A295344 Maximum number of lattice points inside and on a circle of radius n.

1, 5, 14, 32, 52, 81, 116, 157, 208, 258, 319, 384, 457, 540, 623, 716, 812, 914, 1025, 1142, 1268, 1396, 1528, 1669, 1816, 1976, 2131, 2300, 2472, 2650, 2836, 3028, 3228, 3436, 3644, 3859, 4080, 4314, 4548, 4792, 5038, 5289, 5555, 5818, 6092, 6376, 6668, 6952

Offset: 0

Author

Arkadiusz Wesolowski, Nov 20 2017

Keywords

• nonn

Comments

Maximum number of lattice points (i.e., points with integer coordinates) in the plane that can be covered by a circle of radius n.

a(n) >= A000328(n).

Conjecture: sequence contains infinitely many terms that are divisible by 4.

Examples

			For a circle centered at the point (x, y) = (1/2, 1/4) with radius 2, there are 14 lattice points inside and on the circle.
.
.     Center             # Pts in/
.    x      y    Radius  on circle
.  -----  -----  ------  ---------
.    0      0       1         5
.   1/2    1/4      2        14
.   1/2    1/2      3        32
.   1/2    1/2      4        52
.    0      0       5        81
.   1/2    1/3      6       116
.   2/5    1/5      7       157
.   1/2    1/2      8       208
.   1/2    2/9      9       258
.  20/47  19/56    10       319
.   1/2    1/2     11       384
.  11/23   7/20    12       457
.   1/2    1/2     13       540
.  10/21   3/13    14       623
.   1/2    1/2     15       716
.   1/2    1/2     16       812
.   2/5    2/5     17       914
.   3/8    5/14    18      1025
.   1/2    1/6     19      1142
.   9/19   8/17    20      1268
		

References

• B. R. Srinivasan, Lattice Points in a Circle, Proc. Nat. Inst. Sci. India, Part A, 29 (1963), pp. 332-346.

Crossrefs

Cf. A000328, A123690, A291259.

Programs

PARI

L=List([]); for(n=0, 47, if(n>0, j=5, j=1); g=0; h=0; f=ceil(Pi*n^2); for(d=2, floor(f/2), for(c=1, floor(d/2), if(gcd(c, d)==1, for(e=d, d+1, if(e/f<=1/2, a=c/d; b=e/f; if(a+b>=1/2, t=0; for(x=-n, n+1, for(y=-n, n+1, z=(a-x)^2+(b-y)^2; if(z<=n^2,t++))); if(t>j, j=t; if(a>=b, g=a; h=b, g=b; h=a)))))))); print("a("n") = "j", the center of the circle is at point ("g", "h")."); listput(L, j)); print(); print(Vec(L));

Formula

a(n) = Pi*n^2 + O(n), as n goes to infinity.

a(n) = A123690(2*n) for n >= 1.

Extensions

a(10) corrected by Giovanni Resta, Nov 24 2017