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.

A344358 Numbers that are the sum of five fourth powers in five or more ways.

Original entry on oeis.org

59779, 67859, 93394, 108274, 112850, 136915, 142354, 151300, 151475, 161459, 168979, 181219, 183539, 183604, 185299, 187699, 189394, 193379, 195394, 197779, 199090, 199474, 200979, 201874, 202979, 203299, 205859, 211059, 211330, 212419, 213730, 217154, 217810, 217890, 221779, 223090, 223155
Offset: 1

Views

Author

David Consiglio, Jr., May 15 2021

Keywords

Examples

			93394 is a term of this sequence because 93394 = 1^4 + 4^4 + 8^4 + 14^4 + 15^4 = 1^4 + 6^4 + 12^4 + 12^4 + 15^4 = 1^4 + 9^4 + 10^4 + 14^4 + 14^4 = 5^4 + 6^4 + 11^4 + 14^4 + 14^4 = 5^4 + 7^4 + 8^4 + 12^4 + 16^4.
		

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, 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])

A345514 Numbers that are the sum of six cubes in five or more ways.

Original entry on oeis.org

1045, 1169, 1241, 1260, 1377, 1384, 1432, 1440, 1488, 1495, 1530, 1539, 1549, 1556, 1558, 1584, 1586, 1594, 1595, 1602, 1612, 1617, 1640, 1647, 1654, 1657, 1673, 1675, 1677, 1703, 1710, 1712, 1715, 1719, 1729, 1736, 1738, 1745, 1747, 1754, 1764, 1766, 1771
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1169 is a term because 1169 = 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 9^3 = 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 7^3.
		

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, 6):
        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])

A345561 Numbers that are the sum of six fourth powers in four or more ways.

Original entry on oeis.org

6626, 6691, 6866, 9251, 9491, 10115, 10706, 10786, 11555, 12595, 14225, 14691, 14771, 15315, 15330, 15395, 15570, 16051, 16595, 16610, 16660, 16675, 16850, 17090, 17091, 17236, 17316, 17331, 17346, 17860, 17875, 17940, 17955, 18195, 18786, 18851, 18866, 19155
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			6691 is a term because 6691 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4.
		

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 >= 4])
        for x in range(len(rets)):
            print(rets[x])

A345563 Numbers that are the sum of six fourth powers in six or more ways.

Original entry on oeis.org

21251, 37811, 38051, 43251, 43571, 43875, 44115, 44531, 45155, 45651, 45891, 47411, 47586, 48276, 49796, 49971, 52195, 53235, 53315, 54131, 56290, 57395, 57460, 57570, 58035, 58500, 59075, 59330, 59780, 59795, 59811, 59860, 60035, 62180, 62211, 63971, 66340
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			37811 is a term because 37811 = 1^4 + 2^4 + 2^4 + 7^4 + 11^4 + 12^4 = 2^4 + 2^4 + 4^4 + 7^4 + 9^4 + 13^4 = 2^4 + 3^4 + 6^4 + 6^4 + 9^4 + 13^4 = 3^4 + 4^4 + 8^4 + 8^4 + 11^4 + 11^4 = 4^4 + 6^4 + 7^4 + 9^4 + 9^4 + 12^4 = 5^4 + 5^4 + 9^4 + 10^4 + 10^4 + 10^4.
		

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

A345571 Numbers that are the sum of seven fourth powers in five or more ways.

Original entry on oeis.org

6642, 6707, 6772, 6882, 6947, 7922, 7987, 8227, 8962, 9267, 9507, 9747, 10116, 10291, 10722, 10787, 10867, 10932, 10962, 11331, 11411, 11571, 12676, 12851, 12916, 13187, 13252, 13891, 13956, 14131, 14211, 14707, 14772, 14802, 14917, 14932, 14947, 15012, 15092
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			6707 is a term because 6707 = 1^4 + 1^4 + 1^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.
		

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 >= 5])
        for x in range(len(rets)):
            print(rets[x])

A345817 Numbers that are the sum of six fourth powers in exactly five ways.

Original entry on oeis.org

15395, 16610, 18866, 19235, 19410, 20996, 21011, 21316, 21331, 21491, 21620, 23811, 25091, 29700, 29715, 29906, 29955, 30356, 30995, 31235, 31266, 31331, 31506, 32035, 33651, 33795, 33891, 35171, 35411, 35636, 35796, 35971, 37971, 38595, 38675, 39266, 39890
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345562 at term 8 because 21251 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4.

Examples

			16610 is a term because 16610 = 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 5^4 + 6^4 + 11^4 = 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 10^4 = 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 = 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4.
		

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 == 5])
        for x in range(len(rets)):
            print(rets[x])

A345719 Numbers that are the sum of six fifth powers in five or more ways.

Original entry on oeis.org

54827300, 74115800, 74883600, 75609125, 113088250, 120274275, 166078869, 169692136, 174781858, 178736448, 182341225, 185558208, 194939538, 203054589, 218814275, 235067008, 250989825, 251772882, 252721458, 255453233, 258124975, 274616694, 282859667
Offset: 1

Views

Author

David Consiglio, Jr., Jun 24 2021

Keywords

Examples

			74115800 is a term because 74115800 = 1^5 + 4^5 + 21^5 + 21^5 + 29^5 + 34^5 = 1^5 + 8^5 + 14^5 + 23^5 + 32^5 + 32^5 = 4^5 + 11^5 + 13^5 + 22^5 + 24^5 + 36^5 = 5^5 + 6^5 + 19^5 + 20^5 + 24^5 + 36^5 = 6^5 + 25^5 + 25^5 + 25^5 + 29^5 + 30^5.
		

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 >= 5])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-7 of 7 results.