What numbers N are the sum of a sequence of consecutive primes such that the product of the first and last numbers in the sequence is N?
Choosing amongst only the first 9592 primes:
10 = 2 + 3 + 5
39 = 3 + 5 + 7 + 11 + 13
155 = 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31
371 = 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53
I imagine an upper limit can be analytically determined where the product surpasses any possible sum by virtue of the magnitude of the 1st prime exceeding the sqrt of the sum of any possible span of primes up to the 2nd prime of the product (he said, waving his hands frantically...).
program sss
implicit none
integer i,j,k,primes(1000000),pcnt,lim,len,
1 ll,last,l,maxlen,ibeg
integer*16 prod,sum
c
c generate primes from 1 to 10^i
c
primes(1)=2
pcnt=1
2 print*,'k?'
read*,k
if(k.lt.2.or.k.gt.7)go to 2
do 1 j=3,10**k,2
lim=sqrt(1.*j)+1
do i=3,lim,2
if(1.*i*int((1.*j)/i).eq.j)go to 1
enddo
pcnt=pcnt+1
primes(pcnt)=j
c print*,j
c if(j.gt.30)stop
1 enddo
do ibeg=1,pcnt-1
c print*,'ibeg= ',ibeg
maxlen=pcnt-ibeg+1
do len=2,maxlen
last=ibeg+len-1
sum=0
do l=ibeg,last
sum=sum+primes(l)
enddo
prod=primes(ibeg)*primes(last)
if(sum.eq.prod)then
c print*,' sum ',sum
print 3,sum,(primes(ll),ll=ibeg,last)
3 format(i4,' = ',30(i2,1x,'+ '))
endif
enddo
enddo
c do i=1,20
c print*,primes(i)
c enddo
print*,pcnt
end
rabbit-3:~ lord$ sss
k?
5
10 = 2 + 3 + 5 +
39 = 3 + 5 + 7 + 11 + 13 +
155 = 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 +
371 = 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 +
Edited on July 5, 2018, 1:50 pm
Here are some
