2 finite sequences

[Artur]

SF gurus,
I would like pay attention on two (probably finite sequences) published in
NMBRTHRY Digest - 29 Apr 2008 to 12 May 2008 (#2008-36)

Date:    Mon, 12 May 2008 16:13:19 -0400
From:    Zhi-Wei SUN <zwsun at nju.edu.cn>
Subject: Odd numbers in the form p+x(x+1)

Dear number theorists,

  The well-known Goldbach conjecture states that any EVEN integer n>2
is a sum of two primes.

  Here I propose a conjecture on representations of ODD integers.

  CONJECTURE. (1) Any odd integer n>1 can be written as p+x(x+1), where 
p is a prime and x is a nonnegative integer. [The number r(n) of such 
representations is approximately c*sqrt(n)/(log n). Also, for 
n=5,7,9,... we may require x>0 additionally.]

(2) An odd integer n>1 can be written in the form p+x(x+1) with p a 
prime congruent to 1 mod 4, if and only if n is not among the following 
30 multiples of three:
   3, 9, 21, 27, 45, 51, 87, 105, 135, 141, 189, 225, 273, 321, 327, 
471,  525, 627, 741, 861, 975, 1197, 1461, 1557, 1785, 2151, 12285, 
13575, 20997, 49755.
An odd integer n>1 can be written in the form p+x(x+1) with p a prime 
congruent to 3 mod 4, if and only if n is not among the following 15 
multiples of three:
   57, 111, 297, 357, 429, 615, 723, 765, 1185, 1407, 2925, 3597, 4857, 
5385, 5397.

(3) In general, for any a=0,1,2,... and r=1,3,5,..., all sufficiently 
large odd integers can be expressed in the form p+x(x+1) with p a prime 
congruent to r mod 2^a. [For example, if we use N_r to denote the 
largest odd integer not in the form p+x(x+1) with p=r (mod 8), then 
N_1=358245, N_3=172995, N_5=359907, N_7=444045.]

  I have verified the above conjecture for odd integers not exceeding 
1,000,000. Note that p is not allowed to be zero in the conjecture.

  I view my above conjecture as a supplement to the Goldbach conjecture 
(concerning representations of EVEN NUMBERS)
and my previous conjecture concerning representations of NATURAL NUMBERS 
by the form p+x(x+1)/2 with p a prime or zero.


  Zhi-Wei Sun


  http://math.nju.edu.cn/~zwsun







rg> From seqfan-owner at ext.jussieu.fr  Tue May 13 20:37:01 2008
rg> Date: Tue, 13 May 2008 20:36:34 +0200
rg> From: "Robert Gerbicz" <robert.gerbicz at gmail.com>
rg> To: seqfan at ext.jussieu.fr
rg> Subject: Re: number of partitions bench mark
rg> 
rg> 
rg> These errors are also in Maple 9.5 version.
rg> And the list isn't full, for numbpart(101269) and for numbpart(501269) (very
rg> probably not the next terms in the A110375 sequence) it's giving also
rg> incorrect answer.

I ran a complete scan for differences between Maple 9 and PARI-2.3.3
and submitted the next terms in the sequence, so it covers the
range from n=0 to n=27020 completely then:

> From oeis at research.att.com  Wed May 14 10:06:35 2008
> Date: Wed, 14 May 2008 05:06:31 -0400
> From: The On-Line Encyclopedia of Integer Sequences <oeis at research.att.com>
> To: njas at research.att.com
> Cc: mathar at strw.leidenuniv.nl
> Subject: COMMENT R. J. Mathar A110375
> Reply-to: mathar at strw.leidenuniv.nl
> X-Spam-Status: No
> ...
>  Subject: COMMENT FROM R. J. Mathar RE A110375
> 
> 
> %I A110375
> %S A110375 11269, 11566, 12376, 12430, 12700, 12754, 15013, 17589, 17797, 18181, 18421, 18453, 18549, 18597, 18885, 18949, 18997, 20865, 21531, 21721, 21963, 22683, 23421, 23457, 23547, 23691, 23729, 23853, 24015, 24087, 24231, 24339, 24519, 24591, 24627, 24681, 24825, 24933, 25005, 25023, 25059, 25185, 25293, 27020
> %C A110375 [added more terms from comparison Maple 9 and PARI-2.3.3 outputs]
> %O A110375 1
> %K A110375 ,nonn,
> %A A110375 R. J. Mathar (mathar at strw.leidenuniv.nl), May 14 2008






Thanks everybody for the cleanup, code, and extension of A005045!

Regarding the OGF, the apparently closely related A002817
might have the key for a proof:

> Group is dihedral group D_8 of order 8 with cycle index
> (1/8)*(x1^4+2*x4+3*x2^2+2*x1^2*x2); setting all x_i = n gives the
> formula a(n) = (1/8)*(n^4+2*n+3*n^2+2*n^3)

If a similar idea can be applied for A005045 one might
come up with a proof.  I am not enough of a combinatorial
person to do this and don't want to bother Brendan again.