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.

A323622 The first row of the order of square grid cells touched by a circle expanding from the middle of a cell.

Original entry on oeis.org

0, 1, 3, 6, 9, 13, 18, 23, 29, 35, 42, 48, 57, 65, 74, 84, 94, 104, 115, 127, 138, 151, 163, 177, 192, 205, 221, 235, 250, 267, 284, 301, 317, 337, 356, 376, 394, 415, 436, 456, 477, 500, 521, 546, 568, 590, 616, 640, 666, 690, 716, 745, 770, 798, 826, 855, 884, 913, 942, 973, 1003, 1033, 1066
Offset: 0

Views

Author

Rok Cestnik, Jan 20 2019

Keywords

Comments

Related to, but not the same as the case with the circle centered at the corner of a cell, see A232499.

Links

Crossrefs

For the grid read by antidiagonals see A323621.
For the second row of the grid see A323623.
For the diagonal of the grid see A323624.
For the (2,1) diagonal of the grid see A323625.
Cf. A232499.

Programs

  • Python
    N = 12
from math import sqrt
# the distance to the edge of each cell
edges = [[-1 for j in range(N)] for i in range(N)]
edges[0][0] = 0
for i in range(1,N):
    edges[i][0] = i-0.5
    edges[0][i] = i-0.5
for i in range(1,N):
    for j in range(1,N):
        edges[i][j] = sqrt((i-0.5)**2+(j-0.5)**2)
# the values of the distances
values = []
for i in range(N):
    for j in range(N):
        values.append(edges[i][j])
values = list(set(values))
values.sort()
# the cell order
board = [[-1 for j in range(N)] for i in range(N)]
count = 0
for v in values:
    for i in range(N):
        for j in range(N):
            if(edges[i][j] == v):
                board[i][j] = count
    count += 1
# print out the sequence
for i in range(N):
    print(str(board[i][0])+" ", end="")

A323623 The second row of the order of square grid cells touched by a circle expanding from the middle of a cell.

Original entry on oeis.org

1, 2, 4, 7, 10, 14, 19, 24, 30, 36, 43, 49, 58, 66, 75, 85, 95, 105, 116, 128, 139, 152, 164, 178, 193, 206, 222, 236, 251, 268, 285, 302, 318, 338, 357, 377, 395, 416, 437, 457, 478, 501, 522, 547, 569, 591, 617, 641, 667, 691, 717, 746, 771, 799, 827, 856, 885, 914, 943, 974, 1004, 1034, 1067
Offset: 0

Views

Author

Rok Cestnik, Jan 20 2019

Keywords

Comments

Related to, but not the same as the case with the circle centered at the corner of a cell, see A232499.

Links

Crossrefs

For the grid read by antidiagonals see A323621.
For the first row of the grid see A323622.
For the diagonal of the grid see A323624.
For the (2,1) diagonal of the grid see A323625.
Cf. A232499.

Programs

  • Python
    N = 12
from math import sqrt
# the distance to the edge of each cell
edges = [[-1 for j in range(N)] for i in range(N)]
edges[0][0] = 0
for i in range(1,N):
    edges[i][0] = i-0.5
    edges[0][i] = i-0.5
for i in range(1,N):
    for j in range(1,N):
        edges[i][j] = sqrt((i-0.5)**2+(j-0.5)**2)
# the values of the distances
values = []
for i in range(N):
    for j in range(N):
        values.append(edges[i][j])
values = list(set(values))
values.sort()
# the cell order
board = [[-1 for j in range(N)] for i in range(N)]
count = 0
for v in values:
    for i in range(N):
        for j in range(N):
            if(edges[i][j] == v):
                board[i][j] = count
    count += 1
# print out the sequence
for i in range(N):
    print(str(board[i][1])+" ", end="")

A323624 The diagonal of the order of square grid cells touched by a circle expanding from the middle of a cell.

