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 1-5 of 5 results.

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.

Links

Crossrefs

Cf. A345595, A345804, A345844, A345853, A345855, A346347.

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.

Links

Crossrefs

Cf. A345619, A345844, A346327, A346336, A346338, A346347.

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])

A345634 Numbers that are the sum of ten fifth powers in two or more ways.

Original entry on oeis.org

4102, 4133, 4164, 4195, 4226, 4257, 4344, 4375, 4406, 4437, 4468, 4586, 4617, 4648, 4679, 4828, 4859, 4890, 5070, 5101, 5125, 5156, 5187, 5218, 5249, 5312, 5367, 5398, 5429, 5460, 5609, 5640, 5671, 5851, 5882, 6093, 6148, 6179, 6210, 6241, 6390, 6421, 6452
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			4133 is a term because 4133 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 4^5 = 1^5 + 1^5 + 1^5 + 1^5 + 2^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5.

Links

Crossrefs

Cf. A003355, A345595, A345619, A345635, A346347.

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, 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])

A346348 Numbers that are the sum of ten fifth powers in exactly three ways.

Original entry on oeis.org

8194, 21940, 52419, 52450, 52481, 52661, 52692, 52903, 53442, 53473, 53684, 54465, 54520, 54551, 54582, 54762, 54793, 55004, 55691, 55722, 55933, 56714, 57644, 57675, 57886, 58815, 60194, 60225, 60436, 60768, 61217, 62295, 62326, 62537, 63466, 65419, 67969
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345635 at term 19 because 55543 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5.

Examples

			8194 is a term because 8194 = 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5.

Links

Crossrefs

Cf. A345635, A345855, A346338, A346347, A346349.

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, 10):
    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])

A346346 Numbers that are the sum of ten fifth powers in exactly one way.

Original entry on oeis.org

10, 41, 72, 103, 134, 165, 196, 227, 252, 258, 283, 289, 314, 320, 345, 376, 407, 438, 469, 494, 500, 525, 531, 556, 587, 618, 649, 680, 711, 736, 742, 767, 798, 829, 860, 891, 922, 953, 978, 1009, 1033, 1040, 1064, 1071, 1095, 1102, 1126, 1133, 1157, 1164
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A003355 at term 229 because 4102 = 1^5 + 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 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5.

Examples

			10 is a term because 10 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5.

Links

Crossrefs

Cf. A003355, A345853, A346336, A346347.

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, 10):
    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 1-5 of 5 results.

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