N=300000
prime_list=[]
p=3
while p<N:
prime_list.append(p)
p=next_prime(p)
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goldbach_representations=[0 for i in srange(2*N)]
for p in prime_list:
for q in prime_list:
goldbach_representations[p+q]+=1
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goldbach_reps_pairs=[]
for i in srange(len(goldbach_representations)/2):
if goldbach_representations[i]>0:
goldbach_reps_pairs.append([i,goldbach_representations[i]])
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def hl(n):
m=Integer(2*n)
pd= m.prime_factors()
hlf=1
for p in pd[1:]:
hlf=hlf*(p-1)/(p-2)
return(hlf)
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gbc= [] #goldbach corrected by hl(n)
for x in goldbach_reps_pairs:
#print(type(x[0]))
#a=x[0]
#print( hl(a) )
gbc.append([x[0],x[1]/hl(x[0])])
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var('t')
A=list_plot(gbc)
B=plot(1.7*t/(log(t))^2,t, 2,100000,color='yellow')
C=plot(1.7*t/(log(t))^2-2*sqrt(2*t)/log(t),t, 2,100000,color='red')
D=plot(1.7*t/(log(t))^2+2*sqrt(2*t)/log(t),t, 2,100000,color='green')
show(A+B+C+D)
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What happens if we use Li(x) instead of x/log(x) to approximate $\pi(x)$?
def f(n):
return(2*Li(n)*(Li(2*n)-Li(n))/n)
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a=.75
A=list_plot(gbc)
B=plot(a*f(t),t, 2,N,color='yellow')
C=plot(a*f(t)-2*sqrt(2*t)/log(t),t, 2,N,color='red')
D=plot(a*f(t)+2*sqrt(2*t)/log(t),t, 2,N,color='green')
show(A+B+C+D)
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