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 1-7 of 7 results.

A345183 Numbers that are the sum of five third powers in eight or more ways.

Original entry on oeis.org

4392, 4915, 5139, 5256, 5321, 5624, 5643, 5678, 5741, 5769, 5797, 5832, 5860, 5914, 6075, 6112, 6138, 6202, 6462, 6497, 6499, 6560, 6588, 6616, 6642, 6651, 6677, 6833, 6859, 6884, 6947, 7001, 7008, 7038, 7057, 7064, 7099, 7111, 7128, 7155, 7190, 7218, 7316
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Examples

			4915 is a term because 4915 = 1^3 + 2^3 + 7^3 + 12^3 + 12^3  = 1^3 + 3^3 + 7^3 + 9^3 + 14^3  = 1^3 + 8^3 + 8^3 + 11^3 + 11^3  = 2^3 + 4^3 + 6^3 + 6^3 + 15^3  = 3^3 + 3^3 + 5^3 + 7^3 + 15^3  = 3^3 + 3^3 + 10^3 + 11^3 + 11^3  = 4^3 + 6^3 + 6^3 + 8^3 + 14^3  = 8^3 + 8^3 + 8^3 + 9^3 + 11^3.

Links

Crossrefs

Cf. A344801, A344944, A345152, A345180, A345184, A345185, A345517.

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, 5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 8])
for x in range(len(rets)):
    print(rets[x])

A345516 Numbers that are the sum of six cubes in seven or more ways.

Original entry on oeis.org

1710, 1766, 1773, 1981, 1988, 2051, 2105, 2160, 2168, 2196, 2249, 2251, 2259, 2277, 2314, 2322, 2349, 2368, 2375, 2376, 2417, 2424, 2431, 2438, 2457, 2466, 2480, 2492, 2494, 2513, 2520, 2531, 2538, 2539, 2548, 2555, 2557, 2564, 2565, 2574, 2583, 2593, 2611
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1766 is a term because 1766 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 11^3 = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 10^3 = 1^3 + 1^3 + 2^3 + 3^3 + 8^3 + 9^3 = 1^3 + 3^3 + 3^3 + 5^3 + 8^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 9^3 = 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 9^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 10^3.

Links

Crossrefs

Cf. A344811, A345180, A345515, A345517, A345525, A345564, A345769.

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, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 7])
    for x in range(len(rets)):
        print(rets[x])

A345518 Numbers that are the sum of six cubes in nine or more ways.

Original entry on oeis.org

2438, 2457, 2494, 2555, 2593, 2709, 2772, 2889, 2942, 2980, 3033, 3043, 3096, 3104, 3160, 3195, 3215, 3222, 3241, 3250, 3257, 3267, 3276, 3313, 3339, 3374, 3402, 3427, 3430, 3437, 3465, 3467, 3491, 3493, 3528, 3547, 3556, 3582, 3584, 3592, 3608, 3609, 3617
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			2457 is a term because 2457 = 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 12^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 12^3 = 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 11^3 = 1^3 + 5^3 + 5^3 + 7^3 + 7^3 + 9^3 = 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 11^3 = 2^3 + 3^3 + 3^3 + 3^3 + 9^3 + 10^3 = 2^3 + 5^3 + 5^3 + 6^3 + 6^3 + 10^3 = 3^3 + 3^3 + 5^3 + 8^3 + 8^3 + 8^3 = 3^3 + 3^3 + 4^3 + 7^3 + 8^3 + 9^3.

Links

Crossrefs

Cf. A345185, A345476, A345517, A345519, A345527, A345566, A345771.

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, 6):
    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])

A345526 Numbers that are the sum of seven cubes in eight or more ways.

Original entry on oeis.org

1385, 1496, 1515, 1552, 1557, 1585, 1587, 1603, 1613, 1622, 1648, 1655, 1665, 1674, 1681, 1704, 1711, 1718, 1719, 1720, 1737, 1739, 1741, 1746, 1753, 1755, 1765, 1767, 1772, 1774, 1781, 1782, 1793, 1800, 1802, 1805, 1809, 1811, 1818, 1819, 1826, 1828, 1830
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1496 is a term because 1496 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 8^3 = 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 1^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 2^3 + 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 + 7^3.

Links

Crossrefs

Cf. A345485, A345517, A345525, A345527, A345538, A345574, A345780.

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 >= 8])
    for x in range(len(rets)):
        print(rets[x])

A345565 Numbers that are the sum of six fourth powers in eight or more ways.

