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本帖最后由 愚工688 于 2024-4-5 02:10 编辑
我已经解决了偶数哥德巴赫猜想问题。
任意偶数拆分成两个整数,必然可以写成:2A=(A-x)+(A+x)的形式。
偶数哥德巴赫猜想“1+1”的数学原理:
【与A构成“非同余”的变量x与A是组合成“1+1”的主要途径】,
这是针对偶数“1+1”的精确制导工具。
例如:偶数2024040500的“1+1”得出:(今天日期的百倍)
A= 1012020224 , 【与A构成“非同余”的变量x】 = : 675 , 777 , 795 , 1035 , 1233 , 1257 , 1287 , 1497 , 2145 , 2223 , 2313 , 2493 , 3693 , 3843 , 4575 ,……;
【变量x与A组合成的“1+1”】:[ 2024040448 = ] 1012019549 + 1012020899 ; 1012019447 + 1012021001 ; 1012019429 + 1012021019 ; 1012019189 + 1012021259 ; 1012018991 + 1012021457 ; 1012018967 + 1012021481 ; 1012018937 + 1012021511 ; 1012018727 + 1012021721 ; 1012018079 + 1012022369 ; 1012018001 + 1012022447 ; 1012017911 + 1012022537 ; 1012017731 + 1012022717 ; 1012016531 + 1012023917 ; 1012016381 + 1012024067 ; 1012015649 + 1012024799 ; 1012015547 + 1012024901 ; 1012015289 + 1012025159 ;……;
偶数素对数量的波动性变化的规律,不仅仅取决于含有的素因子,还取决于根号内最大素数。因此只有在根号内最大素数不变的情况下才谈得上“素数对呈“线性”增长概念”。
例如以今天日期的十倍的连续偶数的素对数量的计算为例:
inf( 202404050 )≈ 560649.3 , Δ≈,infS(m) = 405469.59 , k(m)= 1.38272
inf( 202404052 )≈ 411707.6 , Δ≈,infS(m) = 405469.59 , k(m)= 1.01538
inf( 202404054 )≈ 811603.4 , Δ≈,infS(m) = 405469.6 , k(m)= 2.00164
inf( 202404056 )≈ 405469.6 , Δ≈,infS(m) = 405469.6 , k(m)= 1
inf( 202404058 )≈ 405669.8 , Δ≈,infS(m) = 405469.61 , k(m)= 1.00049
inf( 202404060 )≈ 1112145.2 , Δ≈,infS(m) = 405469.61 , k(m)= 2.74286
inf( 202404062 )≈ 491110.9 , Δ≈,infS(m) = 405469.61 , k(m)= 1.21121
inf( 202404064 )≈ 409484.2 , Δ≈,infS(m) = 405469.62 , k(m)= 1.0099
inf( 202404066 )≈ 810939.3 , Δ≈,infS(m) = 405469.62 , k(m)= 2
inf( 202404068 )≈ 405584.7 , Δ≈,infS(m) = 405469.63 , k(m)= 1.00028
inf( 202404070 )≈ 601762.2 , Δ≈,infS(m) = 405469.63 , k(m)= 1.48411
inf( 202404072 )≈ 884661 , Δ≈,infS(m) = 405469.63 , k(m)= 2.18182
inf( 202404074 )≈ 461536.2 , Δ≈,infS(m) = 405469.64 , k(m)= 1.13828
inf( 202404076 )≈ 486563.6 , Δ≈,infS(m) = 405469.64 , k(m)= 1.2
inf( 202404078 )≈ 810939.3 , Δ≈,infS(m) = 405469.65 , k(m)= 2
inf( 202404080 )≈ 546317 , Δ≈,infS(m) = 405469.65 , k(m)= 1.34737
inf( 202404082 )≈ 405469.7 , Δ≈,infS(m) = 405469.65 , k(m)= 1
inf( 202404084 )≈ 838902.8 , Δ≈,infS(m) = 405469.66 , k(m)= 2.06897
inf( 202404086 )≈ 408085.6 , Δ≈,infS(m) = 405469.66 , k(m)= 1.00645
inf( 202404088 )≈ 405469.7 , Δ≈,infS(m) = 405469.67 , k(m)= 1
inf( 202404090 )≈ 1326336.4 , Δ≈,infS(m) = 405469.67 , k(m)= 3.27111
inf( 202404092 )≈ 450521.9 , Δ≈,infS(m) = 405469.67 , k(m)= 1.11111
inf( 202404094 )≈ 426934 , Δ≈,infS(m) = 405469.68 , k(m)= 1.05294
inf( 202404096 )≈ 834410.5 , Δ≈,infS(m) = 405469.68 , k(m)= 2.05789
inf( 202404098 )≈ 443450.4 , Δ≈,infS(m) = 405469.69 , k(m)= 1.09367
inf( 202404100 )≈ 549789.4 , Δ≈,infS(m) = 405469.69 , k(m)= 1.35593
