Mean of 10^9 consecutive primes?

[Richard Mathar]

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) ;
}