|
%I #8 Dec 25 2019 08:34:07
%S 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,2,3,3,4,5,6,6,6,6,6,6,6,
%T 6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,
%U 6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,7,7,8,8,9,9,10,10,10,10,11
%N Number of numbers m <= n such that 9 equals the second digit after decimal point of square root of n in decimal representation.
%C For n > 1: if A111862(n)=9 then a(n) = a(n-1) + 1, otherwise a(n) = a(n-1).
%C Lim_{n->infinity} a(n)/n = 1/10.
%D G. Pólya and G. Szegő, Problems and Theorems in Analysis I (Springer 1924, reprinted 1972), Part Two, Chap. 4, Sect. 4, Problem 178.
%e a(10) = 0, a(100) = 10, a(1000) = 106, a(10000) = 994.
%Y Cf. A111890, A111891, A111892, A111893, A111894, A111895, A111896, A111897, A111898, A111859.
%K nonn,base
%O 1,23
%A _Reinhard Zumkeller_, Aug 20 2005
|
