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.

A345575 Numbers that are the sum of seven fourth powers in nine or more ways.

Original entry on oeis.org

19491, 21267, 21332, 23652, 31251, 35427, 36052, 37812, 38067, 39891, 40356, 41732, 41747, 43267, 43876, 43891, 43956, 44131, 44196, 44532, 44547, 44612, 45156, 45171, 45411, 45651, 45652, 45827, 45891, 45892, 45907, 46276, 46451, 46516, 47427, 47667, 47971
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345527, A345566, A345574, A345576, A345584, A345631, A345831.

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])

A345506 Numbers that are the sum of seven cubes in ten or more ways.

Original entry on oeis.org

1704, 1711, 1774, 1800, 1837, 1863, 1889, 1893, 1926, 1938, 1963, 1982, 1989, 2008, 2015, 2019, 2045, 2052, 2053, 2059, 2078, 2097, 2106, 2113, 2143, 2161, 2169, 2171, 2176, 2197, 2204, 2217, 2223, 2224, 2227, 2230, 2234, 2241, 2250, 2260, 2266, 2267, 2276
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345487, A345519, A345527, A345540, A345576, A345782.

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])

A345567 Numbers that are the sum of six fourth powers in ten or more ways.

Original entry on oeis.org

122915, 151556, 161475, 162755, 173075, 183620, 185315, 197795, 199106, 199940, 201875, 201955, 202275, 204275, 204340, 204595, 206115, 207395, 209795, 211075, 212420, 213731, 217620, 217826, 217891, 218515, 221250, 223715, 223955, 224180, 224451, 225875
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A341897, A344196, A345519, A345566, A345576, A345822.

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])

A345585 Numbers that are the sum of eight fourth powers in ten or more ways.

Original entry on oeis.org

17972, 17987, 19492, 19507, 19747, 20116, 20787, 21268, 21283, 21333, 21348, 21413, 21508, 21523, 21588, 21892, 21957, 22067, 22132, 22563, 22628, 23172, 23237, 23252, 23587, 23588, 23603, 23653, 23668, 23733, 23843, 23908, 24277, 24452, 24802, 24948, 25363
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345540, A345576, A345584, A345594, A345618, A345842.

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])

A345832 Numbers that are the sum of seven fourth powers in exactly ten ways.

Original entry on oeis.org

31251, 44547, 45827, 45892, 47667, 47971, 49572, 51092, 53316, 53476, 54531, 54596, 54756, 57411, 58276, 58660, 59781, 59811, 59827, 59861, 59876, 59892, 61076, 64581, 65876, 65891, 66356, 66596, 66676, 67716, 67876, 68131, 68322, 68772, 69171, 69667, 70116
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345576 at term 5 because 45907 = 1^4 + 1^4 + 3^4 + 4^4 + 8^4 + 12^4 + 12^4 = 1^4 + 6^4 + 6^4 + 8^4 + 8^4 + 9^4 + 13^4 = 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 11^4 + 13^4 = 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 11^4 + 13^4 = 2^4 + 2^4 + 4^4 + 7^4 + 7^4 + 7^4 + 14^4 = 2^4 + 3^4 + 6^4 + 6^4 + 7^4 + 7^4 + 14^4 = 2^4 + 4^4 + 6^4 + 7^4 + 9^4 + 11^4 + 12^4 = 2^4 + 5^4 + 5^4 + 10^4 + 10^4 + 10^4 + 11^4 = 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 9^4 + 14^4 = 3^4 + 6^4 + 6^4 + 6^4 + 9^4 + 11^4 + 12^4 = 4^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 + 13^4.

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. A345576, A345782, A345822, A345831, A345842, A346259.

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])

A345643 Numbers that are the sum of seven fifth powers in ten or more ways.

Original entry on oeis.org

134581976, 189642309, 219063107, 235438301, 252277376, 275782407, 281935070, 290928076, 300919884, 308188849, 309631268, 315635200, 322947868, 327287951, 335530174, 342030094, 358852218, 361946949, 379913293, 384699424, 387538625, 391133568
Offset: 1

Views

Author

David Consiglio, Jr., Jun 22 2021

Keywords

Examples

			189642309 is a term because 189642309 = 1^5 + 1^5 + 2^5 + 19^5 + 30^5 + 36^5 + 40^5 = 1^5 + 2^5 + 6^5 + 7^5 + 18^5 + 20^5 + 45^5 = 1^5 + 6^5 + 21^5 + 27^5 + 29^5 + 36^5 + 39^5 = 2^5 + 9^5 + 19^5 + 23^5 + 33^5 + 33^5 + 40^5 = 3^5 + 4^5 + 21^5 + 28^5 + 29^5 + 34^5 + 40^5 = 6^5 + 7^5 + 11^5 + 29^5 + 33^5 + 36^5 + 37^5 = 7^5 + 12^5 + 17^5 + 20^5 + 29^5 + 32^5 + 42^5 = 8^5 + 11^5 + 21^5 + 21^5 + 22^5 + 34^5 + 42^5 = 13^5 + 14^5 + 14^5 + 19^5 + 21^5 + 38^5 + 40^5 = 20^5 + 21^5 + 24^5 + 24^5 + 24^5 + 38^5 + 38^5.

Links

Crossrefs

Cf. A344196, A345576, A345618, A345631, A346259.

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.

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