OFFSET
0,2
COMMENTS
Least integer k such that A002088(k) >= 10^n.
Since A002088(k) = (3*k^2)/(Pi^2) + O(k log k), the digits of a(n) for n even (resp. odd) approach the decimal digits of Pi/sqrt(3)=1.8137993642342... and Pi*sqrt(10/3)=5.7357372095454... resp.
Conjecture: For n even, either a(n) = floor(Pi/sqrt(3)*10^(n/2)) or a(n) = floor(Pi/sqrt(3)*10^(n/2))+1. For n odd, either a(n) = floor(Pi*sqrt(10/3)*10^((n-1)/2)) or a(n) = floor(Pi*sqrt(10/3)*10^((n-1)/2))+1.
LINKS
Chai Wah Wu, Table of n, a(n) for n = 0..30
PROG
(Python)
from sympy import totient
def A343802(n):
s, c = 0, 0
while s < 10**n:
c += 1
s += totient(c)
return c
CROSSREFS
KEYWORD
nonn
AUTHOR
Chai Wah Wu, Apr 29 2021
STATUS
approved