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.

A345582 Numbers that are the sum of eight fourth powers in seven or more ways.

Original entry on oeis.org

8003, 8243, 9043, 9218, 9283, 9523, 10372, 10803, 10868, 10948, 11043, 11412, 11557, 11587, 12083, 12692, 12932, 13188, 13268, 13333, 13508, 13972, 14147, 14212, 14387, 14788, 14883, 14933, 14948, 14963, 15013, 15028, 15093, 15173, 15268, 15317, 15332, 15397
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			8243 is a term because 8243 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 6^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 8^4 = 2^4 + 4^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 3^4 + 4^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4.

Links

Crossrefs

Cf. A345537, A345573, A345581, A345583, A345591, A345615, A345839.

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

A345829 Numbers that are the sum of seven fourth powers in exactly seven ways.

Original entry on oeis.org

16691, 17347, 17971, 20706, 21956, 22547, 22612, 23156, 23587, 23827, 23892, 24436, 25107, 25427, 25716, 25971, 26051, 27812, 29092, 29187, 29332, 29427, 29442, 29636, 29701, 29716, 29956, 29971, 30036, 30132, 30612, 30981, 30996, 31011, 31316, 31331, 31347
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345573 at term 4 because 19491 = 1^4 + 1^4 + 1^4 + 6^4 + 8^4 + 8^4 + 10^4 = 1^4 + 2^4 + 4^4 + 4^4 + 7^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 8^4 + 11^4 = 2^4 + 4^4 + 4^4 + 5^4 + 6^4 + 7^4 + 11^4 = 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 8^4 + 10^4 = 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 11^4 = 3^4 + 5^4 + 7^4 + 8^4 + 8^4 + 8^4 + 8^4.

Examples

			17347 is a term because 17347 = 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 11^4 = 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 11^4 = 1^4 + 4^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 8^4 + 9^4 + 9^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 9^4 + 9^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 9^4 + 9^4.

Links

Crossrefs

Cf. A345573, A345779, A345819, A345828, A345830, A345839, A346284.

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

A345838 Numbers that are the sum of eight fourth powers in exactly six ways.

Original entry on oeis.org

6723, 6788, 6853, 6898, 6963, 7028, 7938, 8068, 8178, 8308, 8483, 8963, 9173, 9348, 9413, 9493, 9668, 9763, 9828, 10003, 10132, 10258, 10277, 10307, 10628, 10708, 10738, 10788, 10933, 10978, 11108, 11123, 11188, 11347, 11363, 11428, 11492, 11652, 11668, 11843
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345581 at term 8 because 8003 = 1^4 + 1^4 + 1^4 + 2^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 9^4 = 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 7^4.

Examples

			6788 is a term because 6788 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.

Links

Crossrefs

Cf. A345581, A345788, A345828, A345837, A345839, A345848, A346331.

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

A345840 Numbers that are the sum of eight fourth powers in exactly eight ways.

Original entry on oeis.org

13268, 14212, 14788, 15667, 16612, 16627, 16707, 16772, 16822, 16852, 16882, 16947, 17363, 17428, 17877, 18117, 18948, 19157, 19237, 19252, 19682, 19828, 20291, 20372, 20612, 20707, 20722, 20772, 20917, 20962, 21253, 21331, 21458, 21478, 21573, 21717, 21763
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345583 at term 4 because 15427 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 6^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 5^4 + 8^4 + 8^4 + 9^4 = 1^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 11^4 = 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 10^4 = 2^4 + 2^4 + 3^4 + 5^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 7^4 + 10^4 = 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 10^4 = 4^4 + 4^4 + 5^4 + 6^4 + 7^4 + 7^4 + 8^4 + 8^4.

Examples

			14212 is a term because 14212 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 8^4 + 10^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 7^4 + 10^4 = 1^4 + 1^4 + 1^4 + 5^4 + 6^4 + 8^4 + 8^4 + 8^4 = 1^4 + 2^4 + 4^4 + 4^4 + 5^4 + 7^4 + 8^4 + 9^4 = 1^4 + 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 8^4 + 9^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 + 10^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 10^4 = 3^4 + 4^4 + 4^4 + 5^4 + 7^4 + 7^4 + 8^4 + 8^4.

