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.

A345579 Numbers that are the sum of eight fourth powers in four or more ways.

Original entry on oeis.org

2933, 2948, 3013, 3173, 3188, 3557, 4148, 4163, 4213, 4228, 4293, 4388, 4403, 4453, 4468, 4643, 4772, 4837, 4883, 5012, 5123, 5188, 5203, 5268, 5333, 5363, 5378, 5398, 5428, 5443, 5508, 5538, 5573, 5603, 5618, 5668, 5683, 5733, 5748, 5858, 5923, 6052, 6163
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345534, A345570, A345578, A345580, A345588, A345612, A345836.

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])

A345826 Numbers that are the sum of seven fourth powers in exactly four ways.

Original entry on oeis.org

2932, 4147, 4212, 4387, 5427, 5602, 5667, 6627, 6692, 6817, 6822, 6837, 6852, 6867, 7012, 7122, 7251, 7316, 7491, 7747, 7857, 8052, 8097, 8162, 8402, 8467, 8532, 8707, 8787, 9027, 9092, 9157, 9172, 9202, 9237, 9252, 9332, 9412, 9442, 9492, 9572, 9652, 9682
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345570 at term 9 because 6642 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

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. A345570, A345776, A345816, A345825, A345827, A345836, A346281.

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])

A345835 Numbers that are the sum of eight fourth powers in exactly three ways.

Original entry on oeis.org

518, 2678, 2693, 2708, 2738, 2758, 2773, 2838, 2853, 2868, 2883, 2918, 2998, 3078, 3108, 3123, 3253, 3302, 3317, 3363, 3382, 3428, 3477, 3492, 3542, 3622, 3732, 3778, 3797, 3893, 3926, 3953, 3973, 3988, 4018, 4053, 4101, 4118, 4133, 4166, 4193, 4243, 4258
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345578 at term 13 because 2933 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

Examples

			2678 is a term because 2678 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4.

Links

Crossrefs

Cf. A345578, A345785, A345825, A345834, A345836, A345845, A346328.

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

A345837 Numbers that are the sum of eight fourth powers in exactly five ways.

Original entry on oeis.org

4228, 4403, 4468, 5443, 5508, 5683, 6613, 6643, 6658, 6708, 6773, 6838, 6868, 6883, 6948, 7013, 7093, 7138, 7203, 7267, 7268, 7332, 7397, 7478, 7507, 7572, 7588, 7828, 7858, 7923, 7988, 8113, 8133, 8228, 8353, 8418, 8533, 8547, 8548, 8612, 8723, 8788, 8852
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345580 at term 11 because 6723 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

Examples

			4403 is a term because 4403 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4.

Links

Crossrefs

Cf. A345580, A345787, A345827, A345836, A345838, A345847, A346330.

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

A345846 Numbers that are the sum of nine fourth powers in exactly four ways.

Original entry on oeis.org

2854, 2919, 2934, 2949, 2964, 3014, 3029, 3094, 3159, 3174, 3204, 3254, 3269, 3429, 3444, 3558, 3573, 3638, 3798, 3813, 3974, 4034, 4134, 4164, 4179, 4182, 4209, 4214, 4274, 4294, 4389, 4439, 4454, 4534, 4614, 4644, 4709, 4773, 4788, 4838, 4884, 4918, 4949
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345588 at term 11 because 3189 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4.

Examples

			2919 is a term because 2919 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 7^4.

Links

Crossrefs

Cf. A345588, A345796, A345836, A345845, A345847, A345856, A346339.

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, 9):
    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])

A345786 Numbers that are the sum of eight cubes in exactly four ways.

Original entry on oeis.org

256, 347, 382, 401, 408, 427, 434, 438, 445, 464, 478, 480, 490, 499, 502, 506, 511, 516, 523, 530, 532, 534, 537, 560, 565, 567, 569, 571, 578, 586, 593, 595, 600, 602, 604, 605, 611, 612, 616, 619, 621, 624, 626, 643, 645, 656, 660, 663, 664, 668, 675, 679
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345534 at term 11 because 471 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3.
Likely finite.

Examples

			347 is a term because 347 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3.

Links

Crossrefs

Cf. A345534, A345776, A345785, A345787, A345796, A345836.

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, 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])

A346329 Numbers that are the sum of eight fifth powers in exactly four ways.

Original entry on oeis.org

391250, 392031, 455750, 519236, 604822, 622281, 672023, 672054, 672265, 673554, 697492, 703978, 707368, 730259, 763292, 857761, 893605, 893636, 893816, 893847, 894027, 894058, 894452, 894628, 896729, 897151, 901380, 903839, 909124, 909597, 910411, 911403
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345612 at term 33 because 926372 = 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 8^5 + 10^5 + 15^5 = 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 10^5 + 15^5 = 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 7^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 6^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5.

Examples

			391250 is a term because 391250 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 1^5 + 1^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 2^5 + 3^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 3^5 + 3^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5.

Links

Crossrefs

Cf. A345612, A345836, A346281, A346328, A346330, A346339.

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, 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])
Showing 1-7 of 7 results.

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

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