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 1-6 of 6 results.

A345521 Numbers that are the sum of seven cubes in three or more ways.

Original entry on oeis.org

222, 229, 248, 255, 262, 281, 283, 285, 318, 346, 370, 374, 377, 379, 381, 396, 400, 407, 412, 419, 426, 433, 437, 438, 444, 451, 463, 470, 472, 475, 477, 489, 494, 496, 501, 503, 505, 507, 510, 522, 529, 533, 536, 559, 564, 566, 568, 570, 577, 578, 584, 585
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			229 is a term because 229 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.

Links

Crossrefs

Cf. A345480, A345512, A345520, A345522, A345533, A345569, A345775.

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, 7):
    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])

A345568 Numbers that are the sum of seven fourth powers in two or more ways.

Original entry on oeis.org

262, 277, 292, 307, 342, 357, 372, 422, 437, 502, 517, 532, 547, 597, 612, 677, 772, 787, 852, 886, 901, 916, 966, 981, 1027, 1046, 1141, 1156, 1221, 1362, 1377, 1396, 1442, 1510, 1525, 1557, 1572, 1587, 1590, 1617, 1637, 1652, 1717, 1765, 1812, 1827, 1892
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			277 is a term because 277 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.

Links

Crossrefs

Cf. A003341, A345520, A345559, A345569, A345577, A345605, A345824.

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, 7):
    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])

A345511 Numbers that are the sum of six cubes in two or more ways.

Original entry on oeis.org

158, 165, 184, 221, 228, 235, 247, 254, 256, 261, 268, 273, 275, 280, 282, 284, 287, 291, 294, 306, 310, 313, 317, 324, 331, 332, 343, 345, 347, 350, 352, 362, 369, 371, 373, 376, 378, 380, 385, 387, 388, 392, 395, 399, 404, 406, 408, 411, 418, 425, 430, 432
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Links

Crossrefs

Cf. A003329, A048930, A343702, A344806, A345512, A345520, A345559.

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, 6):
    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])

A345532 Numbers that are the sum of eight cubes in two or more ways.

Original entry on oeis.org

132, 139, 158, 160, 167, 174, 181, 186, 193, 195, 197, 200, 212, 216, 219, 223, 230, 237, 238, 244, 249, 251, 256, 258, 263, 265, 270, 272, 275, 277, 282, 284, 286, 288, 289, 291, 293, 296, 298, 300, 301, 303, 307, 308, 310, 312, 314, 315, 317, 319, 321, 322
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			139 is a term because 139 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3.

Links

Crossrefs

Cf. A003331, A345489, A345520, A345533, A345541, A345577, A345784.

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, 8):
    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])

A345774 Numbers that are the sum of seven cubes in exactly two ways.

Original entry on oeis.org

131, 159, 166, 173, 185, 192, 211, 236, 243, 257, 264, 269, 274, 276, 288, 290, 292, 295, 299, 300, 302, 307, 309, 311, 314, 320, 321, 325, 332, 333, 337, 339, 340, 344, 348, 351, 353, 355, 358, 359, 360, 363, 372, 384, 385, 386, 388, 389, 393, 395, 398, 403
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345520 at term 8 because 222 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3.
Likely finite.

Examples

			159 is a term because 159 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3.

Links

Crossrefs

Cf. A048930, A345520, A345773, A345775, A345784, A345824.

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, 7):
    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])

A345479 Numbers that are the sum of seven squares in two or more ways.

Original entry on oeis.org

22, 25, 28, 30, 31, 33, 34, 37, 38, 39, 40, 42, 43, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			25 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 4^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2
so 25 is a term.

Links

Crossrefs

Cf. A344806, A345478, A345480, A345489, A345520.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 for x in range(1, 1000)]
for pos in cwr(power_terms, 7):
    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])
Showing 1-6 of 6 results.

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