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.

A345571 Numbers that are the sum of seven fourth powers in five or more ways.

Original entry on oeis.org

6642, 6707, 6772, 6882, 6947, 7922, 7987, 8227, 8962, 9267, 9507, 9747, 10116, 10291, 10722, 10787, 10867, 10932, 10962, 11331, 11411, 11571, 12676, 12851, 12916, 13187, 13252, 13891, 13956, 14131, 14211, 14707, 14772, 14802, 14917, 14932, 14947, 15012, 15092
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			6707 is a term because 6707 = 1^4 + 1^4 + 1^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.
		

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, 7):
        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])

A345607 Numbers that are the sum of seven fifth powers in four or more 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., Jun 20 2021

Keywords

Examples

			1117071 is a term because 1117071 = 1^5 + 2^5 + 6^5 + 7^5 + 7^5 + 14^5 + 14^5 = 2^5 + 2^5 + 4^5 + 6^5 + 10^5 + 12^5 + 15^5 = 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 7^5 + 16^5 = 3^5 + 5^5 + 6^5 + 6^5 + 7^5 + 8^5 + 16^5.
		

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])

A345609 Numbers that are the sum of seven fifth powers in six or more ways.

Original entry on oeis.org

13562501, 14583968, 21555313, 22057487, 22066065, 23089782, 23345024, 24217918, 24401574, 24855016, 24952718, 24993517, 25052501, 25385064, 28608832, 29558618, 30653536, 31613713, 32559143, 33005785, 33533765, 33635825, 33828631, 34267551, 34268332, 35431351
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			14583968 is a term because 14583968 = 1^5 + 4^5 + 14^5 + 16^5 + 19^5 + 21^5 + 23^5 = 2^5 + 4^5 + 14^5 + 14^5 + 20^5 + 22^5 + 22^5 = 4^5 + 5^5 + 10^5 + 15^5 + 20^5 + 21^5 + 23^5 = 6^5 + 8^5 + 9^5 + 15^5 + 15^5 + 20^5 + 25^5 = 6^5 + 8^5 + 14^5 + 14^5 + 14^5 + 16^5 + 26^5 = 6^5 + 10^5 + 12^5 + 12^5 + 16^5 + 16^5 + 26^5.
		

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

A345613 Numbers that are the sum of eight fifth powers in five or more ways.

Original entry on oeis.org

926372, 952653, 993573, 1133343, 1414591, 1431366, 1431397, 1447327, 1597928, 1637020, 1663391, 1697685, 1876624, 1933329, 1992377, 1993376, 1993666, 2033328, 2091879, 2175912, 2182160, 2231110, 2280544, 2280575, 2280786, 2281567, 2283668, 2329602, 2345563
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			952653 is a term because 952653 = 2^5 + 2^5 + 6^5 + 7^5 + 9^5 + 12^5 + 12^5 + 13^5 = 2^5 + 2^5 + 7^5 + 7^5 + 9^5 + 11^5 + 11^5 + 14^5 = 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5 = 3^5 + 4^5 + 4^5 + 6^5 + 10^5 + 10^5 + 13^5 + 13^5 = 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 9^5 + 10^5 + 15^5.
		

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])

A345719 Numbers that are the sum of six fifth powers in five or more ways.

Original entry on oeis.org

54827300, 74115800, 74883600, 75609125, 113088250, 120274275, 166078869, 169692136, 174781858, 178736448, 182341225, 185558208, 194939538, 203054589, 218814275, 235067008, 250989825, 251772882, 252721458, 255453233, 258124975, 274616694, 282859667
Offset: 1

Views

Author

David Consiglio, Jr., Jun 24 2021

Keywords

Examples

			74115800 is a term because 74115800 = 1^5 + 4^5 + 21^5 + 21^5 + 29^5 + 34^5 = 1^5 + 8^5 + 14^5 + 23^5 + 32^5 + 32^5 = 4^5 + 11^5 + 13^5 + 22^5 + 24^5 + 36^5 = 5^5 + 6^5 + 19^5 + 20^5 + 24^5 + 36^5 = 6^5 + 25^5 + 25^5 + 25^5 + 29^5 + 30^5.
		

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, 6):
        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])

A346282 Numbers that are the sum of seven fifth powers in exactly five ways.

Original entry on oeis.org

6768576, 6776120, 7883668, 8625376, 8740709, 10036201, 10604054, 12476826, 12618493, 13006575, 13060213, 13080706, 13174250, 13536416, 13550162, 13662500, 14110656, 15169276, 15247994, 16053313, 16060683, 16374218, 16573507, 16600001, 17735057, 17749152
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345608 at term 16 because 13562501 = 1^5 + 1^5 + 1^5 + 9^5 + 14^5 + 20^5 + 25^5 = 1^5 + 15^5 + 15^5 + 15^5 + 15^5 + 15^5 + 25^5 = 6^5 + 7^5 + 11^5 + 16^5 + 18^5 + 19^5 + 24^5 = 7^5 + 7^5 + 11^5 + 13^5 + 19^5 + 21^5 + 23^5 = 2^5 + 6^5 + 14^5 + 18^5 + 18^5 + 21^5 + 22^5 = 1^5 + 5^5 + 15^5 + 20^5 + 20^5 + 20^5 + 20^5.

Examples

			6768576 is a term 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.
		

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 == 5])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-6 of 6 results.