inf( 202404102 )≈ 837229.3 , Δ≈,infS(m) = 405469.69 , k(m)= 2.06484
inf( 202404104 )≈ 489245.6 , Δ≈,infS(m) = 405469.7 , k(m)= 1.20661
inf( 202404106 )≈ 405469.7 , Δ≈,infS(m) = 405469.7 , k(m)= 1
inf( 202404108 )≈ 897039.2 , Δ≈,infS(m) = 405469.71 , k(m)= 2.21235
inf( 202404110 )≈ 540626.3 , Δ≈,infS(m) = 405469.71 , k(m)= 1.33333
inf( 202404112 )≈ 429320.9 , Δ≈,infS(m) = 405469.71 , k(m)= 1.05882
inf( 202404114 )≈ 901043.8 , Δ≈,infS(m) = 405469.72 , k(m)= 2.22222
inf( 202404116 )≈ 405592.6 , Δ≈,infS(m) = 405469.72 , k(m)= 1.0003
inf( 202404118 )≈ 493112.9 , Δ≈,infS(m) = 405469.73 , k(m)= 1.21615
inf( 202404120 )≈ 1081252.6 , Δ≈,infS(m) = 405469.73 , k(m)= 2.66667
inf( 202404122 )≈ 405469.7 , Δ≈,infS(m) = 405469.73 , k(m)= 1
inf( 202404124 )≈ 442330.6 , Δ≈,infS(m) = 405469.74 , k(m)= 1.09091
inf( 202404126 )≈ 831732.8 , Δ≈,infS(m) = 405469.74 , k(m)= 2.05128
inf( 202404128 )≈ 405469.8 , Δ≈,infS(m) = 405469.75 , k(m)= 1
inf( 202404130 )≈ 541590 , Δ≈,infS(m) = 405469.75 , k(m)= 1.33571
inf( 202404132 )≈ 985243.2 , Δ≈,infS(m) = 405469.75 , k(m)= 2.42988
inf( 202404134 )≈ 417054.6 , Δ≈,infS(m) = 405469.76 , k(m)= 1.02857
inf( 202404136 )≈ 450522 , Δ≈,infS(m) = 405469.76 , k(m)= 1.11111
inf( 202404138 )≈ 829109.2 , Δ≈,infS(m) = 405469.77 , k(m)= 2.04481
inf( 202404140 )≈ 566370.5 , Δ≈,infS(m) = 405469.77 , k(m)= 1.39683
inf( 202404142 )≈ 432886.3 , Δ≈,infS(m) = 405469.77 , k(m)= 1.06762
inf( 202404144 )≈ 810939.6 , Δ≈,infS(m) = 405469.78 , k(m)= 2
inf( 202404146 )≈ 503341.8 , Δ≈,infS(m) = 405469.78 , k(m)= 1.24138
inf( 202404148 )≈ 405469.8 , Δ≈,infS(m) = 405469.79 , k(m)= 1
inf( 202404150 )≈ 1250998.1 , Δ≈,infS(m) = 405469.79 , k(m)= 3.08531
inf( 202404152 )≈ 405748.1 , Δ≈,infS(m) = 405469.79 , k(m)= 1.00069
inf( 202404154 )≈ 406349.3 , Δ≈,infS(m) = 405469.8 , k(m)= 1.00217
time start =09:38:49 ,time end =09:39:40 ,time use =
区间下界计算值,infS(m)=inf(m)/ k(m) 。区间下界计算值,infS(m)是排除了波动系数的影响的素对低位计算值。
很明显的是infS(m)值是线性增大的。
例:
inf( 202404050 )≈ 560649.3 , Δ≈,infS(m) = 405469.59 ,
inf( 202404150 )≈ 1250998.1 , Δ≈,infS(m) = 405469.79 ,
偶数值增大了100,,infS(m)增大了0.2,
那么推测到偶数值增大1000时,infS(m)增大了2.
偶数值增大2000时,infS(m)增大了4.
偶数值增大3000时,infS(m)增大了6.
inf( 202405050 )≈ 1092875 , Δ≈,infS(m) = 405471.59 , k(m)= 2.69532
inf( 202405052 )≈ 409333.2 , Δ≈,infS(m) = 405471.6 , k(m)= 1.00952
inf( 202405054 )≈ 405471.6 , Δ≈,infS(m) = 405471.6 , k(m)= 1
time start =10:04:36 ,time end =10:04:39 ,time use =
inf( 202406050 )≈ 763520.5 , Δ≈,infS(m) = 405473.6 , k(m)= 1.88303
inf( 202406052 )≈ 832462.1 , Δ≈,infS(m) = 405473.6 , k(m)= 2.05306
inf( 202406054 )≈ 405473.6 , Δ≈,infS(m) = 405473.61 , k(m)= 1
验证:实际数据符合推测。
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