20201221, 12:52  #1 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{3}×389 Posts 
"Mixed Sierpinski conjecture base 5" proven!! (if probable primes allowed)
This conjecture is from http://www.kurims.kyotou.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...e5reserve.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 the only mixedremain kvalue 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 20201221 at 13:05 
20201221, 14:08  #2 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
110000101000_{2} Posts 
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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