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 1-6 of 6 results.

A345590 Numbers that are the sum of nine fourth powers in six or more ways.

Original entry on oeis.org

4469, 4484, 5444, 5459, 5524, 5589, 5699, 5764, 6629, 6659, 6674, 6694, 6724, 6739, 6789, 6804, 6854, 6869, 6884, 6899, 6914, 6934, 6949, 6964, 6979, 7014, 7029, 7044, 7094, 7109, 7154, 7219, 7269, 7284, 7334, 7348, 7349, 7413, 7459, 7478, 7494, 7523, 7524
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			4484 is a term because 4484 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4.

Links

Crossrefs

Cf. A345545, A345581, A345589, A345591, A345599, A345623, A345848.

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

A345614 Numbers that are the sum of eight fifth powers in six or more ways.

Original entry on oeis.org

1431397, 2593811, 3329119, 3345410, 3609912, 3800722, 3932480, 4093604, 4096697, 4104553, 4114187, 4129433, 4154031, 4169869, 4377714, 4451412, 4475603, 4484634, 4501409, 4730845, 4756642, 4882770, 4912477, 4915506, 4970823, 5003645, 5112274, 5259111, 5449985
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			2593811 is a term because 2593811 = 1^5 + 1^5 + 4^5 + 9^5 + 13^5 + 13^5 + 13^5 + 17^5 = 1^5 + 1^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 17^5 = 1^5 + 6^5 + 6^5 + 8^5 + 9^5 + 9^5 + 14^5 + 18^5 = 2^5 + 5^5 + 6^5 + 6^5 + 6^5 + 15^5 + 15^5 + 16^5 = 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 12^5 + 13^5 + 18^5 = 4^5 + 4^5 + 4^5 + 6^5 + 11^5 + 11^5 + 13^5 + 18^5.

Links

Crossrefs

Cf. A345581, A345609, A345613, A345615, A345623, A346331.

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

A345622 Numbers that are the sum of nine fifth powers in five or more ways.

Original entry on oeis.org

392063, 392274, 406559, 458875, 519237, 538291, 607947, 663871, 672024, 672055, 672266, 672297, 673586, 673797, 674578, 675390, 680041, 681330, 704582, 704822, 714299, 730260, 732603, 763027, 763324, 765873, 766417, 777820, 780099, 814082, 820887, 825678
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			392274 is a term because 392274 = 1^5 + 1^5 + 4^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 1^5 + 3^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 2^5 + 3^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 3^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 2^5 + 3^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5.

Links

Crossrefs

Cf. A345589, A345613, A345621, A345623, A345637, A346340.

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

A345624 Numbers that are the sum of nine fifth powers in seven or more ways.

Original entry on oeis.org

1431398, 1431429, 1431640, 1439173, 1447570, 1504636, 1531397, 1597929, 1671167, 1696159, 1697686, 1697928, 1778835, 1936454, 1952415, 1969221, 1975049, 2017344, 2092122, 2182161, 2198967, 2208680, 2247917, 2280818, 2283911, 2289343, 2314335, 2329845, 2340319
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1431429 is a term because 1431429 = 1^5 + 2^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5 = 1^5 + 2^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 10^5 + 14^5 + 15^5 = 2^5 + 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 2^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 2^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5.

Links

Crossrefs

Cf. A345591, A345615, A345623, A345625, A345639, A346342.

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 >= 7])
    for x in range(len(rets)):
        print(rets[x])

A346341 Numbers that are the sum of nine fifth powers in exactly six ways.

Original entry on oeis.org

926404, 936607, 952896, 985421, 993574, 993605, 993816, 1075779, 1123321, 1133344, 1134367, 1151406, 1160105, 1166111, 1177144, 1206514, 1209669, 1209847, 1215545, 1225630, 1251130, 1264929, 1265320, 1278611, 1414834, 1422367, 1422609, 1430384, 1431367
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345623 at term 30 because 1431398 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 7^5 + 10^5 + 12^5 + 16^5 = 1^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5 = 1^5 + 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 2^5 + 3^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 + 15^5 = 1^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 1^5 + 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 1^5 + 1^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 1^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5.

Examples

			926404 is a term because 926404 = 2^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 8^5 + 10^5 + 15^5 = 2^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 10^5 + 15^5 = 2^5 + 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 2^5 + 7^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 2^5 + 6^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5 = 1^5 + 1^5 + 4^5 + 4^5 + 7^5 + 11^5 + 12^5 + 12^5 + 12^5.

Links

Crossrefs

Cf. A345623, A345848, A346331, A346340, A346342, A346351.

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

A345638 Numbers that are the sum of ten fifth powers in six or more ways.

Original entry on oeis.org

392095, 392306, 399839, 406802, 407583, 434676, 491643, 492063, 520261, 521106, 538323, 538534, 540927, 553325, 555098, 563526, 582089, 592398, 608190, 611072, 614196, 637833, 639903, 640715, 640895, 640926, 640957, 641106, 643671, 653523, 655327, 656616
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			392306 is a term because 392306 = 1^5 + 1^5 + 1^5 + 3^5 + 3^5 + 8^5 + 9^5 + 10^5 + 10^5 + 10^5 = 1^5 + 1^5 + 2^5 + 4^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 1^5 + 2^5 + 3^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 2^5 + 2^5 + 3^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 2^5 + 3^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 2^5 + 2^5 + 3^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5.

Links

Crossrefs

Cf. A345599, A345623, A345637, A345639, A346351.

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 >= 6])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-6 of 6 results.

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