素数の規則を見つけたい。。。

素数の規則を見つけたい。。。 (701ﾚｽ)
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194: 2023/10/22(日)14:35:28.61 ID:1rLOY4nu(10/11) AAS
cos(2pi*(1-((n/(13*11*17*19*23)^11+1/(2*5*7*3))*2*3*5*7*(11*13*17*19*23)^11)/(2*3*5*7*11*13*17*19*23)^7)) > cos(2pi*(29^2/(2*3*5*7*11*13*17*19*23)^7))

n = 864 (151588688860480401830821308882900152330122196839031250 m + 57736288309081718076562795675036302431140590123061457), m element Z
n = 350 (374207506215585906233798888213787804609215937339780000 m + 142526151711561726909000729894946758001444199618071711), m element Z
n = 69 (1898154017035580683794632041664141037872834464767000000 m + 722958740565892817654351528452628482616021302410508679), m element Z
n = 15 (8731508478363671145455307391655048774215038537928200000 m + 3325610206603106961210017030882091020033697991088339923), m element Z
n = 4 (32743156793863766795457402718706432903306394517230750000 m + 12471038274761651104537563865807841325126367466581274711), m element Z

e^(i*2pi*(1-((864*57736288309081718076562795675036302431140590123061457/(13*11*17*19*23)^11+1/(2*5*7*3))*2*3*5*7*(11*13*17*19*23)^11)/(2*3*5*7*11*13*17*19*23)^7)) ＝e^(-(83 i π)/13752125853422782054092109141856701819388685697236915000000)
省3

202: 2023/11/26(日)00:48:12.61 ID:5ylX1SN5(3/3) AAS
cos(2pi*((2*a+1)/2^3-(3*b+2)/3^3-c/5^3-d/7^3+e/11^3+f/13)) > cos(2pi*(17^2/(2310)^3*1/13))

a = 4 n_1, b = 9 n_2, c = 5 (25 n_3 + 11), d = 343 n_4 + 114, e = 1331 n_5 + 1165, f = 13 n_6 + 11,
a = 4 n_1, b = 9 n_2, c = 125 n_3 + 11, d = 343 n_4 + 176, e = 1331 n_5 + 118, f = 13 n_6 + 6, 　cos(2pi*((2*4+1)/2^3-(3*9+2)/3^3-11/5^3-176/7^3+118/11^3+6/13)) =cos((71 π)/80121541500)
a = 4 n_1, b = 9 n_2, c = 125 n_3 + 92, d = 7 (49 n_4 + 34), e = 1331 n_5 + 402, f = 13 n_6 + 1,
a = 4 n_1, b = 3 (3 n_2 + 1), c = 5 (25 n_3 + 13), d = 343 n_4 + 103, e = 1331 n_5 + 751, f = 13 n_6 + 7,
a = 4 n_1, b = 3 (3 n_2 + 1), c = 125 n_3 + 28, d = 7 (49 n_4 + 46), e = 1331 n_5 + 183, f = 13 n_6 + 4,

264: 2023/12/24(日)23:54:35.61 ID:JbDEdDB5(16/17) AAS
表現できる素数は一定のはずなのでnの値によらず
約(2^n-2^(n-1))*(3^n-3^(n-1))*(5^n-5^(n-1))*(7^n-7^(n-1))*(2*11^2)/(2*3*5*7)^n個は一定になる
(2^2-2)*(3^2-3)*(5^2-5)*(7^2-7)*(2*11^2)/(2*3*5*7)^2≒55個
(2^3-2^2)*(3^3-3^2)*(5^3-5^2)*(7^3-7^2)*(2*11^2)/(2*3*5*7)^3≒55個
(2^4-2^3)*(3^4-3^3)*(5^4-5^3)*(7^4-7^3)*(2*11^2)/(2*3*5*7)^4≒55個

269: 2023/12/25(月)00:16:21.61 ID:cm14oBhI(4/19) AAS
(2^2-2^(1))*(3^2-3^(1))*(5^2-5^(1))*(7^2-7^(1))*(11^2-11^(1))*(13^2)/(2*3*5*7*11)^2≒35個
1から13^2の範囲内には39個の素数があるためほぼ等しい

