A062940 Number of squares (including 0) with n digits.

4, 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

Amarnath Murthy, Jul 07 2001

Keywords

• nonn
• easy
• base

Comments

Sum of first 2n terms = 10^n. - Zak Seidov, Aug 05 2006

a(n)/a(n-1) ~ 10^(1/2). For the sequence giving the number of members of the sequence a(k)=k^r with n digits we have a(n)/a(n-1) ~ 10^(1/r). - Ctibor O. Zizka, Mar 09 2008

Examples

			a(1)=4 because there are 4 one-digit squares: 0,1,4,9. - _Zak Seidov_, Aug 05 2006
a(2)=6 because there are 6 two-digit squares: 16,25,36,49,64,81. - _Zak Seidov_, Aug 05 2006
22 squares (100=10^2, 121=11^2, ..., 961=31^2) have 3 digits, hence a(3)=22.
		

Links

• Harry J. Smith, Table of n, a(n) for n = 1..200

Crossrefs

A variant of A049415. A049415(n) = A017936(n+1) - A017936(n) = A049416(n+1) - A049416(n). Cf. A000290, A062941.

Column k=2 of A216653.

Programs

Maple

r:= proc(n, k) local b; b:= iroot(n, k); b+`if`(b^k r(10^n, 2) -r(10^(n-1), 2) +`if`(n=1, 1, 0):
seq(a(n), n=1..40);  # Alois P. Heinz, Sep 12 2012

PARI

je=[4]; for(n=2, 45, je=concat(je, ceil(sqrt(10^n))-ceil(sqrt(10^(n-1))))); je

PARI

{ default(realprecision, 200); for (n=1, 200, b=ceil(10^(n/2)); if (n>1, a=b - c, a=4); c=b; write("b062940.txt", n, " ", a) ) } \\ Harry J. Smith, Aug 14 2009

Formula

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

a(n) = A017934(n) - A017934(n-1) - (-1)^n, n >= 2. - R. J. Mathar, Mar 17 2008

Extensions

Corrected and extended by Dean Hickerson and Jason Earls, Jul 10 2001

Edited by R. J. Mathar, Aug 07 2008