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.

A345539 Numbers that are the sum of eight cubes in nine or more ways.

Original entry on oeis.org

984, 1080, 1136, 1171, 1185, 1192, 1197, 1204, 1223, 1243, 1262, 1269, 1273, 1280, 1288, 1295, 1299, 1306, 1318, 1325, 1332, 1333, 1337, 1344, 1356, 1360, 1369, 1370, 1374, 1377, 1379, 1386, 1393, 1397, 1400, 1404, 1406, 1412, 1415, 1416, 1419, 1422, 1423
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1080 is a term because 1080 = 1^3 + 1^3 + 1^3 + 2^3 + 4^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3.

Links

Crossrefs

Cf. A345496, A345527, A345538, A345540, A345548, A345584, A345791.

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, 8):
    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])

A345549 Numbers that are the sum of nine cubes in ten or more ways.

Original entry on oeis.org

966, 971, 978, 985, 992, 1004, 1011, 1018, 1022, 1048, 1055, 1056, 1062, 1063, 1074, 1076, 1078, 1081, 1083, 1085, 1088, 1092, 1093, 1095, 1097, 1098, 1100, 1102, 1104, 1107, 1109, 1111, 1112, 1114, 1117, 1118, 1119, 1121, 1123, 1124, 1126, 1128, 1130, 1133
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			971 is a term because 971 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3.

Links

Crossrefs

Cf. A345540, A345548, A345558, A345594, A345802, A346803.

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, 9):
    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])

A345792 Numbers that are the sum of eight cubes in exactly ten ways.

Original entry on oeis.org

1185, 1243, 1288, 1295, 1299, 1386, 1397, 1400, 1412, 1423, 1448, 1449, 1451, 1458, 1460, 1464, 1467, 1475, 1477, 1501, 1503, 1505, 1512, 1513, 1516, 1539, 1540, 1541, 1553, 1558, 1559, 1568, 1577, 1578, 1586, 1588, 1591, 1592, 1594, 1595, 1596, 1600, 1608
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345540 at term 3 because 1262 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 10^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 10^3 = 1^3 + 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 6^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 7^3 + 7^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 6^3 + 9^3 = 1^3 + 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 6^3 + 7^3 = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 10^3 = 2^3 + 2^3 + 4^3 + 4^3 + 6^3 + 6^3 + 7^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 + 7^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 9^3.
Likely finite.

Examples

			1243 is a term because 1243 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 8^3 = 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 + 7^3.

Links

Crossrefs

Cf. A345540, A345782, A345791, A345802, A345842.

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, 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])

A345497 Numbers that are the sum of eight squares in ten or more ways.

Original entry on oeis.org

70, 71, 73, 74, 77, 78, 79, 80, 82, 83, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			71 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 8^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 4^2 + 4^2 + 4^2 + 4^2
   = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 6^2
   = 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2
   = 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2 + 4^2
so 71 is a term.

Links

Crossrefs

Cf. A025427, A025429, A345487, A345496, A345540, A346803.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 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])
  • Python
    def A345397(n): return (70, 71, 73, 74, 77, 78, 79, 80, 82, 83)[n-1] if n<11 else n+74 # Chai Wah Wu, May 09 2024

Formula

From Chai Wah Wu, May 09 2024: (Start)
All integers >= 85 are terms. Proof: since 594 can be written as the sum of 3 positive squares in 10 ways (see A025427) and any integer >= 34 can be written as a sum of 5 positive squares (see A025429), any integer >= 628 can be written as a sum of 8 positive squares in 10 or more ways. Integers from 85 to 627 are terms by inspection.
a(n) = 2*a(n-1) - a(n-2) for n > 12.
G.f.: x*(-x^11 + x^10 - x^9 + x^8 - 2*x^5 + 2*x^4 - x^3 + x^2 - 69*x + 70)/(x - 1)^2. (End)
Showing 1-6 of 6 results.

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

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