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.

A003341 Numbers that are the sum of 7 positive 4th powers.

Original entry on oeis.org

7, 22, 37, 52, 67, 82, 87, 97, 102, 112, 117, 132, 147, 162, 167, 177, 182, 197, 212, 227, 242, 247, 262, 277, 292, 307, 322, 327, 337, 342, 352, 357, 372, 387, 402, 407, 417, 422, 437, 452, 467, 482, 487, 502, 517, 532, 547, 562, 567, 577, 582, 592, 597, 612, 627
Offset: 1

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Author

Keywords

Examples

			From _David A. Corneth_, Aug 04 2020: (Start)
5971 is in the sequence as 5971 = 3^4 + 3^4 + 5^4 + 6^4 + 6^4 + 6^4 + 6^4.
12022 is in the sequence as 12022 = 1^4 + 2^4 + 7^4 + 7^4 + 7^4 + 7^4 + 7^4.
16902 is in the sequence as 16902 = 1^4 + 1^4 + 3^4 + 6^4 + 7^4 + 9^4 + 9^4. (End)
		

Crossrefs

Programs

  • Maple
    N:= 1000:
    S1:= {seq(i^4,i=1..floor(N^(1/4)))}:
    S2:= select(`<=`,{seq(seq(i+j,i=S1),j=S1)},N):
    S4:= select(`<=`,{seq(seq(i+j,i=S2),j=S2)},N):
    S6:= select(`<=`,{seq(seq(i+j,i=S2),j=S4)},N):
    sort(convert(select(`<=`,{seq(seq(i+j,i=S1),j=S6)},N),list)); # Robert Israel, Jul 21 2019
  • Python
    from itertools import combinations_with_replacement as mc
    def aupto(limit):
        qd = [k**4 for k in range(1, int(limit**.25)+2) if k**4 + 6 <= limit]
        ss = set(sum(c) for c in mc(qd, 7))
        return sorted(s for s in ss if s <= limit)
    print(aupto(630)) # Michael S. Branicky, Jul 22 2021

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

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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.
		

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])

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.
		

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

A345813 Numbers that are the sum of six fourth powers in exactly one ways.

Original entry on oeis.org

6, 21, 36, 51, 66, 81, 86, 96, 101, 116, 131, 146, 161, 166, 181, 196, 211, 226, 246, 306, 321, 326, 336, 371, 386, 401, 406, 436, 451, 466, 486, 501, 546, 561, 576, 581, 611, 626, 630, 641, 645, 660, 661, 675, 676, 690, 691, 705, 706, 710, 725, 740, 755, 756
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A003340 at term 20 because 261 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 = 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4.

Examples

			21 is a term because 21 = 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, 6):
        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])

A346278 Numbers that are the sum of seven fifth powers in exactly one way.

Original entry on oeis.org

7, 38, 69, 100, 131, 162, 193, 224, 249, 280, 311, 342, 373, 404, 435, 491, 522, 553, 584, 615, 646, 733, 764, 795, 826, 857, 975, 1006, 1030, 1037, 1061, 1068, 1092, 1123, 1154, 1185, 1216, 1217, 1248, 1272, 1279, 1303, 1334, 1365, 1396, 1427, 1459, 1490
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A003352 at term 123 because 4099 = 1^5 + 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 = 1^5 + 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5.

Examples

			7 is a term because 7 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5.
		

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, 7):
        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])

A345773 Numbers that are the sum of seven cubes in exactly one way.

Original entry on oeis.org

7, 14, 21, 28, 33, 35, 40, 42, 47, 49, 54, 56, 59, 61, 66, 68, 70, 73, 75, 77, 80, 84, 85, 87, 91, 92, 94, 96, 98, 99, 103, 105, 106, 110, 111, 112, 113, 117, 118, 122, 124, 125, 129, 132, 133, 136, 137, 138, 140, 143, 144, 145, 147, 148, 150, 151, 152, 154
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A003330 at term 44 because 131 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.
Likely finite.

Examples

			14 is a term because 14 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3.
		

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, 7):
        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])
Showing 1-6 of 6 results.