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.

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.
		

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, 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])

A345612 Numbers that are the sum of eight fifth powers in four or more 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., Jun 20 2021

Keywords

Examples

			392031 is a term because 392031 = 1^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 2^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 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, 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])

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.
		

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, 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])

A346328 Numbers that are the sum of eight fifth powers in exactly three ways.

Original entry on oeis.org

52417, 54518, 69634, 70954, 84458, 84489, 84700, 85481, 87582, 92233, 101264, 102890, 112574, 117225, 119326, 134473, 143264, 143442, 143506, 149781, 151448, 158719, 159465, 165634, 166998, 167029, 167196, 167240, 168021, 170122, 174773, 183804, 184457
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345611 at term 105 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.

Examples

			52417 is a term because 52417 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5.
		

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, 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])

A346330 Numbers that are the sum of eight fifth powers in exactly five ways.

Original entry on oeis.org

926372, 952653, 993573, 1133343, 1414591, 1431366, 1447327, 1597928, 1637020, 1663391, 1697685, 1876624, 1933329, 1992377, 1993376, 1993666, 2033328, 2091879, 2175912, 2182160, 2231110, 2280544, 2280575, 2280786, 2281567, 2283668, 2329602, 2345563, 2388619
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345613 at term 7 because 1431397 = 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5 = 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 1^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5.

Examples

			926372 is a term 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.
		

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, 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])

A346339 Numbers that are the sum of nine fifth powers in exactly four ways.

Original entry on oeis.org

55542, 120350, 143507, 167241, 182549, 192233, 202890, 326685, 327986, 328247, 329028, 329809, 333257, 351722, 358474, 358968, 359210, 359538, 359813, 365404, 367071, 367313, 374034, 374846, 375627, 376619, 377158, 379259, 381157, 383910, 384765, 390396
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345621 at term 37 because 392063 = 2^5 + 2^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 2^5 + 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 2^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 1^5 + 1^5 + 1^5 + 3^5 + 8^5 + 9^5 + 10^5 + 10^5 + 10^5.

Examples

			55542 is a term because 55542 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5.
		

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 == 4])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-6 of 6 results.