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 1-5 of 5 results.

A345859 Numbers that are the sum of ten fourth powers in exactly seven ways.

Original entry on oeis.org

4485, 5445, 5460, 5525, 5540, 5590, 5605, 5670, 5700, 5715, 5765, 5780, 5830, 5845, 6645, 6710, 6775, 6855, 6900, 6915, 6930, 6935, 6965, 6980, 7175, 7190, 7235, 7255, 7335, 7364, 7415, 7430, 7475, 7479, 7495, 7510, 7604, 7620, 7654, 7669, 7670, 7685, 7715
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345600 at term 16 because 6675 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 8^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 8^4 = 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4.

Examples

			5445 is a term because 5445 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 7^4 + 7^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 7^4 = 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 8^4 = 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4.
		

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, 10):
        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])

A346342 Numbers that are the sum of nine fifth powers in exactly seven ways.

Original entry on oeis.org

1431429, 1439173, 1447570, 1504636, 1597929, 1671167, 1696159, 1697686, 1697928, 1778835, 1936454, 1975049, 2017344, 2092122, 2182161, 2198967, 2208680, 2280818, 2283911, 2289343, 2314335, 2329845, 2340319, 2345806, 2362370, 2388651, 2497771, 2529407, 2530672
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345624 at term 1 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

			1431398 is a term 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.
		

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, 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])

A345639 Numbers that are the sum of ten fifth powers in seven or more ways.

Original entry on oeis.org

555098, 674040, 683166, 707315, 763631, 777852, 778844, 780945, 783224, 893654, 896500, 897668, 920887, 926616, 927819, 928802, 936850, 937631, 944383, 945017, 952897, 953077, 953139, 953350, 953414, 955178, 963131, 975133, 979482, 984133, 985453, 985664
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			674040 is a term because 674040 = 1^5 + 1^5 + 2^5 + 3^5 + 3^5 + 7^5 + 9^5 + 10^5 + 12^5 + 12^5 = 1^5 + 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 = 1^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 8^5 + 8^5 + 9^5 + 14^5 = 1^5 + 3^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 = 2^5 + 2^5 + 2^5 + 2^5 + 4^5 + 6^5 + 8^5 + 10^5 + 11^5 + 13^5 = 2^5 + 3^5 + 3^5 + 4^5 + 4^5 + 6^5 + 6^5 + 9^5 + 9^5 + 14^5 = 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 7^5 + 8^5 + 12^5 + 13^5.
		

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, 10):
        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])

A346351 Numbers that are the sum of ten fifth powers in exactly six ways.

Original entry on oeis.org

392095, 392306, 399839, 406802, 407583, 434676, 491643, 492063, 520261, 521106, 538323, 538534, 540927, 553325, 563526, 582089, 592398, 608190, 611072, 614196, 637833, 639903, 640715, 640895, 640926, 640957, 641106, 643671, 653523, 655327, 656616, 664895
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345638 at term 15 because 555098 = 2^5 + 2^5 + 2^5 + 2^5 + 2^5 + 2^5 + 2^5 + 3^5 + 7^5 + 14^5 = 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 7^5 + 8^5 + 10^5 + 13^5 = 1^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 7^5 + 10^5 + 13^5 = 1^5 + 2^5 + 5^5 + 7^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 13^5 = 4^5 + 4^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 11^5 + 12^5 = 3^5 + 3^5 + 4^5 + 4^5 + 6^5 + 7^5 + 9^5 + 9^5 + 11^5 + 12^5 = 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 + 11^5.

Examples

			392095 is a term because 392095 = 2^5 + 2^5 + 2^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 1^5 + 1^5 + 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 7^5 + 10^5 + 12^5 = 2^5 + 2^5 + 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 2^5 + 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 2^5 + 2^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 1^5 + 1^5 + 1^5 + 2^5 + 3^5 + 8^5 + 9^5 + 10^5 + 10^5 + 10^5.
		

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, 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])

A346353 Numbers that are the sum of ten fifth powers in exactly eight ways.

Original entry on oeis.org

944383, 953139, 953414, 985453, 1118585, 1151438, 1185375, 1198879, 1206546, 1209912, 1216569, 1217172, 1218912, 1223321, 1225398, 1234631, 1241834, 1251195, 1251406, 1252123, 1259685, 1265563, 1265594, 1267937, 1275375, 1281736, 1295418, 1297697, 1298088
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345640 at term 8 because 1192180 = 5^5 + 5^5 + 5^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 10^5 + 16^5 = 2^5 + 5^5 + 5^5 + 5^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 16^5 = 3^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 6^5 + 8^5 + 13^5 + 15^5 = 3^5 + 4^5 + 4^5 + 4^5 + 6^5 + 7^5 + 7^5 + 7^5 + 13^5 + 15^5 = 2^5 + 2^5 + 2^5 + 3^5 + 8^5 + 8^5 + 9^5 + 9^5 + 12^5 + 15^5 = 1^5 + 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 12^5 + 13^5 + 14^5 = 1^5 + 2^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 12^5 + 13^5 + 13^5 = 1^5 + 2^5 + 2^5 + 2^5 + 4^5 + 11^5 + 11^5 + 12^5 + 12^5 + 13^5 = 6^5 + 9^5 + 9^5 + 10^5 + 11^5 + 11^5 + 11^5 + 11^5 + 11^5 + 11^5.

Examples

			944383 is a term because 944383 = 4^5 + 4^5 + 4^5 + 6^5 + 7^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 + 10^5 + 12^5 + 14^5 = 2^5 + 4^5 + 5^5 + 5^5 + 7^5 + 7^5 + 7^5 + 10^5 + 12^5 + 14^5 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 6^5 + 9^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 4^5 + 6^5 + 7^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5 = 1^5 + 1^5 + 2^5 + 6^5 + 7^5 + 10^5 + 11^5 + 11^5 + 12^5 + 12^5.
		

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, 10):
        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])
Showing 1-5 of 5 results.