Mean of 10^9 consecutive primes?

[zak seidov]

Right now I'm trying to calculate 
sum of first 10^9 primes... using Mathematica... spent
already hours...

My requests
Is this sum known?
Can anyone calculate it to compare with my value?

Just for control:

n=106333207,s=112600741085770645

(n may be correct up to =/- 1 because
interrupting Mathematica may cause it(?))

Thanks,
Zak


--- Richard Mathar <mathar at strw.leidenuniv.nl> wrote:

> 
> zs> From seqfan-owner at ext.jussieu.fr  Thu Sep 28
> 08:50:40 2006
> zs> Date: Wed, 27 Sep 2006 23:49:48 -0700 (PDT)
> zs> From: zak seidov <zakseidov at yahoo.com>
> zs> Subject: Mean of 10^9 consecutive primes?
> zs> To: seqfan at ext.jussieu.fr
> zs> ....
> zs> Can anyone make this for 10^9 (12?)
> zs> consecutive primes -
> zs> and submit it to Neil.
> zs> 
> zs> Thanks, Zak
> zs> 
> zs> %I A000001
> zs> %S A000001 
> zs>
>
11860710,19524458,30466003,57980974,63924288,90876871,98124660,100711080,107813124,130902871,133345096,137765645,149928192,187600902
> zs> ,214934436,223349020
> zs> %N A000001 Arithmetic mean of 1000000
> consecutive
> zs> primes.
> zs> %C A000001 Corresponding indices of the first
> primes
> zs> are:
> zs>
>
275775,740092,1383476,2948575,3280201,4764532,5159226,5299723,5684491,6926061,7056669,7292768,7940227,9929283,11358606,11796712
> zs> Cf. A102655 Numbers which are the arithmetic
> mean of
> zs> four successive 
> zs> primes.
> zs> A122040 Numbers which are equal to arithmetic
> mean of
> zs> six successive 
> zs> primes.
> zs> A122480 Least k>2 such that n is equal to
> zs> arithmetic mean of k 
> zs> consecutive primes.
> zs> A122808 Least number that is the arithmetic mean
> of n
> zs> (but no fewer) 
> zs> consecutive primes.
> zs> %e A000001 
> zs>
>
(p(275775)+p(275776)+...+p(275775+999998)+p(275775+999999))/1000000=11860710,
> zs> p(n)=nth prime.
> zs> %Y A000001 A102655, A122040,A122480, A122808.
> zs> %O A000001 1
> zs> %K A000001 ,nonn,
> zs> %A A000001 Zak Seidov (zakseidov at yahoo.com), Sep
> 27
> zs> 2006
> 
> For those who have more patience or faster computers
> than I do, here the PARI
> program that would attack the job.... and could also
> be used to pull out some
> of the other sequences..
> 
> 
> \\ emit numbers with integer mean of nsucc
> successive primes
> \\ nsucc = number of primes in each sum to be tested
> for integer mean
> aMeanPri(nsucc)={
> 	local(pstart,psum,indx) ;
> 	\\ pstart and pend the current first and last
> primes in the list of nsucc
> 	pstart=2 ;
> 	pend=pstart ;
> 	\\ psum the current sum over nsucc primes
> 	psum=pstart ;
> 	for(i=2,nsucc,
> 		pend=nextprime(pend+1) ;
> 		psum += pend ;
> 	) ;
> 	indx=1 ;
> 	while(1,
> 		if(psum % nsucc ==0,
> 			print(psum/nsucc," ",indx),
> 		) ;
> 		\\ delete the first prime from the sum, add the
> next one...
> 		psum -= pstart ;
> 		pstart=nextprime(pstart+1) ;
> 		pend=nextprime(pend+1) ;
> 		psum += pend ;
> 		indx++ ;
> 	) ;
> }
> 
> \\ test: compute A122040 
> A122040()={
> 	aMeanPri(6) ;
> }
> 
> {
> 	\\A122040() ;
> 	aMeanPri(1000000) ;
> }
> 


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