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.

A345827 Numbers that are the sum of seven fourth powers in exactly five ways.

Original entry on oeis.org

6642, 6707, 6772, 6882, 6947, 7922, 7987, 8227, 8962, 9267, 9507, 9747, 10116, 10291, 10722, 10867, 10932, 10962, 11331, 11411, 11571, 12676, 12851, 12916, 13187, 13252, 13891, 13956, 14131, 14211, 14707, 14772, 14802, 14917, 14932, 14947, 15012, 15092, 15316
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345571 at term 16 because 10787 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 8^4 + 9^4 = 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 9^4 = 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 8^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 8^4.

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])

A345608 Numbers that are the sum of seven fifth powers in five or more ways.

Original entry on oeis.org

6768576, 6776120, 7883668, 8625376, 8740709, 10036201, 10604054, 12476826, 12618493, 13006575, 13060213, 13080706, 13174250, 13536416, 13550162, 13562501, 13662500, 14110656, 14583968, 15169276, 15247994, 16053313, 16060683, 16374218, 16573507, 16600001
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			6776120 is a term because 6776120 = 2^5 + 4^5 + 7^5 + 12^5 + 17^5 + 18^5 + 20^5 = 3^5 + 6^5 + 6^5 + 12^5 + 14^5 + 18^5 + 21^5 = 4^5 + 6^5 + 8^5 + 11^5 + 13^5 + 16^5 + 22^5 = 4^5 + 7^5 + 7^5 + 7^5 + 16^5 + 19^5 + 20^5 = 5^5 + 6^5 + 6^5 + 8^5 + 16^5 + 19^5 + 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])

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])

A346283 Numbers that are the sum of seven fifth powers in exactly six ways.

Original entry on oeis.org

13562501, 14583968, 21555313, 22057487, 22066065, 23089782, 23345024, 24217918, 24401574, 24855016, 24952718, 24993517, 25052501, 25385064, 29558618, 30653536, 31613713, 32559143, 33005785, 33533765, 33635825, 33828631, 34267551, 34268332, 35431351, 35736040
Offset: 1

Views

Author

David Consiglio, Jr., Jul 12 2021

Keywords

Comments

Differs from A345609 at term 15 because 28608832 = 3^5 + 4^5 + 4^5 + 8^5 + 10^5 + 24^5 + 29^5 = 2^5 + 12^5 + 12^5 + 16^5 + 18^5 + 24^5 + 28^5 = 6^5 + 6^5 + 14^5 + 14^5 + 22^5 + 22^5 + 28^5 = 7^5 + 8^5 + 13^5 + 14^5 + 17^5 + 26^5 + 27^5 = 2^5 + 8^5 + 11^5 + 19^5 + 22^5 + 23^5 + 27^5 = 6^5 + 6^5 + 12^5 + 14^5 + 24^5 + 24^5 + 26^5 = 7^5 + 7^5 + 8^5 + 16^5 + 24^5 + 25^5 + 25^5.

Examples

			13562501 is a term 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.
		

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])

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])

A346360 Numbers that are the sum of six fifth powers in exactly five 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, 287677700
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345719 at term 25 because 287718651 = 10^5 + 11^5 + 20^5 + 22^5 + 30^5 + 48^5 = 8^5 + 10^5 + 21^5 + 27^5 + 27^5 + 48^5 = 3^5 + 6^5 + 25^5 + 30^5 + 30^5 + 47^5 = 9^5 + 10^5 + 13^5 + 26^5 + 37^5 + 46^5 = 6^5 + 9^5 + 14^5 + 31^5 + 35^5 + 46^5 = 10^5 + 11^5 + 12^5 + 23^5 + 41^5 + 44^5.

Examples

			54827300 is a term because 54827300 = 4^5 + 7^5 + 21^5 + 22^5 + 23^5 + 33^5 = 5^5 + 10^5 + 15^5 + 20^5 + 28^5 + 32^5 = 1^5 + 14^5 + 16^5 + 19^5 + 28^5 + 32^5 = 4^5 + 11^5 + 13^5 + 22^5 + 29^5 + 31^5 = 5^5 + 6^5 + 19^5 + 20^5 + 29^5 + 31^5.
		

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])
Showing 1-6 of 6 results.