A141492 - OEIS

%I #15 Aug 06 2021 11:06:15

%S 3,21,144,1085,8685,72381,620420,5428680,48254941,434294481,

%T 3948131653,36191206824,334072678386,3102103442165,28952965460216,

%U 271434051189531,2554673422960304,24127471216847323,228576043106974645,2171472409516259137,20680689614440563220

%N a(n) is the floor of the reciprocal of the difference between the 10^n-th root of 10^n and 1.

%C Conjecture: This sequence converges to the number of primes < 10^n or Pi(10^n).

%C From _Jon E. Schoenfield_, Aug 05 2021: (Start)

%C Define f(x) = 1/(x^(1/x) - 1). As x increases, f(x) -> 1/y - 1/2 + y/12 - y^3/720 + y^5/30240 + ... where y = log(x)/x. So if we let x = 10^n, then we have (see Formula section) a(n) = floor(f(10^n)) and, as n increases, f(10^n) = 1/((10^n)^(1/10^n) - 1) -> 10^n/(log(10)*n) - 1/2 + (log(10)/12)*n/10^n - ...

%C Conjecture: a(n) = floor(10^n/(log(10)*n) - 1/2) for n >= 1. (End)

%F a(n) = floor(1/((10^n)^(1/10^n) - 1)).

%o (PARI) for(x=1,n,y=1/((10^x)^(1/10^x)-1);print1(floor(y)","))

%Y Cf. A006880.

%K nonn

%O 1,1

%A _Cino Hilliard_, Aug 10 2008

%E Name corrected and a(20)-a(21) from _Jon E. Schoenfield_, Aug 05 2021