Original entry on oeis.org

0, 2, 5, 10, 16, 22, 32, 40, 50, 62, 73, 88, 101, 118, 134, 152, 170, 189, 210, 230, 253, 275, 299, 325, 351, 381, 406, 435, 465, 495, 527, 561, 593, 628, 663, 699, 737, 775, 813, 853, 895, 935, 981, 1021, 1068, 1113, 1156, 1205, 1253, 1302, 1352, 1401, 1454, 1502, 1557, 1609, 1664, 1723
Offset: 0

Views

Author

Rok Cestnik, Jan 20 2019

Keywords

Comments

Related to, but not the same as the case with the circle centered at the corner of a cell, see A232499.

Links

Crossrefs

For the grid read by antidiagonals see A323621.
For the first row of the grid see A323622.
For the second row of the grid see A323623.
For the (2,1) diagonal of the grid see A323625.
Cf. A232499.

Programs

  • Python
    N = 12
from math import sqrt
# the distance to the edge of each cell
edges = [[-1 for j in range(N)] for i in range(N)]
edges[0][0] = 0
for i in range(1,N):
    edges[i][0] = i-0.5
    edges[0][i] = i-0.5
for i in range(1,N):
    for j in range(1,N):
        edges[i][j] = sqrt((i-0.5)**2+(j-0.5)**2)
# the values of the distances
values = []
for i in range(N):
    for j in range(N):
        values.append(edges[i][j])
values = list(set(values))
values.sort()
# the cell order
board = [[-1 for j in range(N)] for i in range(N)]
count = 0
for v in values:
    for i in range(N):
        for j in range(N):
            if(edges[i][j] == v):
                board[i][j] = count
    count += 1
# print out the sequence
for i in range(N):
    print(str(board[i][i])+",", end="")

A323625 The (2,1) diagonal of the order of square grid cells touched by a circle expanding from the middle of a cell.

Original entry on oeis.org

0, 4, 11, 21, 34, 49, 68, 90, 112, 137, 167, 198, 230, 265, 305, 345, 388, 432, 480, 529, 583, 635, 692, 752, 812, 876, 941, 1010, 1079, 1151, 1225, 1305, 1383, 1462, 1545, 1630, 1720, 1811, 1901, 1995, 2092, 2190, 2287, 2391, 2499, 2606, 2715, 2827, 2941, 3056, 3174, 3295, 3421, 3541, 3668, 3792, 3923, 4058, 4193, 4333, 4466, 4609, 4754, 4899, 5042, 5194, 5344, 5498, 5654, 5813, 5972, 6133
Offset: 0

Views

Author

Rok Cestnik, Jan 20 2019

Keywords

Comments

Related to, but not the same as the case with the circle centered at the corner of a cell, see A232499.

Links

Crossrefs

For the grid read by antidiagonals see A323621.
For the first row of the grid see A323622.
For the second row of the grid see A323623.
For the (1,1) diagonal of the grid see A323624.
Cf. A232499.

Programs

  • Python
    N = 24
from math import sqrt
# the distance to the edge of each cell
edges = [[-1 for j in range(N)] for i in range(N)]
edges[0][0] = 0
for i in range(1,N):
    edges[i][0] = i-0.5
    edges[0][i] = i-0.5
for i in range(1,N):
    for j in range(1,N):
        edges[i][j] = sqrt((i-0.5)**2+(j-0.5)**2)
# the values of the distances
values = []
for i in range(N):
    for j in range(N):
        values.append(edges[i][j])
values = list(set(values))
values.sort()
# the cell order
board = [[-1 for j in range(N)] for i in range(N)]
count = 0
for v in values:
    for i in range(N):
        for j in range(N):
            if(edges[i][j] == v):
                board[i][j] = count
    count += 1
# print out the sequence
for i in range(int(round(N/2))):
    print(str(board[2*i][i])+" ", end="")
Showing 1-4 of 4 results.

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