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Number of numbers m <= n such that 0 equals the second digit after decimal point of square root of n in decimal representation.
11

%I #7 Dec 24 2019 21:42:29

%S 1,1,1,2,2,2,2,2,3,3,3,3,4,4,4,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,

%T 6,7,7,7,7,7,8,8,8,8,9,9,9,9,10,10,10,10,10,10,10,10,10,10,10,10,10,

%U 10,10,11,11,11,11,11,12,12,12,12,12,13,13,13,13,13,13,13,14,14,14,14,14,14

%N Number of numbers m <= n such that 0 equals the second digit after decimal point of square root of n in decimal representation.

%C For n > 1: if A111862(n)=4 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) = 3, a(100) = 15, a(1000) = 104, a(10000) = 1006.

%Y Cf. A111891, A111892, A111893, A111894, A111895, A111896, A111897, A111898, A111899, A111850.

%K nonn,base

%O 1,4

%A _Reinhard Zumkeller_, Aug 20 2005