求factors的那个利用了库，但是其实很简单，可以自己实现
因为，候选的素因子是很少的，比如，低于100
否则得到的结果，不是最佳的Jacobi sum的参数选择
按照Jacobi算法的要求，
\(N = \prod p^{p_i} \enspace S = 2 \prod_{q-1|N } q^{v_i}\), p, q都是素数，且\(v_i\)满足如果\(q | N\), \(v_i = p_i + 1 \)否则\(v_i = 1\)
using Primes
using Printf
function allMul(x, y)
[i * j
for i in x
for j in y]
end
function calc(n, factors)
factorList =
primeList = []
for (p, k) in factors
numpower =
num = 1
for i in 1:k
num *= p
push!(numpower, num)
end
factorList = allMul(factorList, numpower)
end
sort!(factorList)
return factorList
end
function filterPrime(n, factors, factorList)
k = factors
s = big(2)^big(k+2)
primeList = [(2, k + 2)]
for x in factorList
z = x + 1
if Primes.isprime(z) && (z > 2)
k = factors
s *= big(z)^big(k + 1)
push!(primeList, (z, k + 1))
end
end
return log10(s), s, primeList
end
numbers = [
210,420,840,1260,1680,2520,3360,3780,5040
,7560,8400,10080,12600,15120,25200,30240,42840,45360,55440,60480,75600,85680,100800,110880,128520,131040,166320,
196560,257040,332640,393120,514080,655200,665280,786240,831600,917280,982800,1081080,1179360,1285200,1310400,1441440,1663200,1965600,2162160,2751840,2827440,
3326400,3341520,3603600,3931200,4324320,5654880,6652800,6683040,7207200,8648640,10810800,12972960,14414400,18378360,21621600,36756720,43243200,64864800,73513440,
86486400,113097600,122522400,129729600,147026880,172972800,183783600,216216000,220540320,245044800,273873600,294053760,302702400,367567200,514594080,551350800,
735134400,821620800,1029188160,1074427200,1102701600,1470268800,1643241600,2058376320,2148854400,2205403200,2572970400,2940537600,3491888400,3675672000,4108104000
]
for n in numbers
factors = Primes.factor(n)
factorList = calc(n, factors)
digits, s, primeList = filterPrime(n, factors, factorList)
println(n, " factors = ", factors)
println( " factorList = ", factorList)
println("primes = ", primeList)
println(" s = ", s)
@printf(" digits = %.3f\n", digits)
println("========================================================");
end
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