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.

A345831 Numbers that are the sum of seven fourth powers in exactly nine ways.

Original entry on oeis.org

19491, 21267, 21332, 23652, 35427, 36052, 37812, 38067, 39891, 40356, 41732, 41747, 43267, 43876, 43891, 43956, 44131, 44196, 44532, 44612, 45156, 45171, 45411, 45651, 45652, 45891, 46276, 46451, 46516, 47427, 48036, 48052, 48532, 48707, 49747, 49956, 49987
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345575 at term 5 because 31251 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 10^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 10^4 + 11^4 = 1^4 + 4^4 + 4^4 + 4^4 + 5^4 + 6^4 + 13^4 = 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 5^4 + 6^4 + 11^4 + 11^4 = 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 10^4 + 11^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 10^4 + 12^4 = 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 + 10^4 = 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 + 11^4 = 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 + 11^4.

Examples

			21267 is a term because 21267 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 2^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 2^4 + 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 11^4 = 3^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 + 8^4.

Links

Crossrefs

Cf. A345575, A345781, A345821, A345830, A345832, A345841, A346286.

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 == 9])
    for x in range(len(rets)):
        print(rets[x])

A345576 Numbers that are the sum of seven fourth powers in ten or more ways.

Original entry on oeis.org

31251, 44547, 45827, 45892, 45907, 47667, 47971, 48292, 49572, 49812, 50052, 51092, 52372, 53316, 53476, 54531, 54596, 54756, 54996, 57411, 58036, 58116, 58276, 58516, 58660, 58756, 59331, 59781, 59796, 59811, 59827, 59861, 59876, 59892, 60036, 60371, 60436
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			44547 is a term because 44547 = 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 11^4 + 13^4 = 1^4 + 2^4 + 2^4 + 6^4 + 7^4 + 7^4 + 14^4 = 1^4 + 2^4 + 6^4 + 6^4 + 9^4 + 11^4 + 12^4 = 1^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 + 13^4 = 2^4 + 2^4 + 8^4 + 9^4 + 9^4 + 9^4 + 12^4 = 2^4 + 4^4 + 6^4 + 6^4 + 9^4 + 9^4 + 13^4 = 2^4 + 4^4 + 7^4 + 7^4 + 8^4 + 11^4 + 12^4 = 3^4 + 3^4 + 4^4 + 4^4 + 7^4 + 12^4 + 12^4 = 3^4 + 6^4 + 6^4 + 7^4 + 8^4 + 11^4 + 12^4 = 4^4 + 4^4 + 8^4 + 8^4 + 9^4 + 11^4 + 11^4.

Links

Crossrefs

Cf. A345506, A345567, A345575, A345585, A345643, A345832.

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 >= 10])
    for x in range(len(rets)):
        print(rets[x])

A345822 Numbers that are the sum of six fourth powers in exactly ten ways.

Original entry on oeis.org

122915, 151556, 161475, 162755, 173075, 183620, 185315, 199106, 199940, 201875, 202275, 204275, 204340, 204595, 206115, 207395, 209795, 211075, 213731, 217826, 217891, 218515, 221250, 223955, 224180, 225875, 226595, 227186, 228035, 236195, 237796, 237890
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345567 at term 8 because 197795 = 1^4 + 2^4 + 5^4 + 6^4 + 16^4 + 19^4 = 1^4 + 2^4 + 7^4 + 11^4 + 12^4 + 20^4 = 1^4 + 2^4 + 10^4 + 12^4 + 17^4 + 17^4 = 2^4 + 4^4 + 7^4 + 9^4 + 13^4 + 20^4 = 2^4 + 11^4 + 13^4 + 14^4 + 15^4 + 16^4 = 3^4 + 6^4 + 6^4 + 9^4 + 13^4 + 20^4 = 3^4 + 6^4 + 7^4 + 14^4 + 15^4 + 18^4 = 4^4 + 9^4 + 11^4 + 12^4 + 15^4 + 18^4 = 7^4 + 7^4 + 14^4 + 14^4 + 15^4 + 16^4.

Examples

			151556 is a term because 151556 = 1^4 + 2^4 + 2^4 + 9^4 + 11^4 + 19^4 = 1^4 + 2^4 + 3^4 + 7^4 + 16^4 + 17^4 = 1^4 + 8^4 + 11^4 + 12^4 + 13^4 + 17^4 = 2^4 + 3^4 + 7^4 + 8^4 + 11^4 + 19^4 = 3^4 + 3^4 + 3^4 + 4^4 + 12^4 + 19^4 = 3^4 + 4^4 + 11^4 + 11^4 + 14^4 + 17^4 = 3^4 + 4^4 + 13^4 + 13^4 + 13^4 + 16^4 = 4^4 + 6^4 + 9^4 + 9^4 + 9^4 + 19^4 = 4^4 + 7^4 + 11^4 + 11^4 + 11^4 + 18^4 = 4^4 + 8^4 + 9^4 + 13^4 + 13^4 + 17^4.

Links

Crossrefs

Cf. A341898, A345567, A345772, A345821, A345832, A346365.

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, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 10])
    for x in range(len(rets)):
        print(rets[x])

A345842 Numbers that are the sum of eight fourth powers in exactly ten ways.

