and having nodes of outdegree 0, 1, or 2 [g.f. f(x)
defined by
f = 1+x*f+(x*f)^2].

I do not know whether the author (Paul D Hanna) had a reason
to write
x^2/(x+x^2+x^4) instead of x/(1+x+x^3). Probably not.

Best wishes,
Emeric

P.S. Sorry if this is a duplicate.