cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A053411 Circle numbers (version 1): a(n)= number of points (i,j), i,j integers, contained in a circle of diameter n, centered at the origin.

Original entry on oeis.org

1, 1, 5, 9, 13, 21, 29, 37, 49, 69, 81, 97, 113, 137, 149, 177, 197, 225, 253, 293, 317, 349, 377, 421, 441, 489, 529, 577, 613, 665, 709, 749, 797, 861, 901, 973, 1009, 1085, 1129, 1201, 1257, 1313, 1373, 1457, 1517, 1597, 1653, 1741, 1793, 1885, 1961
Offset: 0

Views

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 10 2000

Keywords

Comments

a(n)/(n/2)^2 -> Pi.

Links

Crossrefs

Cf. A053414, A053415, A053416, A053417.
Bisections: A000328 and A036704.

Programs

  • Mathematica
    a[n_] := (m = Ceiling[n/2]; Sum[Boole[i^2 + j^2 <= n^2/4], {i, -m, m}, {j, -Ceiling @ Sqrt[ m^2 - i^2 ], Ceiling @ Sqrt[ m^2 - i^2 ]}]); Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Jun 06 2013 *)

A051233 Number of unit squares at least 50% covered by a circle inscribed in an integer square of size n X n.

Original entry on oeis.org

1, 4, 9, 12, 21, 32, 37, 52, 69, 80, 97, 112, 137, 156, 177, 208, 225, 256, 293, 316, 349, 384, 421, 448, 489, 540, 577, 616, 665, 716, 749, 812, 861, 912, 973, 1020, 1085, 1124, 1201, 1264, 1313, 1396, 1457, 1528, 1597, 1664, 1741, 1804, 1885, 1976, 2053
Offset: 1

Views

Author

Joe K. Crump (joecr(AT)carolina.rr.com)

Keywords

Comments

From Robert G. Wilson v, Mar 20 2017: (Start)
For n odd, the center of the circle is in the middle of the center square and thus a(2n-1) == 1 (mod 4).
For n even, the center of the circle is at the four corners of the center 4 squares and thus a(2n) == 0 (mod 4). (End)

Examples

			a(2)=4 because an inscribed circle in a 2 X 2 grid covers at least 50% of each of the unit squares within it.

Links

Crossrefs

Cf. A124623, A120883, A036704.

Formula

Conjecture: a(n) <= A124623(n) with equality in most cases. - Sean A. Irvine, Sep 03 2021

Extensions

Data corrected by Sean A. Irvine, Sep 02 2021

A053456 Open disk numbers (version 1): a(n) is the number of points (i,j), i,j, integers, contained in an open disk of diameter n, centered at (0,0).

Original entry on oeis.org

0, 1, 1, 9, 9, 21, 25, 37, 45, 69, 69, 97, 109, 137, 145, 177, 193, 225, 249, 293, 305, 349, 373, 421, 437, 489, 517, 577, 609, 665, 697, 749, 793, 861, 889, 973, 1005, 1085, 1125, 1201, 1245, 1313, 1369, 1457, 1513, 1597, 1649, 1741, 1789, 1885, 1941
Offset: 0

Views

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 13 2000

Keywords

Crossrefs

Cf. A053411, A053457, A053458, A053459.
Bisections: A051132, A036704.
Cf. A000796 (Pi).

Formula

a(n)/(n/2)^2->Pi.

A124623 Number of unit squares having center within inscribed circle of an n X n integer square.

Original entry on oeis.org

1, 4, 9, 12, 21, 32, 37, 52, 69, 80, 97, 112, 137, 156, 177, 208, 225, 256, 293, 316, 349, 384, 421, 448, 489, 540, 577, 616, 665, 716, 749, 812, 861, 912, 973, 1020, 1085, 1124, 1201, 1264, 1313, 1396, 1457, 1528, 1597, 1664, 1741, 1804, 1885, 1976, 2053, 2128
Offset: 1

Views

Author

William A. Berry (waberr2(AT)uky.edu), Dec 21 2006

Keywords

Comments

From Robert G. Wilson v, Mar 22 2017: (Start)
For n odd, the center of the circle is in the middle of the center square and thus a(2n-1) == 1 (mod 4).
For n even, the center of the circle is at the four corners of the center 4 squares and thus a(2n) == 0 (mod 4).
a(n) ~ n*Pi/4. (End)

Links

Crossrefs

Cf. A051233.

Programs

  • Mathematica
    f[n_] := 4*Length[ Select[ Flatten[ Table[ If[ OddQ@ n, x^2 + y^2, x(x -1) + y(y -1) + 1/2], {x, n/2}, {y, n/2}]], 4# < n^2 &]] + If[ OddQ@ n, 2(n -1) + 1, 0]; Array[f, 52] (* Robert G. Wilson v, Mar 22 2017 *)

Formula

a(n) = n^2 - 4*k(n); k(n) = number of exterior centers per quadrant.
a(2n-1) = A036704(n-1). - Robert G. Wilson v, Mar 28 2017
a(2n) = 4*A120883(n-1). - Robert G. Wilson v, Mar 28 2017
Showing 1-4 of 4 results.

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