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 1-7 of 7 results.

A345577 Numbers that are the sum of eight fourth powers in two or more ways.

Original entry on oeis.org

263, 278, 293, 308, 323, 343, 358, 373, 388, 423, 438, 453, 503, 518, 533, 548, 563, 583, 598, 613, 628, 678, 693, 758, 773, 788, 803, 853, 868, 887, 902, 917, 932, 933, 967, 982, 997, 1028, 1043, 1047, 1062, 1108, 1127, 1142, 1157, 1172, 1222, 1237, 1283
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			278 is a term because 278 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.

Links

Crossrefs

Cf. A003342, A345532, A345568, A345578, A345586, A345610, A345834.

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, 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])

A345824 Numbers that are the sum of seven fourth powers in exactly two 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 26 2021

Keywords

Comments

Differs from A345568 at term 61.

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. A345568, A345774, A345814, A345823, A345825, A345834, A346279.

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])

A345835 Numbers that are the sum of eight fourth powers in exactly three ways.

Original entry on oeis.org

518, 2678, 2693, 2708, 2738, 2758, 2773, 2838, 2853, 2868, 2883, 2918, 2998, 3078, 3108, 3123, 3253, 3302, 3317, 3363, 3382, 3428, 3477, 3492, 3542, 3622, 3732, 3778, 3797, 3893, 3926, 3953, 3973, 3988, 4018, 4053, 4101, 4118, 4133, 4166, 4193, 4243, 4258
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345578 at term 13 because 2933 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

Examples

			2678 is a term because 2678 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4.

Links

Crossrefs

Cf. A345578, A345785, A345825, A345834, A345836, A345845, A346328.

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, 8):
    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])

A345844 Numbers that are the sum of nine fourth powers in exactly two ways.

Original entry on oeis.org

264, 279, 294, 309, 324, 339, 344, 359, 374, 389, 404, 424, 439, 454, 469, 504, 549, 564, 579, 584, 614, 629, 644, 664, 679, 694, 709, 759, 789, 804, 819, 839, 854, 869, 884, 888, 903, 918, 933, 934, 948, 949, 968, 983, 998, 1013, 1014, 1029, 1044, 1048, 1059
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345586 at term 17 because 519 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

Examples

			279 is a term because 279 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.

Links

Crossrefs

Cf. A345586, A345794, A345834, A345843, A345845, A345854, A346337.

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, 9):
    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])

A345784 Numbers that are the sum of eight cubes in exactly two ways.

Original entry on oeis.org

132, 139, 158, 160, 167, 174, 181, 186, 193, 195, 197, 200, 212, 216, 219, 238, 244, 251, 258, 265, 272, 277, 288, 296, 298, 300, 301, 303, 307, 314, 315, 317, 321, 322, 327, 328, 329, 333, 334, 336, 338, 340, 341, 348, 350, 352, 356, 359, 360, 361, 363, 366
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345532 at term 16 because 223 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3.
Likely finite.

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. A345532, A345774, A345783, A345785, A345794, A345834.

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])

A345833 Numbers that are the sum of eight fourth powers in exactly one ways.

Original entry on oeis.org

8, 23, 38, 53, 68, 83, 88, 98, 103, 113, 118, 128, 133, 148, 163, 168, 178, 183, 193, 198, 213, 228, 243, 248, 258, 328, 338, 353, 368, 403, 408, 418, 433, 468, 483, 488, 498, 568, 578, 593, 608, 632, 643, 647, 648, 658, 662, 663, 673, 677, 692, 707, 708, 712
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A003342 at term 26 because 263 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4.

Examples

			23 is a term because 23 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4.

Links

Crossrefs

Cf. A003342, A345783, A345823, A345834, A345843, A346326.

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, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 1])
    for x in range(len(rets)):
        print(rets[x])

A346327 Numbers that are the sum of eight fifth powers in exactly two ways.

Original entry on oeis.org

4100, 4131, 4162, 4193, 4342, 4373, 4404, 4584, 4615, 4826, 5123, 5154, 5185, 5365, 5396, 5607, 6146, 6177, 6388, 7169, 7224, 7255, 7286, 7466, 7497, 7708, 8247, 8278, 8489, 9270, 10348, 10379, 10590, 11371, 11875, 11906, 11937, 12117, 12148, 12359, 12898
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345610 at term 128 because 52417 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5.

Examples

			4100 is a term because 4100 = 1^5 + 1^5 + 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 = 1^5 + 1^5 + 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5.

Links

Crossrefs

Cf. A345610, A345834, A346279, A346326, A346328, A346337.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**5 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])
Showing 1-7 of 7 results.

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

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