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.

A003340 Numbers that are the sum of 6 positive 4th powers.

Original entry on oeis.org

6, 21, 36, 51, 66, 81, 86, 96, 101, 116, 131, 146, 161, 166, 181, 196, 211, 226, 246, 261, 276, 291, 306, 321, 326, 336, 341, 356, 371, 386, 401, 406, 421, 436, 451, 466, 486, 501, 516, 531, 546, 561, 576, 581, 596, 611, 626, 630, 641, 645, 660, 661, 675, 676, 690
Offset: 1

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Author

Keywords

Examples

			From _David A. Corneth_, Aug 04 2020: (Start)
13090 is in the sequence as 13090 = 4^4 + 4^4 + 5^4 + 6^4 + 8^4 + 9^4.
17539 is in the sequence as 17539 = 2^4 + 3^4 + 4^4 + 5^4 + 9^4 + 10^4.
23732 is in the sequence as 23732 = 3^4 + 5^4 + 5^4 + 7^4 + 10^4 + 10^4. (End)
		

Crossrefs

Programs

  • Mathematica
    Select[Range[1000], AnyTrue[PowersRepresentations[#, 6, 4], First[#]>0&]&] (* Jean-François Alcover, Jul 18 2017 *)
  • Python
    from itertools import combinations_with_replacement as combs_with_rep
    def aupto(limit):
        qd = [k**4 for k in range(1, int(limit**.25)+2) if k**4 + 5 <= limit]
        ss = set(sum(c) for c in combs_with_rep(qd, 6))
        return sorted(s for s in ss if s <= limit)
    print(aupto(700)) # Michael S. Branicky, Jun 21 2021

A344238 Numbers that are the sum of five fourth powers in two or more ways.

Original entry on oeis.org

260, 275, 340, 515, 884, 1555, 2595, 2660, 2675, 2690, 2705, 2755, 2770, 2835, 2930, 2945, 3010, 3185, 3299, 3314, 3379, 3554, 3923, 3970, 3985, 4050, 4115, 4145, 4160, 4210, 4225, 4290, 4355, 4400, 4465, 4594, 4769, 4834, 5075, 5090, 5155, 5265, 5330, 5395, 5440, 5505, 5570, 5699, 6370, 6545, 6580, 6595, 6610
Offset: 1

Views

Author

David Consiglio, Jr., May 12 2021

Keywords

Examples

			340 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4
    = 2^4 + 3^4 + 3^4 + 3^4 + 3^4
so 340 is a term of this sequence.
		

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, 50)]
    for pos in cwr(power_terms, 5):
        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])

A345560 Numbers that are the sum of six fourth powers in three or more ways.

Original entry on oeis.org

2676, 2851, 2916, 4131, 4226, 4241, 4306, 4371, 4481, 4850, 5346, 5411, 5521, 5586, 5651, 6561, 6611, 6626, 6691, 6756, 6771, 6801, 6821, 6836, 6851, 6866, 6931, 7106, 7235, 7475, 7491, 7666, 7841, 7906, 7971, 8146, 8211, 8321, 8386, 8451, 8531, 8706, 9011
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			2851 is a term because 2851 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.
		

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, 6):
        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.
		

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, 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])

A345814 Numbers that are the sum of six fourth powers in exactly two ways.

Original entry on oeis.org

261, 276, 291, 341, 356, 421, 516, 531, 596, 771, 885, 900, 965, 1140, 1361, 1509, 1556, 1571, 1636, 1811, 2180, 2596, 2611, 2661, 2691, 2706, 2721, 2741, 2756, 2771, 2786, 2836, 2931, 2946, 2961, 3011, 3026, 3091, 3186, 3201, 3220, 3266, 3285, 3300, 3315
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345559 at term 25 because 2676 = 1^4 + 1^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4.

Examples

			276 is a term because 276 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 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, 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])

A345507 Numbers that are the sum of six fifth powers in two or more ways.

Original entry on oeis.org

4098, 4129, 4340, 5121, 7222, 11873, 20904, 36865, 51447, 51478, 51509, 51689, 51720, 51931, 52470, 52501, 52712, 53493, 54571, 54602, 54813, 55594, 57695, 59222, 59253, 59464, 60245, 62346, 63146, 66997, 67586, 68253, 68284, 68495, 68906, 68937, 69148, 69276
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			4129 is a term because 4129 = 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 4^5 = 2^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5.
		

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, 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])

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

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, 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])
Showing 1-7 of 7 results.