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.

A345855 Numbers that are the sum of ten fourth powers in exactly three ways.

Original entry on oeis.org

520, 535, 550, 600, 615, 680, 775, 790, 855, 1030, 1144, 1159, 1224, 1365, 1380, 1399, 1445, 1540, 1555, 1605, 1635, 1685, 1700, 1768, 1795, 1815, 1830, 1860, 1875, 1895, 1989, 2070, 2164, 2229, 2244, 2439, 2485, 2580, 2595, 2645, 2675, 2680, 2695, 2710, 2755
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345596 at term 21 because 1620 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4.

Examples

			535 is a term because 535 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

Links

Crossrefs

Cf. A345596, A345805, A345845, A345854, A345856, A346348.

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

A346338 Numbers that are the sum of nine fifth powers in exactly three ways.

Original entry on oeis.org

52418, 52449, 52660, 53441, 54519, 54550, 54761, 55690, 57643, 60193, 62294, 69224, 69635, 69666, 69877, 70658, 70955, 70986, 71197, 71325, 71978, 72759, 73001, 74079, 76031, 77410, 78730, 84162, 84459, 84490, 84521, 84701, 84732, 84943, 85185, 85482, 85513
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345620 at term 8 because 55542 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5.

Examples

			52418 is a term 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.

Links

Crossrefs

Cf. A345620, A345845, A346328, A346337, A346339, A346348.

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

A345635 Numbers that are the sum of ten fifth powers in three or more ways.

Original entry on oeis.org

8194, 21940, 52419, 52450, 52481, 52661, 52692, 52903, 53442, 53473, 53684, 54465, 54520, 54551, 54582, 54762, 54793, 55004, 55543, 55574, 55691, 55722, 55785, 55933, 56566, 56714, 57644, 57675, 57886, 58667, 58815, 60194, 60225, 60436, 60768, 61217, 62295
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			21940 is a term because 21940 = 1^5 + 2^5 + 2^5 + 5^5 + 5^5 + 5^5 + 5^5 + 5^5 + 5^5 + 5^5 = 1^5 + 3^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 = 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5.

Links

Crossrefs

Cf. A345596, A345620, A345634, A345636, A346348.

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])

A346347 Numbers that are the sum of ten fifth powers in exactly two 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., Jul 13 2021

Keywords

Comments

Differs from A345634 at term 67 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.

Examples

			4102 is a term 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.

Links

Crossrefs

Cf. A345634, A345854, A346337, A346346, A346348.

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])

A346349 Numbers that are the sum of ten fifth powers in exactly four ways.

Original entry on oeis.org

55543, 55574, 55785, 56566, 58667, 63318, 72349, 73002, 85186, 86506, 87287, 87529, 88310, 103134, 111498, 113599, 114591, 118250, 119031, 120351, 120382, 120593, 121374, 123475, 128126, 134475, 134878, 135201, 137157, 142008, 142219, 143000, 143211, 143506
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345636 at term 92 because 200009 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 6^5 + 6^5 + 6^5 + 9^5 + 10^5 = 1^5 + 3^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 8^5 + 8^5 + 10^5 = 1^5 + 3^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 8^5 + 10^5 = 2^5 + 2^5 + 4^5 + 4^5 + 6^5 + 8^5 + 8^5 + 8^5 + 8^5 + 9^5 = 1^5 + 2^5 + 3^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5.

Examples

			55543 is a term 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.

Links

Crossrefs

Cf. A345636, A345856, A346339, A346348, A346350.

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 == 4])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-5 of 5 results.

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