primes and all k tuplets normalize merits allways < 1, for every big numbers

[hal1se]

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

[hal1se]

[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)


___________

[Cybertronic]

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


Maintread: https://matheplanet.de/matheplanet/n...232720&start=0



Stand: 08.12.2020


regards


Norman