N=100000
P=[]
for i in srange(3,N):
if is_prime(i):
P.append(i)
len(P)
9591 9591 |
r=[0 for i in srange(2*N)]
for p in P:
for q in P:
r[p+q]+=1
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list_plot(r)
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list_plot(r[:N])
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print(r[:100])
[0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 3, 0, 2, 0, 3, 0, 4, 0, 4, 0, 4, 0, 5, 0, 6, 0, 5, 0, 4, 0, 6, 0, 4, 0, 7, 0, 8, 0, 3, 0, 6, 0, 8, 0, 6, 0, 7, 0, 10, 0, 8, 0, 6, 0, 10, 0, 6, 0, 7, 0, 12, 0, 5, 0, 10, 0, 12, 0, 4, 0, 10, 0, 12, 0, 9, 0, 10, 0, 14, 0, 8, 0, 9, 0, 16, 0, 9, 0, 8, 0, 18, 0, 8, 0, 9, 0, 14, 0, 6, 0] [0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 3, 0, 2, 0, 3, 0, 4, 0, 4, 0, 4, 0, 5, 0, 6, 0, 5, 0, 4, 0, 6, 0, 4, 0, 7, 0, 8, 0, 3, 0, 6, 0, 8, 0, 6, 0, 7, 0, 10, 0, 8, 0, 6, 0, 10, 0, 6, 0, 7, 0, 12, 0, 5, 0, 10, 0, 12, 0, 4, 0, 10, 0, 12, 0, 9, 0, 10, 0, 14, 0, 8, 0, 9, 0, 16, 0, 9, 0, 8, 0, 18, 0, 8, 0, 9, 0, 14, 0, 6, 0] |
rbetter=[]
for i in srange(N):
if r[i]>0:
rbetter.append([i,r[i]])
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list_plot(rbetter)
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print(rbetter[:50])
[[6, 1], [8, 2], [10, 3], [12, 2], [14, 3], [16, 4], [18, 4], [20, 4], [22, 5], [24, 6], [26, 5], [28, 4], [30, 6], [32, 4], [34, 7], [36, 8], [38, 3], [40, 6], [42, 8], [44, 6], [46, 7], [48, 10], [50, 8], [52, 6], [54, 10], [56, 6], [58, 7], [60, 12], [62, 5], [64, 10], [66, 12], [68, 4], [70, 10], [72, 12], [74, 9], [76, 10], [78, 14], [80, 8], [82, 9], [84, 16], [86, 9], [88, 8], [90, 18], [92, 8], [94, 9], [96, 14], [98, 6], [100, 12], [102, 16], [104, 10]] [[6, 1], [8, 2], [10, 3], [12, 2], [14, 3], [16, 4], [18, 4], [20, 4], [22, 5], [24, 6], [26, 5], [28, 4], [30, 6], [32, 4], [34, 7], [36, 8], [38, 3], [40, 6], [42, 8], [44, 6], [46, 7], [48, 10], [50, 8], [52, 6], [54, 10], [56, 6], [58, 7], [60, 12], [62, 5], [64, 10], [66, 12], [68, 4], [70, 10], [72, 12], [74, 9], [76, 10], [78, 14], [80, 8], [82, 9], [84, 16], [86, 9], [88, 8], [90, 18], [92, 8], [94, 9], [96, 14], [98, 6], [100, 12], [102, 16], [104, 10]] |
list_plot(rbetter[:50])
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Experimentally, it appears that numbers congruent to 0 mod 3 have more representations.
r0=[]
r1=[]
r2=[]
for s in rbetter:
if s[0]%3==0:
r0.append(s)
if s[0]%3==1:
r1.append(s)
if s[0]%3==2:
r2.append(s)
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A=list_plot(r0,color='red')
B=list_plot(r1,color='blue')
C=list_plot(r2,color='darkgreen')
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show(A+B+C)
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show(A+C+B)
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We've established a heuristic to suggest that the number of representations of $n$ as a sum of two primes is inflated by a factor $HL(n)$ where
\[ HL(n) = \prod_{p|n} \frac{p-1}{p-2} \]
is a product over the odd primes dividing $n$.
