Part (in %) of Prime numbers in set of natural numbers.

[Mblacob]

If we know the index [n] of given Prime Number P[n] in the set of Prime numbers, then P[n] + 1 is the number of all positive integers in the set with n Prime numbers.
It implies that ratio [tex]n/(P[n] + 1)[/tex] gives the part (in percentage) of Prime numbers in this set.
Function [tex]f(n, P) = n/(P[n] + 1)[/tex]\infty, where n = 1,2,3,...\rightarrow\infty and P[n] the Prime Number, converges to 0.05……
Example:
n = 4, P[4] = 7 \Rightarrow0.5
n = 46, P[46] = 199 \Rightarrow 0.23
n = 1053, P[1053] = 8423 \Rightarrow 0.125
n = 21944, P[21944] = 248779 \Rightarrow 0.088206447
n = 41191865, P[41191865] = 800934961 \Rightarrow 0.051429725
n = 50409172, P[50409172] = 990919439 \Rightarrow 0.05087111
n = 50848100, P[50848100] = 1000011601 \Rightarrow 0.05084751
Looks like around 5% of positive integers are Prime Numbers!?

[Guest]

Given results is a good empirical proven of Prime Number Theorem.

n P[n] 1/LN(P[n]) 1/(LN(P[n]) -1.08366) n/P[n]
======= ============ =========== ================== ===========
46 199 0.188917894 0.237549732 0.231155779
1053 8423 0.110635118 0.125706133 0.12501484
21944 248779 0.080487301 0.088178288 0.088206802
2610944 43112609 0.056885002 0.060621984 0.06056103
41191865 800934961 0.048777418 0.05149959 0.051429725
50409172 990919439 0.048276193 0.05094118 0.05087111
50848100 1000011601 0.048254915 0.050917489 0.05084751

Where 1/LN(P[n]) and 1/(LN(P[n]) -1.08366) Prime Number Theorem formulas.