(2^2-2^(1))*(3^2-3^(1))*(5^2-5^(1))*(7^2-7^(1))*(11^2-11^(1))*(13^2-13^(1))*(17^2)/(2*3*5*7*11*13)^2≒55個
1から17^2の範囲内には61個の素数があるためほぼ等しい

348: 2024/01/04(木)01:13:55.61 ID:HQkE/6B8(3/5) AAS
(3 4)^2 (3 5)^2 *C = (3 5)^2 x1 + ((3 4)^2 (-2 Pi x1 + 2 Pi x2 + I (3 5)^2 Log[E^(((2 I) Pi x1)/(4 5)^2 + ((2 I) Pi x2)/(3 5)^2)]))/(2 Pi)
32400 C = (16200 i log(e^((i π x1)/200 + (2 i π x2)/225)))/π + 81 x1 + 144 x2=0 ←n=2 a=3,b=4,c=5のときC=0のため3^2+4^2＝5^2

(3 4)^3 (3 5)^3 *C = (3 5)^3 x1 + ((3 4)^3 (-2 Pi x1 + 2 Pi x2 + I (3 5)^3 Log[E^(((2 I) Pi x1)/(4 5)^3 + ((2 I) Pi x2)/(3 5)^3)]))/(2 Pi)
5832000 C - 918 x1 = 0　←n=3 a=3,b=4,c=5のときC≠0のため3^3+4^3≠5^3

411: 2024/01/16(火)20:26:38.61 ID:CGru1Z9S(2/2) AAS
(2*3*5*7*11*13*17)/(2πi)*ln(e^(i*2π*(1/2+a/3+b/5+c/7+d/11+x/13+y/17)))=19
a,b,c,d,x,yに分母の素因数を持たない数を入れて式を満たす組み合わせは一通りだけある

433: 2024/01/21(日)17:44:01.61 ID:h+lG8rsE(9/12) AAS
(2*3*5*7*11)*((2^N1*3^N2*5^N3*7^N4*11^N5+ab)*(1/2+2/3+3/5+1/7+1/11)mod1)=ab
(2*3*5*7*11)*((2^N1*3^N2*5^N3*7^N4*11^N5+a)*(1/2+2/3+3/5+1/7+1/11)mod1)=a
(2*3*5*7*11)*((2^N1*3^N2*5^N3*7^N4*11^N5+b)*(1/2+2/3+3/5+1/7+1/11)mod1)=b

2*3*5*7*11+13*17=2531は素数

(2*3*5*7*11)*((1*13/2+2*13/3+3*13/5+1*13/7+1*13/11)mod1)=(2*3*5*7*11)*((1/2+2/3+4/5+6/7+2/11)mod1)=13
(2*3*5*7*11)*((1*17/2+2*17/3+3*17/5+1*17/7+1*17/11)mod1)=(2*3*5*7*11)*((1/2+1/3+1/5+3/7+6/11)mod1)=17
(2*3*5*7*11)*((1*13*17/2+2*13*17/3+3*13*17/5+1*13*17/7+1*13*17/11)mod1)=(2*3*5*7*11)*((1/2+1/3+3/5+4/7+1/11)mod1)=13*17
省10

677: 2024/11/12(火)19:42:38.61 ID:5PtRFVCd(1) AAS
prime[61]^2*product((1-1/prime(n)),n=1,60)≒7859.86 ←
primepi[prime[61]^2]=7842

prime[k+1]^2*product((1-1/prime(n)),n=1,k)≒prime[k+1]^2未満の素数の数

prime[k+1]/log(prime[k+1])≒prime[k+1]^2*product((1-1/prime(n)),n=1,k)
ζ(1)=lim[k→∞] 1/product((1-1/prime(n)),n=1,k)≒prime[k+1]*log(prime[k+1])=log((prime[k+1])^(prime[k+1]))
ζ(1)=∞＝log(無限の素数^無限の素数)

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