Original entry on oeis.org

17972, 17987, 19492, 19507, 19747, 20116, 21283, 21333, 21413, 21508, 21588, 22067, 22563, 23237, 23252, 23587, 23588, 23603, 23653, 24277, 24452, 24802, 24948, 25603, 26228, 27347, 27683, 27813, 27893, 27973, 28532, 28852, 28853, 28933, 29108, 29173, 29491
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345585 at term 7 because 20787 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 8^4 + 9^4 + 10^4 = 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 6^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 8^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 8^4 + 10^4 = 2^4 + 4^4 + 4^4 + 5^4 + 6^4 + 6^4 + 7^4 + 11^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 8^4 + 10^4 = 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 6^4 + 11^4 = 3^4 + 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 + 8^4.

Examples

			17987 is a term because 17987 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 6^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 6^4 + 6^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 5^4 + 7^4 + 11^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 6^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 + 10^4 = 2^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 + 10^4 = 2^4 + 4^4 + 5^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 = 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 + 10^4 = 3^4 + 5^4 + 6^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4.

Links

Crossrefs

Cf. A345585, A345792, A345832, A345841, A345852, A346335.

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 == 10])
    for x in range(len(rets)):
        print(rets[x])

A345782 Numbers that are the sum of seven cubes in exactly ten ways.

Original entry on oeis.org

1704, 1711, 1800, 1837, 1863, 1926, 1938, 1963, 2008, 2019, 2045, 2053, 2059, 2078, 2113, 2143, 2161, 2171, 2176, 2217, 2223, 2250, 2260, 2266, 2276, 2286, 2295, 2304, 2313, 2315, 2331, 2350, 2354, 2357, 2374, 2404, 2412, 2413, 2442, 2444, 2446, 2447, 2511
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345506 at term 3 because 1774 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 12^3 = 1^3 + 1^3 + 1^3 + 2^3 + 6^3 + 6^3 + 11^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 9^3 + 10^3 = 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 9^3 + 9^3 = 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 9^3 + 9^3 = 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 11^3 = 2^3 + 2^3 + 2^3 + 4^3 + 7^3 + 7^3 + 10^3 = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 11^3 = 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 7^3 + 9^3 = 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 9^3.
Likely finite.

Examples

			1711 is a term because 1711 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 7^3 + 9^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 9^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 10^3 = 1^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3 + 8^3 = 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 + 8^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 9^3 = 4^3 + 4^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3.

Links

Crossrefs

Cf. A345506, A345772, A345781, A345792, A345832.

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, 7):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 10])
    for x in range(len(rets)):
        print(rets[x])

A346259 Numbers that are the sum of seven fifth powers in exactly ten ways.

Original entry on oeis.org

134581976, 189642309, 219063107, 235438301, 252277376, 275782407, 300919884, 308188849, 309631268, 315635200, 327287951, 335530174, 342030094, 358852218, 379913293, 384699424, 387538625, 391133568, 395423876, 405307926, 421322507, 423673757, 425588250
Offset: 1

Views

Author

David Consiglio, Jr., Jul 12 2021

Keywords

Comments

Differs from A345643 at term 7 because 281935070 = 17^5 + 17^5 + 18^5 + 21^5 + 23^5 + 26^5 + 48^5 = 7^5 + 17^5 + 20^5 + 23^5 + 24^5 + 32^5 + 47^5 = 7^5 + 13^5 + 13^5 + 26^5 + 30^5 + 36^5 + 45^5 = 1^5 + 13^5 + 21^5 + 21^5 + 33^5 + 37^5 + 44^5 = 6^5 + 7^5 + 13^5 + 31^5 + 34^5 + 36^5 + 43^5 = 4^5 + 8^5 + 16^5 + 29^5 + 31^5 + 41^5 + 41^5 = 6^5 + 8^5 + 12^5 + 28^5 + 37^5 + 38^5 + 41^5 = 3^5 + 6^5 + 15^5 + 32^5 + 35^5 + 38^5 + 41^5 = 7^5 + 24^5 + 25^5 + 32^5 + 34^5 + 37^5 + 41^5 = 13^5 + 20^5 + 21^5 + 34^5 + 35^5 + 36^5 + 41^5 = 8^5 + 24^5 + 26^5 + 31^5 + 31^5 + 40^5 + 40^5.

Examples

			134581976 is a term because 134581976 = 1^5 + 14^5 + 17^5 + 18^5 + 26^5 + 31^5 + 39^5 = 1^5 + 1^5 + 10^5 + 12^5 + 19^5 + 35^5 + 38^5 = 8^5 + 11^5 + 12^5 + 17^5 + 27^5 + 33^5 + 38^5 = 3^5 + 12^5 + 12^5 + 21^5 + 28^5 + 32^5 + 38^5 = 4^5 + 11^5 + 13^5 + 22^5 + 24^5 + 36^5 + 36^5 = 5^5 + 6^5 + 19^5 + 20^5 + 24^5 + 36^5 + 36^5 = 1^5 + 4^5 + 21^5 + 21^5 + 29^5 + 34^5 + 36^5 = 1^5 + 8^5 + 14^5 + 23^5 + 32^5 + 32^5 + 36^5 = 6^5 + 25^5 + 25^5 + 25^5 + 29^5 + 30^5 + 36^5 = 12^5 + 20^5 + 21^5 + 26^5 + 28^5 + 34^5 + 35^5.

Links

Crossrefs

Cf. A345643, A345832, A346286, A346335, A346365.

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 == 10])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-6 of 6 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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