A049415 Number of squares (of positive integers) with n digits.

3, 6, 22, 68, 217, 683, 2163, 6837, 21623, 68377, 216228, 683772, 2162278, 6837722, 21622777, 68377223, 216227767, 683772233, 2162277661, 6837722339, 21622776602, 68377223398, 216227766017, 683772233983, 2162277660169

Offset: 1

Author

Ulrich Schimke (ulrschimke(AT)aol.com)

Keywords

• nonn
• easy
• base

Comments

a(n) + A180426(n) + A180429(n) + A180347(n) = A052268(n).

Lim_{n->infinity} a(2n)/10^n = 1 - 1/sqrt(10);

lim_{n->infinity} a(2n-1)/10^n = 1/sqrt(10) - 1/10. - Robert G. Wilson v, Aug 29 2012

Examples

			22 squares (100=10^2, 121=11^2, ...., 961=31^2) have 3 digits, hence a(3)=22.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Crossrefs

A049415(n) = A017936(n+1) - A017936(n) = A049416(n+1) - A049416(n).

Cf. A062940.

Programs

Magma

[Ceiling(Sqrt(10^n))-Ceiling(Sqrt(10^(n-1))) : n in [1..30]]; // Vincenzo Librandi, Oct 01 2011

Mathematica

f[n_] := Ceiling[Sqrt[10^n - 1]] - Ceiling[Sqrt[10^(n - 1)]]; f[1] = 3; Array[f, 24] (* Robert G. Wilson v, Aug 29 2012 *)

Formula

a(n) = ceiling(sqrt(10^n)) - ceiling(sqrt(10^(n-1))).

From Jon E. Schoenfield, Nov 30 2019: (Start)

a(2n) = floor(10^n * (1 - 1/sqrt(10))), so each even-indexed term a(2n) is given by the first n digits (after the decimal point) of 1 - 1/sqrt(10) = 0.68377223398316...;

a(2n-1) = ceiling(10^n * (1/sqrt(10) - 1/10)), so each odd-indexed term a(2n-1) is given by the first n digits (after the decimal point) of 1/sqrt(10) - 1/10 = 0.21622776601683..., plus 1. (End)

Extensions

More terms from Dean Hickerson, Jul 10 2001