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 1-5 of 5 results.

A345551 Numbers that are the sum of ten cubes in three or more ways.

Original entry on oeis.org

197, 225, 232, 239, 246, 251, 253, 258, 260, 265, 267, 272, 277, 279, 281, 284, 286, 288, 291, 293, 295, 298, 300, 302, 303, 305, 307, 309, 310, 312, 314, 316, 317, 319, 321, 323, 324, 326, 328, 329, 330, 335, 336, 338, 340, 342, 343, 344, 345, 347, 349, 351
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			225 is a term because 225 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

Links

Crossrefs

Cf. A345510, A345542, A345550, A345552, A345596, A345805.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 3])
    for x in range(len(rets)):
        print(rets[x])

A345795 Numbers that are the sum of nine cubes in exactly three ways.

Original entry on oeis.org

231, 238, 245, 250, 259, 271, 276, 278, 280, 285, 287, 290, 292, 294, 297, 299, 301, 302, 309, 311, 313, 315, 316, 318, 322, 327, 334, 335, 337, 339, 341, 346, 350, 353, 357, 362, 365, 379, 386, 387, 388, 391, 393, 394, 395, 397, 398, 405, 412, 418, 420, 421
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345542 at term 1 because 224 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.
Likely finite.

Examples

			231 is a term because 231 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.

Links

Crossrefs

Cf. A345542, A345785, A345794, A345796, A345805, A345845.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 9):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 3])
    for x in range(len(rets)):
        print(rets[x])

A345855 Numbers that are the sum of ten fourth powers in exactly three ways.

Original entry on oeis.org

520, 535, 550, 600, 615, 680, 775, 790, 855, 1030, 1144, 1159, 1224, 1365, 1380, 1399, 1445, 1540, 1555, 1605, 1635, 1685, 1700, 1768, 1795, 1815, 1830, 1860, 1875, 1895, 1989, 2070, 2164, 2229, 2244, 2439, 2485, 2580, 2595, 2645, 2675, 2680, 2695, 2710, 2755
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345596 at term 21 because 1620 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4.

Examples

			535 is a term because 535 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

Links

Crossrefs

Cf. A345596, A345805, A345845, A345854, A345856, A346348.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1, 1000)]
for pos in cwr(power_terms, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 3])
    for x in range(len(rets)):
        print(rets[x])

A345804 Numbers that are the sum of ten cubes in exactly two ways.

Original entry on oeis.org

73, 80, 99, 134, 136, 141, 148, 155, 160, 162, 167, 169, 174, 176, 183, 186, 188, 190, 192, 193, 195, 199, 202, 204, 206, 209, 211, 212, 213, 214, 216, 218, 221, 223, 228, 230, 235, 240, 244, 247, 249, 254, 262, 266, 269, 270, 273, 274, 290, 292, 297, 304
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345550 at term 22 because 197 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3.
Likely finite.

Examples

			80 is a term because 80 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3.

Links

Crossrefs

Cf. A345550, A345794, A345803, A345805, A345854.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 2])
    for x in range(len(rets)):
        print(rets[x])

A345806 Numbers that are the sum of ten cubes in exactly four ways.

Original entry on oeis.org

225, 232, 251, 258, 265, 272, 284, 286, 291, 307, 310, 314, 321, 323, 328, 342, 347, 356, 363, 366, 373, 375, 377, 380, 389, 391, 398, 399, 405, 412, 419, 422, 424, 427, 434, 438, 441, 445, 450, 451, 458, 459, 461, 464, 469, 471, 476, 478, 481, 484, 488, 489
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345552 at term 9 because 288 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3.
Likely finite.

Examples

			232 is a term because 232 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

Links

Crossrefs

Cf. A345552, A345796, A345805, A345807, A345856.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 4])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-5 of 5 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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