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Old 2020-12-21, 12:52   #1
sweety439
 
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Default "Mixed Sierpinski conjecture base 5" proven!! (if probable primes allowed)

This conjecture is from http://www.kurims.kyoto-u.ac.jp/EMIS...rs/i61/i61.pdf, this article is about the mixed Sierpinski (base 2) theorem, which is that for every odd k<78557, there is a prime either of the form k*2^n+1 or of the form 2^n+k, we generalized this theorem (may be only conjectures to other bases) to other prime bases (since the dual form for composite bases is more complex when gcd(k,b) > 1 (see thread https://mersenneforum.org/showthread.php?t=21954), we only consider prime bases), we conjectured that for every k<the CK for the Sierpinski conjecture base b (for prime b) which is not divisible by b, there is a prime either of the form k*b^n+1 or of the form b^n+k

For base b=5, the remain k for the Sierpinski conjecture is {6436, 7528, 10918, 26798, 29914, 31712, 36412, 41738, 44348, 44738, 45748, 51208, 58642, 60394, 62698, 64258, 67612, 67748, 71492, 74632, 76724, 83936, 84284, 90056, 92906, 93484, 105464, 126134, 139196, 152588} (see http://www.primegrid.com/forum_thread.php?id=5087 and http://www.noprimeleftbehind.net/cru...e5-reserve.htm), and we have these primes in the dual form (5^n+k):

Code:
5^24+6436
5^36+7528
5^144+10918
5^1505+26798
5^4+29914
5^458+36412
5^3+41738
5^9+44348
5^485+44738
5^12+45748
5^12+51208
5^46+58642
5^12+60394
5^2+62698
5^2+64258
5^10+67612
5^41+67748
5^13+71492
5^74+74632
5^7+76724
5^3+83936
5^21+84284
5^181+90056
5^23+92906
5^4+93484
5^11+105464
5^11+126134
5^1+139196
5^15+152588
None of them exceed 1052 digits and all are easily verified to be prime.

the only mixed-remain k-value is 31712, and after the searching, I found that 5^50669+31712 is probable prime (35417 digits), thus, the "mixed Sierpinski conjecture base 5" is now a theorem, in the weaker case that probable primes are allowed in place of proven primes.

Last fiddled with by sweety439 on 2020-12-21 at 13:05
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Old 2020-12-21, 14:08   #2
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"99(4^34019)99 palind"
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Note that the weight of b^n+k is the same as that of k*b^n+1, if gcd(k,b)=1
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