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.

A343989 Numbers that are the sum of five positive cubes in five or more ways.

Original entry on oeis.org

1765, 1980, 2043, 2104, 2195, 2250, 2430, 2449, 2486, 2491, 2493, 2547, 2584, 2592, 2738, 2745, 2764, 2817, 2888, 2915, 2953, 2969, 2979, 3095, 3096, 3133, 3142, 3186, 3188, 3214, 3240, 3249, 3275, 3277, 3310, 3312, 3366, 3403, 3422, 3459, 3464, 3466, 3483, 3492, 3520, 3529, 3583, 3608, 3627, 3653, 3664, 3671
Offset: 1

Views

Author

David Consiglio, Jr., May 06 2021

Keywords

Examples

			2043 = 1^3 + 4^3 + 5^3 +  5^3 + 12^3
     = 2^3 + 2^3 + 3^3 + 10^3 + 10^3
     = 2^3 + 3^3 + 4^3 +  6^3 + 12^3
     = 4^3 + 5^3 + 5^3 +  9^3 + 10^3
     = 4^3 + 6^3 + 6^3 +  6^3 + 11^3
so 2043 is a term.

Links

Crossrefs

Cf. A343987, A343988, A344034, A344358, A344798, A345174, A345514.

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(1,50)]
for pos in cwr(power_terms,5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k,v in keep.items() if v >= 5])
for x in range(len(rets)):
    print(rets[x])

A344940 Numbers that are the sum of five fourth powers in six or more ways.

Original entry on oeis.org

151300, 197779, 211059, 217154, 225890, 236194, 236675, 243235, 246674, 250834, 286114, 288579, 300835, 302130, 302210, 303235, 309059, 317795, 320195, 334819, 334899, 335443, 336210, 338914, 346835, 356899, 363379, 366995, 373234, 375619, 389875, 391154
Offset: 1

Views

Author

David Consiglio, Jr., Jun 03 2021

Keywords

Examples

			151300 is a term because 151300 = 3^4 + 3^4 + 3^4 + 12^4 + 19^4  = 3^4 + 11^4 + 11^4 + 14^4 + 17^4  = 3^4 + 13^4 + 13^4 + 13^4 + 16^4  = 6^4 + 9^4 + 9^4 + 9^4 + 19^4  = 7^4 + 11^4 + 11^4 + 11^4 + 18^4  = 8^4 + 9^4 + 13^4 + 13^4 + 17^4.

Links

Crossrefs

Cf. A344358, A344904, A344941, A344942, A345174, A345563, A345864.

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, 5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 6])
for x in range(len(rets)):
    print(rets[x])

A345148 Numbers that are the sum of four third powers in six or more ways.

Original entry on oeis.org

6883, 12411, 13104, 13923, 14112, 14581, 14896, 14904, 15561, 15876, 16317, 16640, 17208, 17479, 17992, 18739, 18865, 18928, 19035, 19080, 19376, 19665, 19712, 19763, 19880, 20007, 20384, 20755, 20979, 21203, 21231, 21420, 21707, 21896, 22409, 22617, 22743
Offset: 1

Views

Author

David Consiglio, Jr., Jun 09 2021

Keywords

Examples

			6883 is a term because 6883 = 2^3 + 2^3 + 2^3 + 18^3  = 2^3 + 4^3 + 14^3 + 14^3  = 3^3 + 7^3 + 7^3 + 17^3  = 3^3 + 10^3 + 13^3 + 13^3  = 4^3 + 10^3 + 10^3 + 15^3  = 7^3 + 8^3 + 8^3 + 16^3.

Links

Crossrefs

Cf. A025371, A343987, A344904, A345083, A345149, A345150, A345174.

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, 4):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 6])
for x in range(len(rets)):
    print(rets[x])

A345180 Numbers that are the sum of five third powers in seven or more ways.

