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Search: id:A062940
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| A062940 |
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Number of squares (including 0) with n digits. |
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+0
4
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| 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
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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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
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FORMULA
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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
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EXAMPLE
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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
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PROGRAM
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(PARI) je=[]; for(n=2, 45, je=concat(je, ceil(sqrt(10^n))-ceil(sqrt(10^(n-1))))); je
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CROSSREFS
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A variant of A049415. A049415(n) = A017936(n+1)-A017936(n) = A049416(n+1)-A049416(n). Cf. A000290, A062941.
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KEYWORD
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nonn,easy,base,new
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AUTHOR
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Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Jul 07 2001
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EXTENSIONS
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Corrected and extended by Dean Hickerson (dean(AT)math.ucdavis.edu) and Jason Earls (zevi_35711(AT)yahoo.com), Jul 10, 2001
Edited by R. J. Mathar, Aug 07 2008
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