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素数上界估计

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发表于 2024-10-4 09:21 | 显示全部楼层 |阅读模式
素数上界估计

x=2^2时,因为√2^2=2.所以在2^2中筛去所有2的合数和自然数1,
g2^2=1.414 213 562^2=2,h2^2=1.414 213 562^2=2.
π(2^2)=4-2=2.

x=2^3时,因为√2^3=2.828 427 125..所以在2^3中筛去所有2的合数和自然数1,
g3^3=1.587 401 052^3=4.h3^3=1.587 401 052^3=4.
这时候:
g2^3=1.414 213 562^3=2.828 427 1,去掉小数=2.但是g(2^3)=4.
所以g(2^3)=2+2≈g2^3+b^3≈1.414 213 562^3+1.259 921 050^3=4.
π(2^3)<8-2.828&#8239;427&#8239;1=5.171&#8239;572&#8239;9.

x=2^4时,因为√2^4=4.所以在2^4中筛去所有2和3的合数和自然数1,
g4^4=1.778&#8239;279&#8239;410^4=10,h4^4=1.565&#8239;084&#8239;580^4=6.
这时候:
g3^4=1.587&#8239;401&#8239;052^4=6.349&#8239;604&#8239;2,去掉小数=6.但是g(2^4)=10.
所以g(2^4)=6+4≈g3^4+b^4≈1.587&#8239;401&#8239;052^4+1.414&#8239;213&#8239;562=10.
π(2^4)<16-6.349&#8239;604&#8239;2=9.650&#8239;396.

x=2^5时,因为√2^5=5.656&#8239;854&#8239;249.所以在2^5中筛去所有2,3,5的合数和自然数1,
g5^5=1.838&#8239;416&#8239;287^5=21,h5^5=1.615&#8239;394&#8239;266^5=11.
这时候:
g4^5=1.778&#8239;279&#8239;410^5=17.782&#8239;794&#8239;1,去掉小数=17,但是g(2^5)=21.
所以g(2^5)=17+4≈g4^5+b^5≈1.778&#8239;279&#8239;410^5+1.319&#8239;507&#8239;911=21.
π(2^5)<32-17.782&#8239;794&#8239;1=14.217&#8239;205&#8239;9.

x=2^6时,√2^6=8.所以在2^6中筛去所有2,3,5,7的合数和自然数1,
g6^6=1.892&#8239;894&#8239;046^6=46,h6^6=1.618&#8239;870&#8239;407^6=18,
这时候:
g5^6=1.838&#8239;416&#8239;287^6=38.606&#8239;742,去掉小数=38,但是g(2^6)=46.
所以g(2^6)=38+8≈g5^6+b^6≈1.838&#8239;416&#8239;287^6+1.414&#8239;213&#8239;562^6=46.
π(2^6)<64-38.606&#8239;742=25.393&#8239;258.

x=2^7时,√2^7=11.313&#8239;708&#8239;50.所以在2^7中筛去所有2,3,5,7,11的合数和自然数1,
g7^7=1.922&#8239;314&#8239;906^7=97,h7^7=1.633&#8239;246&#8239;253^7=31.
这时候:
g6^7=1.892&#8239;894&#8239;046^7=87.073&#8239;126,去掉小数=87,但是g(2^7)=97.
所以g(2^7)=87+10≈g6^7+b^7≈1.892&#8239;894&#8239;046^7+1.389&#8239;495&#8239;494^7=97.
π(2^7)<128-87.073&#8239;126=40.926&#8239;874.

x=2^8时,√2^8=16.所以在2^8中筛去所有2,3,5,7,11,13的合数和自然数1,
g8^8=1.941&#8239;640&#8239;942^8=202,h8^8=1.646&#8239;452&#8239;553^8=54.
这时候:
g7^8=1.922&#8239;314&#8239;906^8=186.464&#8239;55,去掉小数=186,但是g(2^8)=202.
所以g(2^8)=186+16≈g7^8+b^8≈1.922&#8239;314&#8239;906^8+1.414&#8239;213&#8239;562^8=202.
π(2^8)<256-186.464&#8239;55=69.535&#8239;45.

