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[求助] 找几个素数

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发表于 2025-3-2 11:27:11 | 显示全部楼层 |阅读模式

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素数p满足:个位数是1,3,7,9结尾,其余数位全是偶数,并且满足p=k^2+1。举例如下:
46240001
24206401
22848401
846401
608401
401
好像10^8以内只有以上几个,10^5,10^7这些数里面好像没有
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2025-3-2 11:50:36 | 显示全部楼层
A359813
Number of primes < 10^n with exactly one odd decimal digit.

3, 12, 45, 171, 619, 2560, 10774, 46708, 202635, 904603, 4073767, 18604618, 85445767, 395944114, 1837763447, 8600149593
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2025-3-2 12:01:37 | 显示全部楼层
  1. Select[(2*Range@1000000)^2,And@@EvenQ[IntegerDigits[#]]&&PrimeQ[#+1]&]+1
复制代码


{5,401,462401,608401,846401,22848401,24206401,46240001,202208401,624000401,802022401,4462240001,6206288401,8060448401,8686240001,22224846401,22266608401,42444240401,60624288401,66460840001,68602086401,88684840001,88804000001,202428006401,202680040001,222048288401,222802880401,240884640001,242064000001,280666848401,408602208401,482080262401,600284048401,608400000001,620628840001,624068400401,646062288401,826644640001,826862862401,868624000001,888080064401,2244004000001,2440468840001,2822400000001}
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2025-3-2 12:09:37 | 显示全部楼层
2
5
401
462401
608401
846401
22848401
24206401
46240001
202208401
624000401
802022401
4462240001
6206288401
8060448401
8686240001
22224846401
22266608401
42444240401
60624288401
66460840001
68602086401
88684840001
88804000001
202428006401
202680040001
222048288401
222802880401
240884640001
242064000001
280666848401
408602208401
482080262401
600284048401
608400000001
620628840001
624068400401
646062288401
826644640001
826862862401
868624000001
888080064401
2244004000001
2440468840001
2822400000001
4008084080401
4048224480401
4406640640001
4480080224401
6860208640001
8048682480401
8862648080401
8868484000001
20068428848401
20248020048401
20268004000001
20448664880401
20620862640401
22020682464401
22082668608401
22240844640401
22448644000001
22626004622401
24226280880401
24800400000001
28044226662401
28646044840001
40284662880401
40806288480401
40880422688401
42224004000001
42260400640001
48266088864401
60808804000001
62406840040001
64642886406401
68260644000001
80442602240401
82424246288401
82640462862401
82644826446401
84640000000001
84680484840001
86602008240401
88020422886401
88886806880401
204662208240401
220668648206401
226646400848401
226824082062401
226864446480401
228460806606401
228462620400401
228620448040001
240002064000001
240008260840001
242268602400401
244200628686401
246464880640001
248440644000001
260844464462401
266022624040001
266668246800401
268468880400401
282262848462401
282266208608401
284866208880401
286286400000001
400064802624401
400088004840001
422264688846401
422804068840001
426042624640001
426266400288401
444206206440001
466848620622401
484000000000001
484000880000401
600240200040001
606626062848401
606682220046401
622002602400401
648222802448401
668686846640401
682264848040001
806084086224401
806446404000001
808048846440001
808822224040001
822268242048401
826204888688401
848086224486401
862608648040001
866242624000001
868662444686401
888040000000001
888648024040001
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2025-3-2 12:24:10 | 显示全部楼层
容易看出除了p=5以外,其余都是01结尾,这个容易证明。
由于$p=k^2+1$不是5的倍数,由此得出\(k=0,1,4\pmod 5\).
由于k必须是偶数,容易得出其中\(k=1,4\pmod 5\)时p的十位数是奇数,不满足条件,由此得出10|k.
然后在根据p的百位数是偶数,得出20|k, 所以\(p=400x^2+1\), 再设\(x=10y+d\)
穷举
  1. (12:29) gp > 400*(10*y+0)^2+1
  2. %267 = 40000*y^2 + 1
  3. (12:29) gp > 400*(10*y+1)^2+1
  4. %268 = 40000*y^2 + 8000*y + 401
  5. (12:29) gp > 400*(10*y+2)^2+1
  6. %269 = 40000*y^2 + 16000*y + 1601
  7. (12:29) gp > 400*(10*y+3)^2+1
  8. %270 = 40000*y^2 + 24000*y + 3601
  9. (12:29) gp > 400*(10*y+4)^2+1
  10. %271 = 40000*y^2 + 32000*y + 6401
  11. (12:29) gp > 400*(10*y+5)^2+1
  12. %272 = 40000*y^2 + 40000*y + 10001
  13. (12:30) gp > 400*(10*y+6)^2+1
  14. %273 = 40000*y^2 + 48000*y + 14401
  15. (12:30) gp > 400*(10*y+7)^2+1
  16. %274 = 40000*y^2 + 56000*y + 19601
  17. (12:30) gp > 400*(10*y+8)^2+1
  18. %275 = 40000*y^2 + 64000*y + 25601
  19. (12:30) gp > 400*(10*y+9)^2+1
  20. %276 = 40000*y^2 + 72000*y + 32401
复制代码

可以直到d=2,3,7,8时,百位数为6,千位数为奇数淘汰。
d=5时万位数是奇数淘汰。
d=0时,\(p=40000y^2 + 1\)有可能,可以继续类似上面递归
d=1,4,6,9时,p的末三位401有可能。

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毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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