n=30
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list(n.factor())
[(2, 1), (3, 1), (5, 1)] [(2, 1), (3, 1), (5, 1)] |
def HL(n):
m=2*n
factor_list=list(m.factor()) # this list of factors/powers *will* include 2, so we can know to ignore it!
hl=1
for i in factor_list[1:]:
p=i[0]
hl=hl*(p-1)/(p-2)
return(hl)
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hl_adjusted=[0]
for i in rbetter:
hl_adjusted.append([i[0],i[1]/HL(i[0])])
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list_plot(hl_adjusted)
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A=list_plot(rbetter,color='red')
B=list_plot(hl_adjusted,color='blue')
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show(A+B)
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C=plot(Li(x),x,2,10000000,color='red')
D=plot(x/log(x),x,2,10000000)
show(C+D)
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show(B)
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E=plot(2*x/(log(x))^2,x,2,N,color='red')
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show(B+E)
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integral(1/(x*log(x)),x)
log(log(x)) log(log(x)) |
F=plot(2*4/5/log(x)*(x/log(x)-log(log(x))),x,5,N,color='red')
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show(B+F)
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G=plot(2*4/5*x/(log(x))^2,x,2,N,color='red')
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show(B+G)
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N=10000000
P=[]
for i in srange(3,N):
if is_prime(i):
P.append(i)
len(P)
664578 664578 |
prime_diff_dict={}
for i in srange(1,len(P)):
diff = P[i]-P[i-1]
if prime_diff_dict.has_key(diff):
prime_diff_dict[diff]+=1
else:
prime_diff_dict.update({diff:1})
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len(prime_diff_dict)
74 74 |
list_plot(prime_diff_dict)
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list_plot([[2*i,prime_diff_dict[2*i]] for i in srange(1,20)])
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HL(2)
1 1 |
list_plot([[2*i,prime_diff_dict[2*i]/HL(2*i)] for i in srange(1,20)])
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It appears to be the case that if we divide the number of occurences of 2*i as a consecutive difference by HL(2*i), then we smooth out the curve immensely!
Can we heuristically explain why 6 appears more often as a consecutive difference than 2 or 4?
twin_prime_cores=[]
for i in srange(2,len(P)):
if P[i]-P[i-1]==2:
twin_prime_cores.append((P[i]-1)/6)
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len(twin_prime_cores)
58980 58980 |
twin_prime_cores[:100]
[1, 2, 3, 5, 7, 10, 12, 17, 18, 23, 25, 30, 32, 33, 38, 40, 45, 47, 52, 58, 70, 72, 77, 87, 95, 100, 103, 107, 110, 135, 137, 138, 143, 147, 170, 172, 175, 177, 182, 192, 205, 213, 215, 217, 220, 238, 242, 247, 248, 268, 270, 278, 283, 287, 298, 312, 313, 322, 325, 333, 338, 347, 348, 352, 355, 357, 373, 378, 385, 390, 397, 425, 432, 443, 448, 452, 455, 465, 467, 495, 500, 520, 528, 542, 543, 550, 555, 560, 562, 565, 577, 578, 588, 590, 593, 597, 612, 628, 637, 642] [1, 2, 3, 5, 7, 10, 12, 17, 18, 23, 25, 30, 32, 33, 38, 40, 45, 47, 52, 58, 70, 72, 77, 87, 95, 100, 103, 107, 110, 135, 137, 138, 143, 147, 170, 172, 175, 177, 182, 192, 205, 213, 215, 217, 220, 238, 242, 247, 248, 268, 270, 278, 283, 287, 298, 312, 313, 322, 325, 333, 338, 347, 348, 352, 355, 357, 373, 378, 385, 390, 397, 425, 432, 443, 448, 452, 455, 465, 467, 495, 500, 520, 528, 542, 543, 550, 555, 560, 562, 565, 577, 578, 588, 590, 593, 597, 612, 628, 637, 642] |
Questions regarding twin-prime cores:
Is there an $n_0$ so that every $n>n_0$ is a sum of twin-prime cores?
How does the number of representations grow?
Are there striations present?
Can we come up with a correction factor which helps to explain the striations, akin to HL(n) for Goldbach?
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