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Search: id:A057655
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| A057655 |
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The circle problem: number of points (x,y) in square lattice with x^2+y^2 <= n. |
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+0
9
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| 1, 5, 9, 9, 13, 21, 21, 21, 25, 29, 37, 37, 37, 45, 45, 45, 49, 57, 61, 61, 69, 69, 69, 69, 69, 81, 89, 89, 89, 97, 97, 97, 101, 101, 109, 109, 113, 121, 121, 121, 129, 137, 137, 137, 137, 145, 145, 145, 145, 149, 161, 161, 169, 177, 177, 177
(list; graph; listen)
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OFFSET
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0,2
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REFERENCES
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J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, p. 106.
F. Fricker, Einfuehrung in die Gitterpunktlehre, Birkhaeuser, Boston, 1982.
P. de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 5.
E. Kraetzel, lattice Points, Kluwer, Dordrecht, 1988.
C. D. Olds, A. Lax and G. P. Davidoff, The Geometry of Numbers, Math. Assoc. Amer., 2000, p. 51.
W. Sierpinski, Elementary Theory of Numbers, Elsevier, North-Holland, 1988.
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LINKS
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T. D. Noe, Table of n, a(n) for n = 0..1000
F. Richman, Count Gaussian integers of norm less than and equal to n
W. Sierpi\'{n}ski, Elementary Theory of Numbers, Warszawa 1964.
F. Richman, Counting Gaussian integers in a disk
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FORMULA
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a(n) = 1 + 4*{ [n/1] - [n/3] + [n/5] - [n/7] + ... }. - Gauss
a(n) = 1 + 4*Sum_{ k = 0 .. [sqrt(n)] } [ sqrt(n-k^2) ]. - Liouville (?)
a(n) - Pi*n = O(sqrt(n)) (Gauss). a(n) - Pi*n = O(n^c), c = 23/73 + epsilon ~ 0.3151 (Huxley). If a(n) - Pi*n = O(n^c) then c > 1/4 (Landau, Hardy). It is conjectured that a(n) - Pi*n = O(n^(1/4 + epsilon)) for all epsilon >0.
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MATHEMATICA
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f[n_] := 1 + 4Sum[ Floor@ Sqrt[n - k^2], {k, 0, Sqrt[n]}]; Table[ f[n], {n, 0, 60}] (from Robert G. Wilson v (rgwv(at)rgwv.com), Jun 16 2006)
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PROGRAM
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(PARI) a(n)=sum(x=-n, n, sum(y=-n, n, if((sign(x^2+y^2-n)+1)*sign(x^2+y^2-n), 0, 1)))
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CROSSREFS
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Partial sums of A004018. Cf. A057656, A057961, A057962. For another version see A000328.
A014198(n) + 1.
Sequence in context: A105643 A073168 A127500 this_sequence A141124 A046255 A068388
Adjacent sequences: A057652 A057653 A057654 this_sequence A057656 A057657 A057658
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KEYWORD
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nonn,easy,nice
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com), Oct 15 2000
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