OFFSET
0,1
COMMENTS
It seems nearly certain that, for all n >= 0, a(n) = ceiling(z - 1/2 - 1/(12*z)) where z = 6/(Pi^2 - (floor(Pi*10^n)/10^n)^2). - Jon E. Schoenfield, Aug 31 2019
LINKS
EXAMPLE
floor((10^0)*sqrt(Sum_{m=1..7} 6/m^2)) = 3.
floor((10^1)*sqrt(Sum_{m=1..23} 6/m^2)) = 31.
floor((10^2)*sqrt(Sum_{m=1..600} 6/m^2)) = 314.
floor((10^3)*sqrt(Sum_{m=1..1611} 6/m^2)) = 3141.
floor((10^4)*sqrt(Sum_{m=1..10307} 6/m^2)) = 31415.
floor((10^5)*sqrt(Sum_{m=1..359863} 6/m^2)) = 314159.
PROG
(PARI) a(n) = {my(k = 1); t = floor(10^(n)*Pi); while(floor(10^(n)*sqrt(sum(m = 1, k, 6/m^2))) != t, k++); k; } \\ Jinyuan Wang, Aug 30 2019
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Zachary Russ, Aug 28 2019
EXTENSIONS
a(6)-a(19) from Jon E. Schoenfield, Aug 31 2019
STATUS
approved