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.

A345586 Numbers that are the sum of nine fourth powers in two or more ways.

Original entry on oeis.org

264, 279, 294, 309, 324, 339, 344, 359, 374, 389, 404, 424, 439, 454, 469, 504, 519, 534, 549, 564, 579, 584, 599, 614, 629, 644, 664, 679, 694, 709, 759, 774, 789, 804, 819, 839, 854, 869, 884, 888, 903, 918, 933, 934, 948, 949, 968, 983, 998, 1013, 1014
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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.
		

Crossrefs

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])

A345834 Numbers that are the sum of eight fourth powers in exactly two ways.

Original entry on oeis.org

263, 278, 293, 308, 323, 343, 358, 373, 388, 423, 438, 453, 503, 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, 1302
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345577 at term 14 because 518 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

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.
		

Crossrefs

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])

A345845 Numbers that are the sum of nine fourth powers in exactly three ways.

Original entry on oeis.org

519, 534, 599, 774, 1143, 1364, 1539, 1604, 1619, 1814, 2579, 2644, 2659, 2679, 2694, 2709, 2724, 2739, 2754, 2759, 2774, 2789, 2819, 2834, 2839, 2869, 2884, 2899, 2994, 2999, 3079, 3109, 3124, 3139, 3303, 3318, 3333, 3334, 3363, 3364, 3379, 3383, 3398, 3463
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345587 at term 26 because 285.

Examples

			534 is a term because 534 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.
		

Crossrefs

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 == 3])
        for x in range(len(rets)):
            print(rets[x])

A345794 Numbers that are the sum of nine cubes in exactly two ways.

Original entry on oeis.org

72, 133, 140, 147, 159, 161, 166, 168, 175, 182, 185, 187, 189, 194, 196, 198, 201, 203, 205, 208, 213, 217, 220, 222, 227, 239, 243, 246, 252, 261, 265, 266, 273, 289, 296, 304, 306, 308, 323, 325, 328, 329, 330, 336, 342, 344, 349, 351, 352, 354, 356, 358
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345541 at term 25 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

			133 is a term because 133 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3.
		

Crossrefs

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 == 2])
        for x in range(len(rets)):
            print(rets[x])

A345843 Numbers that are the sum of nine fourth powers in exactly one ways.

Original entry on oeis.org

9, 24, 39, 54, 69, 84, 89, 99, 104, 114, 119, 129, 134, 144, 149, 164, 169, 179, 184, 194, 199, 209, 214, 229, 244, 249, 259, 274, 329, 354, 369, 384, 409, 419, 434, 449, 484, 489, 499, 514, 569, 594, 609, 624, 633, 648, 649, 659, 663, 674, 678, 689, 693, 708
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A003343 at term 28 because 264 = 1^4 + 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 + 1^4 + 4^4.

Examples

			24 is a term because 24 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4.
		

Crossrefs

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 == 1])
        for x in range(len(rets)):
            print(rets[x])

A345854 Numbers that are the sum of ten fourth powers in exactly two ways.

Original entry on oeis.org

265, 280, 295, 310, 325, 340, 345, 355, 360, 375, 390, 405, 420, 425, 440, 455, 470, 485, 505, 565, 580, 585, 595, 630, 645, 660, 665, 695, 710, 725, 745, 760, 805, 820, 835, 840, 870, 885, 889, 900, 904, 919, 920, 934, 935, 949, 950, 964, 965, 969, 984, 999
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345595 at term 20 because 520 = 1^4 + 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 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

Examples

			280 is a term because 280 = 1^4 + 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 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.
		

Crossrefs

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 == 2])
        for x in range(len(rets)):
            print(rets[x])

A346337 Numbers that are the sum of nine fifth powers in exactly two ways.

Original entry on oeis.org

4101, 4132, 4163, 4194, 4225, 4343, 4374, 4405, 4436, 4585, 4616, 4647, 4827, 4858, 5069, 5124, 5155, 5186, 5217, 5366, 5397, 5428, 5608, 5639, 5850, 6147, 6178, 6209, 6389, 6420, 6631, 7170, 7201, 7225, 7256, 7287, 7318, 7412, 7467, 7498, 7529, 7709, 7740
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345619 at term 306 because 52418 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5.

Examples

			4101 is a term because 4101 = 1^5 + 1^5 + 1^5 + 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5.
		

Crossrefs

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, 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])
Showing 1-7 of 7 results.