 mersenneforum.org primes and all k tuplets normalize merits allways < 1, for every big numbers
 Register FAQ Search Today's Posts Mark Forums Read 2019-07-19, 10:04 #1 hal1se   Jul 2018 3×13 Posts primes and all k tuplets normalize merits allways < 1, for every big numbers please search duckduckgo: prime gap wiki first result: https://en.wikipedia.org/wiki/Prime_gap my internet zone block wikipedia, a few years (?) https://www.wikiwand.com/en/Prime_gap prime gap=pn/ln(gn). ln: logarthim natural maximal gap not important! for examle: arround exp(100e6) average prime gap = 100e6 arround exp(100e12) average prime gap = 100e12 so gap not important! what is mean merit: how many jump average prime gap. for example: Merit_______gn______digits______pn______________________________Date____Discoverer 37,005294___26054___306_________1780005161 × 719#/30 − 17768____2017____Dana Jacobsen 37 averge prime jump. 1780005161 * (719#/30) − 17768 = =~ln(5,8832005256790573761517e+305)=704,06055428410137312528606256147 so: gn/ln(pn)=26054/704,06055428410137312528606256147=~37,005339727.. real merit. wikipedi result=~37,005294 near, but microscopic wrong. for example: 41.938784 8350 87 p(Gapcoin) 2017 Gapcoin p(Gapcoin)=2,93703234068022590158723766104419463425709075574811762098588798217895728858676728143227e+86 ln(2,937032340680225901587237661e+86)=199,09971766111134682619409162342 merit=~8350/199,09971766111134682619409162342=~41,93878373.. i think: merit may be important, but normalize merit very important! normalize merit: gn/(ln(pn))^2 above example's normalize merit: 8350/(199,09971766111134682619409162342)/(199,09971766111134682619409162342)= = normalize merit for last example=0,21064210549.. i think: normalize merit > 1 imposible if pn>exp(3). so: prime gap allways < (average prime gap)^2 _________ twin prime gap: p(n) and p(n)+2 a twin prime: t(m) p(n+k) and p(n+k)+2 next twin prime : t(m+1) twin prime gap = p(n+k) - (p(n)+2) so: 3 5 and 5 7 twins gap= 5-5=0 so: 5 7 and 11 13 twins gap= 11-7=4 twin gap merit: (p(n+k) - (p(n)+2)) / (ln(p(n)+2))^2 / 1,32.. 1,32.. twin fix number (fix number only for big number twins: so only p(n) > exp(13) twins) normalize merit for twin prime gap: (1,32)*((p(n+k) - (p(n)+2))) /( (ln(p(n)+2))^3 ) i think normalize merit twin gap allways < 1 if p(n) > exp(13) so: twin prime gap allways < (average twin prime gap)^(3/2) _________ 6tuplet normalize gap merit: 6tuplet =30k+{7,11,13,17,19,23} first five 6tuplet: (7, 11, 13, 17, 19, 23) (97, 101, 103, 107, 109, 113) (16057, 16061, 16063, 16067, 16069, 16073) (19417, 19421, 19423, 19427, 19429, 19433) (43777, 43781, 43783, 43787, 43789, 43793) if 6tuplet's first prime p(n) then: 6tuplet gap: p(n+k) - p(n+16) for example first 6tuplet gap: 97-23=74 6tuplet merit:(17,3..)*(p(n+k) - p(n+16))/((ln( p(n+16))^6) 17,3...: 6tuplet fix number, if p(n)>exp(32) normalize merit 6tuplet gap: (17,3..)*(p(n+k) - p(n+16))/((ln( p(n+16))^7) if you can look: halogram view to 6tuplet distribition: first 6tuplet out of regularity: (7, 11, 13, 17, 19, 23) other 6 tuplets very regularly, so normalize 6tuplet merit>1 imposible (if 6tuplet's first prime p(n)>exp(32)) so: 6tuplets gap allways < (average 6tuplet gap)^(7/6) ___________ k=2 to k=9, so all k tuplet very regularly if we look many big integers. primes and all k tuplets normalize merits allways < 1, for ever big numbers Last fiddled with by hal1se on 2019-07-19 at 10:16   2019-07-19, 11:27 #2 hal1se   Jul 2018 3·13 Posts [QUOTE=hal1se;521914] please forgive, my brain fault again. important correction: so: twin prime gap allways < (average twin prime gap)^(3) _________ so: 6tuplets gap allways < (average 6tuplet gap)^(7) ___________ Last fiddled with by hal1se on 2019-07-19 at 11:28   2020-12-08, 13:17   #3
Cybertronic

Jan 2007
DEUTSCHLAND !

24×17 Posts My prime k-tuplet document

Hello members, I just inform you about my "smallest n-digit prime k-tuplet" document.

Stand: 08.12.2020

regards

Norman
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