1,1

Sum of first 2n terms = 10^n. - Zak Seidov (zakseidov(AT)yahoo.com), 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 (ctibor.zizka(AT)seznam.cz), Mar 09 2008

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

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

a(n)=A017934(n)-A017934(n-1)-(-1)^n, n>=2. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 17 2008

a(1)=4 because there are 4 one-digit squares: 0,1,4,9; - Zak Seidov (zakseidov(AT)yahoo.com), Aug 05 2006

a(2)=6 because there are 6 two-digit squares: 16,25,36,49,64,81. - Zak Seidov (zakseidov(AT)yahoo.com), Aug 05 2006

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

(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) ) } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Aug 14 2009]

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

Sequence in context: A083157 A151519 A061595 this_sequence A075813 A004032 A107952

Adjacent sequences: A062937 A062938 A062939 this_sequence A062941 A062942 A062943

nonn,easy,base

Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Jul 07 2001

Corrected and extended by Dean Hickerson (dean.hickerson(AT)yahoo.com) and Jason Earls (zevi_35711(AT)yahoo.com), Jul 10, 2001

Edited by R. J. Mathar, Aug 07 2008