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 1-3 of 3 results.

A136513 Number of unit square lattice cells inside half-plane (two adjacent quadrants) of origin centered circle of diameter n.

Original entry on oeis.org

0, 0, 2, 2, 6, 8, 12, 16, 26, 30, 38, 44, 56, 60, 74, 82, 96, 108, 128, 138, 154, 166, 188, 196, 220, 238, 262, 278, 304, 324, 344, 366, 398, 416, 452, 468, 506, 526, 562, 588, 616, 644, 686, 714, 754, 780, 824, 848, 894, 930, 976, 1008, 1056, 1090, 1134, 1170
Offset: 1

Views

Author

Glenn C. Foster (gfoster(AT)uiuc.edu), Jan 02 2008

Keywords

Examples

			a(3) = 2 because a circle centered at the origin and of radius 3/2 encloses (-1,1) and (1,1) in the upper half plane.

Links

Crossrefs

Alternating merge of A136514 and A136515.
Cf. A136483, A136485.

Programs

  • Magma
    A136513:= func< n | n eq 1 select 0 else 2*(&+[Floor(Sqrt((n/2)^2-j^2)): j in [1..Floor(n/2)]]) >;
[A136513(n): n in [1..100]]; // G. C. Greubel, Jul 27 2023
  • Mathematica
    Table[2*Sum[Floor[Sqrt[(n/2)^2 -k^2]], {k,Floor[n/2]}], {n,100}]
  • PARI
    a(n) = 2*sum(k=1, n\2, sqrtint((n/2)^2-k^2)); \\ Michel Marcus, Jul 27 2023
  • SageMath
    def A136513(n): return 2*sum(isqrt((n/2)^2-k^2) for k in range(1,(n//2)+1))
[A136513(n) for n in range(1,101)] # G. C. Greubel, Jul 27 2023

Formula

Lim_{n -> oo} a(n)/(n^2) -> Pi/8.
a(n) = 2 * Sum_{k=1..floor(n/2)} floor(sqrt((n/2)^2 - k^2)).
a(n) = 2 * A136483(n).
a(n) = (1/2) * A136485(n).
a(n) = [x^(n^2)] (theta_3(x^4) - 1)^2 / (2 * (1 - x)). - Ilya Gutkovskiy, Nov 24 2021

A136484 Number of unit square lattice cells inside quadrant of origin centered circle of diameter 2n+1.

Original entry on oeis.org

0, 1, 3, 6, 13, 19, 28, 37, 48, 64, 77, 94, 110, 131, 152, 172, 199, 226, 253, 281, 308, 343, 377, 412, 447, 488, 528, 567, 612, 654, 703, 750, 796, 847, 902, 957, 1013, 1068, 1129, 1187, 1252, 1313, 1378, 1446, 1511, 1582, 1650, 1725, 1800, 1877, 1955, 2034
Offset: 0

Views

Author

Glenn C. Foster (gfoster(AT)uiuc.edu), Jan 02 2008

Keywords

Comments

Number of unit square lattice cells inside quadrant of origin centered circle of radius n+1/2.

Examples

			a(2) = 3 because a circle of radius 2+1/2 in the first quadrant encloses (2,1), (1,1), (1,2).

Links

Crossrefs

Cf. A019683, A136483, A136486, A136515.

Programs

  • Magma
    A136484:= func< n | n eq 0 select 0 else (&+[Floor(Sqrt((n+1/2)^2-j^2)): j in [1..n]]) >;
[A136484(n): n in [0..100]]; // G. C. Greubel, Jul 29 2023
  • Mathematica
    Table[Sum[Floor[Sqrt[(n+1/2)^2 - k^2]], {k,n}], {n,0,100}]
  • SageMath
    def A136484(n): return sum(floor(sqrt((n+1/2)^2-k^2)) for k in range(1, n+1))
[A136484(n) for n in range(101)] # G. C. Greubel, Jul 29 2023

Formula

a(n) = Sum_{k=1..n} floor(sqrt((n+1/2)^2 - k^2)).
a(n) = (1/2) * A136515(n).
a(n) = (1/4) * A136486(n).
a(n) = A136483(2*n+1).
Lim_{n -> oo} a(n)/(n^2) -> Pi/16 (A019683).

A136486 Number of unit square lattice cells enclosed by origin centered circle of diameter 2n+1.

Original entry on oeis.org

0, 4, 12, 24, 52, 76, 112, 148, 192, 256, 308, 376, 440, 524, 608, 688, 796, 904, 1012, 1124, 1232, 1372, 1508, 1648, 1788, 1952, 2112, 2268, 2448, 2616, 2812, 3000, 3184, 3388, 3608, 3828, 4052, 4272, 4516, 4748, 5008, 5252, 5512, 5784, 6044, 6328, 6600
Offset: 0

Views

Author

Glenn C. Foster (gfoster(AT)uiuc.edu), Jan 02 2008

Keywords

Comments

a(n) is the number of complete squares that fit inside the circle with radius n+1/2, drawn on squared paper.

Examples

			a(1) = 4 because a circle centered at the origin and of radius 1+1/2 encloses (-1,-1), (-1,1), (1,-1), (1,1).

Links

Crossrefs

Cf. A003881, A136484, A136485, A136515.

Programs

  • Magma
    A136486:= func< n | n eq 0 select 0 else 4*(&+[Floor(Sqrt((n+1/2)^2-j^2)): j in [1..n]]) >;
[A136486(n): n in [0..100]]; // G. C. Greubel, Jul 30 2023
  • Mathematica
    Table[4*Sum[Floor[Sqrt[(n + 1/2)^2 - k^2]], {k,n}], {n, 0, 100}]
  • SageMath
    def A136486(n): return 4*sum(floor(sqrt((n+1/2)^2-k^2)) for k in range(1, n+1))
[A136486(n) for n in range(101)] # G. C. Greubel, Jul 30 2023

Formula

a(n) = 4*Sum_{k=1..n} floor(sqrt((n+1/2)^2 - k^2)).
a(n) = 4 * A136484(n).
a(n) = 2 * A136515(n).
a(n) = A136485(2*n+1).
Lim_{n -> oo} a(n)/(n^2) -> Pi/4 (A003881).
Showing 1-3 of 3 results.

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