x=2^9时,√2^9=22.627&#8239;417&#8239;00.所以在2^9中筛去所有不大于19的合数和自然数1,
g9^9=1.953&#8239;863&#8239;559^9=415,h9^9=1.662&#8239;464&#8239;633^9=97
这时候:
g8^9=1.941&#8239;640&#8239;942^9=392.211&#8239;47,去掉小数=392,但是g(2^9)=415.
所以g(2^9)=392+23≈g8^9+b^9≈1.941&#8239;640&#8239;942^9+1.416&#8239;782&#8239;203^9=415.
π(2^9)<512-392.211&#8239;47=119.788&#8239;53

x=2^10时,√2^10=32.所以在2^10中筛去所有不大于31的合数和自然数1,
g10^10=1.963&#8239;559&#8239;019^10=852,h10^10=1.673&#8239;219&#8239;220^10=172.
这时候:
g9^10=1.953&#8239;863&#8239;559^10=810.853&#8239;38,去掉小数=810,但是g(2^10)=852
所以g(2^10)=810+42≈g9^10+b^10≈1.953&#8239;863&#8239;559^10+1.453&#8239;198&#8239;460^10=852
π(2^10)<2^10-810.853&#8239;38=213.146&#8239;62

x=2^11时,√2^11=53.366&#8239;656&#8239;26.所以在2^11中筛去所有不大于53的合数和自然数1,
g11^11=1.970&#8239;482&#8239;986^11=1739,h11^11=1.684&#8239;068&#8239;923^11=309
这时候:
g10^11=1.963&#8239;559&#8239;019^11=1672.9523,去掉小数=1672,但是g(2^11)=1739
所以g(2^11)=1672+67≈g10^11+b^11≈1.963&#8239;559&#8239;019^11+1.465&#8239;570&#8239;789^11=1739
π(2^11)<2^11-1672.9523=375.0477

x=2^12时,√2^12=64,所以在2^12中筛去所有不大于61的合数和自然数1,
g12^12=1.975&#8239;460&#8239;687^12=3532,h12^12=1.695&#8239;404&#8239;216^12=564
这时候:
g11^12=1.970&#8239;482&#8239;986^12=3426.6699,去掉小数=3426,但是g(2^12)=3532
所以g(2^12)=3426+106≈g11^12+b^12≈1.970&#8239;482&#8239;986^12+1.474&#8239;943&#8239;855^12=3532
π(2^12)<2^12-3426.6699=669.3301

x=2^13,√2^13=90.509&#8239;667&#8239;99,所以在2^13中筛去所有不大于89的合数和自然数1,
g13^13=1.979&#8239;476&#8239;859^13=7164,h13^13=1.704&#8239;871&#8239;999^13=1028
这时候:
g12^13=1.975&#8239;460&#8239;687^13=6977.327,去掉小数=6977,但是g(2^13)=7164
所以g(2^13)=6977+187≈g12^13+b^13≈1.975&#8239;460&#8239;687^13+1.495&#8239;398&#8239;865^13=7164
π(2^13)<2^13-6977.327=1214.673

x=2^14,√2^14=128,所以在2^14中筛去所有不大于127的合数和自然数1,
g14^14=1.982&#8239;468&#8239;624^14=14&#8239;484,h14^14=1.714&#8239;733&#8239;710^14=1900
这时候:
g13^14=1.979&#8239;476&#8239;859^14=14&#8239;180.972,去掉小数=14&#8239;180,但是g(2^14)=14&#8239;484
所以g(2^14)=14&#8239;180+304≈g13^14+b^14≈1.979&#8239;476&#8239;859^14+1.504&#8239;347&#8239;308=14&#8239;484
π(2^14)<2^14-14&#8239;180.972=2203.028