Links

Crossrefs

Cf. A345583, A345790, A345830, A345839, A345841, A345850, A346333.

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 == 8])
    for x in range(len(rets)):
        print(rets[x])

A345849 Numbers that are the sum of nine fourth powers in exactly seven ways.

Original entry on oeis.org

6739, 6854, 6979, 7029, 7044, 7094, 7109, 7269, 7284, 7844, 7909, 7939, 8004, 8149, 8194, 8244, 8309, 8389, 8434, 8628, 8739, 8868, 8979, 9059, 9189, 9254, 9414, 9509, 9524, 9668, 9684, 9734, 9814, 9829, 9843, 9844, 9908, 9909, 9924, 9989, 10019, 10038, 10084
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345591 at term 2 because 6804 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

Examples

			6804 is a term because 6804 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

Links

Crossrefs

Cf. A345591, A345799, A345839, A345848, A345850, A345859, A346342.

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

A345789 Numbers that are the sum of eight cubes in exactly seven ways.

Original entry on oeis.org

902, 908, 921, 938, 958, 963, 982, 991, 996, 1003, 1008, 1010, 1017, 1019, 1028, 1029, 1033, 1047, 1055, 1058, 1061, 1062, 1070, 1087, 1091, 1094, 1096, 1097, 1104, 1108, 1111, 1113, 1115, 1116, 1118, 1120, 1122, 1123, 1127, 1134, 1141, 1143, 1145, 1152, 1153
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345537 at term 7 because 970 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 7^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 8^3 = 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 9^3 = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3 + 7^3 = 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 8^3.
Likely finite.

Examples

			908 is a term because 908 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3.

Links

Crossrefs

Cf. A345537, A345779, A345788, A345790, A345799, A345839.

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

A346332 Numbers that are the sum of eight fifth powers in exactly seven ways.

Original entry on oeis.org

4104553, 4915506, 6011150, 6027989, 6323394, 6563733, 6622231, 6776363, 6785394, 7982834, 8181481, 8288806, 8658144, 8710484, 8773477, 8932244, 8996669, 9252219, 9253706, 9311478, 9904983, 9976120, 10045233, 10053008, 10193511, 10359767, 10514944, 10541225
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345615 at term 13 because 8625619 = 2^5 + 5^5 + 5^5 + 9^5 + 10^5 + 12^5 + 12^5 + 24^5 = 1^5 + 3^5 + 8^5 + 9^5 + 11^5 + 11^5 + 12^5 + 24^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 16^5 + 23^5 = 1^5 + 3^5 + 3^5 + 4^5 + 11^5 + 17^5 + 18^5 + 22^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 16^5 + 22^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 16^5 + 19^5 + 20^5 = 3^5 + 10^5 + 12^5 + 12^5 + 16^5 + 18^5 + 18^5 + 20^5 = 3^5 + 8^5 + 14^5 + 14^5 + 14^5 + 18^5 + 18^5 + 20^5.

Examples

			4104553 is a term because 4104553 = 1^5 + 1^5 + 2^5 + 3^5 + 3^5 + 5^5 + 7^5 + 21^5 = 3^5 + 3^5 + 4^5 + 6^5 + 8^5 + 14^5 + 16^5 + 19^5 = 3^5 + 3^5 + 3^5 + 7^5 + 9^5 + 12^5 + 18^5 + 18^5 = 3^5 + 4^5 + 4^5 + 4^5 + 11^5 + 11^5 + 18^5 + 18^5 = 1^5 + 1^5 + 4^5 + 7^5 + 10^5 + 16^5 + 16^5 + 18^5 = 7^5 + 11^5 + 11^5 + 13^5 + 14^5 + 15^5 + 16^5 + 16^5 = 6^5 + 12^5 + 12^5 + 13^5 + 13^5 + 15^5 + 16^5 + 16^5.

Links

Crossrefs

Cf. A345615, A345839, A346284, A346331, A346333, 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, 8):
    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])
Showing 1-7 of 7 results.

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