Original entry on oeis.org

58035, 59780, 87746, 88595, 96195, 96450, 102371, 106451, 106515, 108035, 108275, 108290, 108771, 112370, 112931, 115251, 122835, 122850, 122915, 124691, 125971, 132546, 133395, 133571, 133586, 134675, 134931, 136931, 138275, 138595, 143650, 144755, 144835
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			59780 is a term because 59780 = 1^4 + 1^4 + 1^4 + 5^4 + 12^4 + 14^4 = 1^4 + 1^4 + 6^4 + 6^4 + 9^4 + 15^4 = 1^4 + 2^4 + 9^4 + 10^4 + 11^4 + 13^4 = 1^4 + 4^4 + 7^4 + 7^4 + 8^4 + 15^4 = 1^4 + 7^4 + 7^4 + 9^4 + 10^4 + 14^4 = 2^4 + 5^4 + 6^4 + 11^4 + 11^4 + 13^4 = 3^4 + 7^4 + 8^4 + 10^4 + 11^4 + 13^4 = 5^4 + 6^4 + 7^4 + 7^4 + 11^4 + 14^4.

Links

Crossrefs

Cf. A344944, A345517, A345564, A345566, A345574, A345722, A345820.

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 >= 8])
    for x in range(len(rets)):
        print(rets[x])

A345770 Numbers that are the sum of six cubes in exactly eight ways.

Original entry on oeis.org

1981, 2105, 2168, 2277, 2368, 2376, 2431, 2466, 2538, 2557, 2583, 2646, 2665, 2672, 2746, 2753, 2763, 2765, 2880, 2881, 2916, 2961, 2970, 2977, 2979, 2987, 3007, 3040, 3042, 3049, 3068, 3088, 3141, 3159, 3169, 3185, 3248, 3278, 3311, 3312, 3367, 3384, 3393
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345517 at term 8 because 2438 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 13^3 = 1^3 + 2^3 + 4^3 + 5^3 + 8^3 + 12^3 = 1^3 + 5^3 + 5^3 + 9^3 + 9^3 + 9^3 = 2^3 + 2^3 + 2^3 + 7^3 + 7^3 + 12^3 = 2^3 + 2^3 + 3^3 + 4^3 + 10^3 + 11^3 = 2^3 + 3^3 + 6^3 + 9^3 + 9^3 + 9^3 = 2^3 + 4^3 + 5^3 + 8^3 + 9^3 + 10^3 = 4^3 + 4^3 + 5^3 + 5^3 + 9^3 + 11^3 = 6^3 + 7^3 + 7^3 + 8^3 + 8^3 + 8^3.

Examples

			2105 is a term because 2105 = 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 11^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 11^3 = 1^3 + 2^3 + 6^3 + 7^3 + 7^3 + 8^3 = 1^3 + 4^3 + 4^3 + 4^3 + 8^3 + 9^3 = 1^3 + 4^3 + 5^3 + 5^3 + 5^3 + 10^3 = 2^3 + 3^3 + 4^3 + 5^3 + 8^3 + 9^3 = 3^3 + 3^3 + 3^3 + 7^3 + 7^3 + 9^3 = 5^3 + 5^3 + 5^3 + 5^3 + 7^3 + 8^3.

Links

Crossrefs

Cf. A345184, A345517, A345769, A345771, A345780, A345820.

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, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 8])
    for x in range(len(rets)):
        print(rets[x])

A344812 Numbers that are the sum of six squares in eight or more ways.

Original entry on oeis.org

78, 81, 84, 86, 87, 89, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 107, 108, 109, 110, 111, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			81 = 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 5^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 3^2 + 3^2 + 5^2 + 6^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 8^2
   = 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 7^2
   = 1^2 + 4^2 + 4^2 + 4^2 + 4^2 + 4^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 7^2
   = 2^2 + 2^2 + 4^2 + 4^2 + 4^2 + 5^2
   = 2^2 + 3^2 + 3^2 + 3^2 + 5^2 + 5^2
   = 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 6^2
so 81 is a term.

Links

Crossrefs

Cf. A344801, A344811, A345476, A345485, A345517.

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, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 8])
    for x in range(len(rets)):
        print(rets[x])

Formula

Conjectures from Chai Wah Wu, Jan 05 2024: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 27.
G.f.: x*(-x^26 + x^25 - x^21 + x^20 - 2*x^7 + x^6 + x^5 - x^4 - x^3 - 75*x + 78)/(x - 1)^2. (End)
Showing 1-7 of 7 results.

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

Content is available under The OEIS End-User License Agreement.