x=2^15,√2^15=181.019&#8239;336&#8239;0,所以在2^15中筛去所有不大于181的合数和自然数1,
g15^15=1.984&#8239;941&#8239;278^15=29&#8239;256,h15^15=1.723&#8239;337&#8239;350^15=3512
这时候:
g14^15=1.982&#8239;468&#8239;624^15=28&#8239;714.08,去掉小数=28&#8239;714,但是g(2^15)=29&#8239;256
所以g(2^15)=28&#8239;714+542≈g14^15+b^15≈1.982&#8239;468&#8239;624^15+1.521&#8239;481&#8239;300^15=29&#8239;256
π(2^15)<2^15-28&#8239;714.08=4053.92

x=2^16,√2^16=256,所以在2^16中筛去所有不大于251的合数和自然数1,
g16^16=1.986&#8239;897&#8239;623^16=58&#8239;994`,h16^16=1.731&#8239;736&#8239;891^16=6542
这时候:
g15^16=1.984&#8239;941&#8239;278^16=58&#8239;071.442,去掉小数=58&#8239;071,但是g(2^16)=58&#8239;994
所以g(2^16)=58&#8239;071+923≈g15^16+b^16≈1.984&#8239;941&#8239;278^16+1.532&#8239;234&#8239;040^16=58&#8239;994
π(2^16)<2^16-58&#8239;071.442=7464.558

x=2^17,√2^17=362.038&#8239;672&#8239;0,所以在2^17中筛去所有不大于359的合数和自然数1,
g17^17=1.988&#8239;488&#8239;710^17=118&#8239;821,h17^17=1.739&#8239;725&#8239;298^17=12251
这时候:
g16^17=1.986&#8239;897&#8239;623^17=117&#8239;215.04,去掉小数=117&#8239;215,但是g(2^17)=118&#8239;821
所以g(2^17)=58&#8239;071+923≈g16^17+b^17≈1.986&#8239;897&#8239;623^17+1.532&#8239;234&#8239;040^17=118&#8239;821
π(2^17)<2^17-117&#8239;215.04=13&#8239;856.96

x=2^18,√2^18=512,所以在2^18中筛去所有不大于509的合数和自然数1,
g18^18=1.989&#8239;822&#8239;861^18=239&#8239;144,h18^18=1.747&#8239;101&#8239;808^18=23000
这时候:
g17^18=1.988&#8239;488&#8239;710^18=236&#8239;274.22,去掉小数=236&#8239;274,但是g(2^18)=239&#8239;144
所以g(2^18)=236&#8239;274+2870≈g17^18+b^18≈1.988&#8239;488&#8239;710^18+1.556&#8239;340&#8239;252^18=
239&#8239;144
π(2^18)<2^18-236&#8239;274.22=25&#8239;869.78

x=2^19,√2^19=724.077&#8239;343&#8239;9,所以在2^19中筛去所有不大于719的合数和自然数1,
g19^19=1.990&#8239;927&#8239;381^19=480&#8239;898,h19^19=1.754&#8239;175&#8239;898^19=43390
这时候:
g18^19=1.989&#8239;822&#8239;861^19=475&#8239;854.20,去掉小数=475&#8239;854,但是g(2^19)=480&#8239;898
所以g(2^19)=475&#8239;854+5044≈g18^19+b^19≈1.989&#8239;822&#8239;861^19+1.566&#8239;328&#8239;676^19=
480&#8239;898
π(2^19)<2^19-475&#8239;854.20=48&#8239;433.80

x=2^20,√2^20=1024,所以在2^20中筛去所有不大于1021的合数和自然数1,
g20^20=1.991&#8239;871&#8239;138^20=966&#8239;551,h20^20=1.760&#8239;748&#8239;439^20=82025
这时候:
g19^20=1.990&#8239;927&#8239;381^20=957&#8239;433.0,去掉小数=957&#8239;433,但是g(2^20)=966&#8239;551
所以g(2^20)=957&#8239;433+9118≈g19^20+b^20≈1.990&#8239;927&#8239;381^20+1.577&#8239;593&#8239;032^20=
966&#8239;551
π(2^20)<2^20-957&#8239;433.0=91&#8239;143

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