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.

A345570 Numbers that are the sum of seven fourth powers in four or more ways.

Original entry on oeis.org

2932, 4147, 4212, 4387, 5427, 5602, 5667, 6627, 6642, 6692, 6707, 6772, 6817, 6822, 6837, 6852, 6867, 6882, 6947, 7012, 7122, 7251, 7316, 7491, 7747, 7857, 7922, 7987, 8052, 8097, 8162, 8227, 8402, 8467, 8532, 8707, 8787, 8962, 9027, 9092, 9157, 9172, 9202
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			4147 is a term because 4147 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4.

Links

Crossrefs

Cf. A345522, A345561, A345569, A345571, A345579, A345607, A345826.

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

A345816 Numbers that are the sum of six fourth powers in exactly four ways.

Original entry on oeis.org

6626, 6691, 6866, 9251, 9491, 10115, 10706, 10786, 11555, 12595, 14225, 14691, 14771, 15315, 15330, 15570, 16051, 16595, 16660, 16675, 16850, 17090, 17091, 17236, 17316, 17331, 17346, 17860, 17875, 17940, 17955, 18195, 18786, 18851, 19155, 19170, 19475, 19490
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345561 at term 16 because 15395 = 1^4 + 1^4 + 1^4 + 6^4 + 8^4 + 10^4 = 1^4 + 2^4 + 5^4 + 8^4 + 8^4 + 9^4 = 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 10^4 = 3^4 + 5^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 11^4.

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.

Links

Crossrefs

Cf. A344355, A345561, A345766, A345815, A345817, A345826, A346359.

Programs

  • Mathematica
    Select[Range[20000],Count[PowersRepresentations[#,6,4],?(#[[1]]>0&)]==4&] (* _Harvey P. Dale, Mar 11 2023 *)
  • 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])

A345825 Numbers that are the sum of seven fourth powers in exactly three ways.

Original entry on oeis.org

2677, 2692, 2757, 2852, 2867, 2917, 2997, 3107, 3172, 3301, 3476, 3541, 3972, 4132, 4227, 4242, 4257, 4307, 4322, 4372, 4437, 4452, 4482, 4497, 4562, 4627, 4737, 4756, 4851, 4866, 4867, 4931, 4996, 5077, 5106, 5107, 5122, 5187, 5252, 5282, 5317, 5347, 5362
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345569 at term 7 because 2932 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

Examples

			2692 is a term because 2692 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4.

Links

Crossrefs

Cf. A345569, A345775, A345815, A345824, A345826, A345835, A346280.

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

A345827 Numbers that are the sum of seven fourth powers in exactly five ways.

Original entry on oeis.org

6642, 6707, 6772, 6882, 6947, 7922, 7987, 8227, 8962, 9267, 9507, 9747, 10116, 10291, 10722, 10867, 10932, 10962, 11331, 11411, 11571, 12676, 12851, 12916, 13187, 13252, 13891, 13956, 14131, 14211, 14707, 14772, 14802, 14917, 14932, 14947, 15012, 15092, 15316
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345571 at term 16 because 10787 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 8^4 + 9^4 = 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 9^4 = 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 8^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 8^4.

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.

Links

Crossrefs

Cf. A345571, A345777, A345817, A345826, A345828, A345837, A346282.

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])

A345836 Numbers that are the sum of eight fourth powers in exactly four ways.

Original entry on oeis.org

2933, 2948, 3013, 3173, 3188, 3557, 4148, 4163, 4213, 4293, 4388, 4453, 4643, 4772, 4837, 4883, 5012, 5123, 5188, 5203, 5268, 5333, 5363, 5378, 5398, 5428, 5538, 5573, 5603, 5618, 5668, 5733, 5748, 5858, 5923, 6052, 6163, 6227, 6292, 6548, 6578, 6628, 6693
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345579 at term 10 because 4228 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4.

Examples

			2948 is a term because 2948 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

Links

Crossrefs

Cf. A345579, A345786, A345826, A345835, A345837, A345846, A346329.

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

A345776 Numbers that are the sum of seven cubes in exactly four ways.

Original entry on oeis.org

470, 496, 503, 603, 634, 653, 659, 685, 690, 692, 711, 712, 747, 751, 754, 761, 766, 773, 775, 777, 780, 783, 787, 792, 794, 812, 813, 829, 831, 836, 842, 843, 859, 867, 871, 875, 883, 885, 890, 892, 899, 901, 904, 906, 907, 911, 913, 918, 919, 927, 930, 936
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345522 at term 5 because 627 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 7^3 = 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3.
Likely finite.

Examples

			496 is a term because 496 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3.

Links

Crossrefs

Cf. A345522, A345766, A345775, A345777, A345786, A345826.

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

A346281 Numbers that are the sum of seven fifth powers in exactly four ways.

Original entry on oeis.org

893604, 1117071, 1182534, 1414559, 1629244, 1933328, 2280543, 2586035, 2867074, 3050644, 3055295, 3055977, 3256432, 3329360, 3369543, 3436058, 3551890, 3576363, 3896969, 3914877, 3930849, 4055954, 4087746, 4088690, 4093572, 4096665, 4098161, 4104068, 4104310
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345607 at term 92 because 6768576 = 4^5 + 6^5 + 6^5 + 6^5 + 9^5 + 12^5 + 23^5 = 1^5 + 3^5 + 4^5 + 8^5 + 11^5 + 17^5 + 22^5 = 6^5 + 12^5 + 13^5 + 14^5 + 15^5 + 15^5 + 21^5 = 8^5 + 10^5 + 12^5 + 12^5 + 16^5 + 18^5 + 20^5 = 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 18^5 + 20^5.

Examples

			893604 is a term because 893604 = 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 10^5 + 15^5 = 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5.

Links

Crossrefs

Cf. A345607, A345826, A346280, A346282, A346329, A346359.

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 == 4])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-7 of 7 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.