Original entry on oeis.org

4392, 4472, 4544, 4600, 4915, 4957, 5076, 5113, 5120, 5132, 5139, 5165, 5174, 5256, 5321, 5347, 5354, 5384, 5391, 5410, 5445, 5474, 5481, 5507, 5543, 5617, 5624, 5643, 5678, 5715, 5741, 5760, 5769, 5797, 5832, 5834, 5860, 5895, 5914, 5923, 5984, 5986, 6049
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Examples

			4472 is a term because 4472 = 1^3 + 4^3 + 4^3 + 4^3 + 15^3  = 2^3 + 2^3 + 9^3 + 11^3 + 11^3  = 2^3 + 3^3 + 4^3 + 5^3 + 15^3  = 2^3 + 3^3 + 7^3 + 11^3 + 12^3  = 3^3 + 3^3 + 6^3 + 10^3 + 13^3  = 3^3 + 4^3 + 5^3 + 8^3 + 14^3  = 5^3 + 5^3 + 7^3 + 10^3 + 12^3.

Links

Crossrefs

Cf. A344800, A344942, A345150, A345174, A345181, A345183, A345516.

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, 5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 7])
for x in range(len(rets)):
    print(rets[x])

A345515 Numbers that are the sum of six cubes in six or more ways.

Original entry on oeis.org

1377, 1488, 1586, 1595, 1647, 1673, 1677, 1710, 1738, 1764, 1766, 1773, 1799, 1829, 1836, 1837, 1862, 1881, 1890, 1911, 1953, 1955, 1981, 1988, 2007, 2011, 2014, 2018, 2025, 2044, 2051, 2070, 2079, 2097, 2105, 2107, 2108, 2142, 2153, 2160, 2168, 2170, 2177
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1488 is a term because 1488 = 1^3 + 1^3 + 1^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3.

Links

Crossrefs

Cf. A344810, A345174, A345514, A345516, A345524, A345563, A345768.

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 >= 6])
    for x in range(len(rets)):
        print(rets[x])

A345175 Numbers that are the sum of five third powers in exactly six ways.

Original entry on oeis.org

2430, 2979, 3214, 3249, 3312, 3492, 3520, 3737, 3753, 3788, 3816, 3842, 3942, 3968, 4121, 4185, 4213, 4267, 4355, 4411, 4418, 4446, 4453, 4456, 4465, 4482, 4509, 4563, 4626, 4663, 4670, 4723, 4753, 4896, 4905, 4924, 4938, 4941, 4950, 4960, 4976, 4987, 4994
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Comments

Differs from A345174 at term 20 because 4392 = 1^3 + 1^3 + 10^3 + 10^3 + 11^3 = 1^3 + 2^3 + 2^3 + 9^3 + 14^3 = 1^3 + 8^3 + 9^3 + 10^3 + 10^3 = 2^3 + 2^3 + 3^3 + 5^3 + 15^3 = 2^3 + 3^3 + 5^3 + 8^3 + 14^3 = 2^3 + 8^3 + 8^3 + 8^3 + 12^3 = 3^3 + 6^3 + 7^3 + 8^3 + 13^3 = 5^3 + 5^3 + 5^3 + 9^3 + 13^3.

Examples

			2430 is a term because 2430 = 1^3 + 2^3 + 2^3 + 5^3 + 12^3  = 1^3 + 3^3 + 4^3 + 7^3 + 11^3  = 2^3 + 2^3 + 6^3 + 6^3 + 11^3  = 2^3 + 3^3 + 3^3 + 9^3 + 10^3  = 3^3 + 5^3 + 8^3 + 8^3 + 8^3  = 3^3 + 4^3 + 7^3 + 8^3 + 9^3.

Links

Crossrefs

Cf. A294740, A343988, A344941, A345149, A345174, A345181, A345768.

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, 5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 6])
for x in range(len(rets)):
    print(rets[x])

A344799 Numbers that are the sum of five squares in six or more ways.

Original entry on oeis.org

77, 80, 83, 85, 86, 88, 91, 92, 94, 98, 99, 100, 101, 103, 104, 106, 107, 109, 110, 112, 113, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 130, 131, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149
Offset: 1

Views

Author

Sean A. Irvine, May 28 2021

Keywords

Links

Crossrefs

Cf. A025371, A294740, A344798, A344800, A344810, A345174.
Showing 1-7 of 7 results.

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