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 1-4 of 4 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])

A345486 Numbers that are the sum of seven squares in nine or more ways.

Original entry on oeis.org

69, 70, 78, 79, 81, 82, 85, 87, 88, 90, 91, 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, 132, 133, 134, 135, 136
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			70 is a term because 70 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 8^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 5^2 + 5^2 = 1^2 + 1^2 + 2^2 + 4^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 4^2 + 5^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 7^2 = 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 6^2 = 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2.

Links

Crossrefs

Cf. A345476, A345485, A345487, A345496, A345527.

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, 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])

Formula

Conjectures from Chai Wah Wu, Jan 05 2024: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 13.
G.f.: x*(-x^12 + x^11 - x^10 + x^9 - x^8 - x^7 + 2*x^6 - x^5 + x^4 - 7*x^3 + 7*x^2 - 68*x + 69)/(x - 1)^2. (End)

A345495 Numbers that are the sum of eight squares in eight or more ways.

Original entry on oeis.org

56, 59, 62, 64, 65, 67, 68, 70, 71, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			59 is a term because 59 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 6^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2 = 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 = 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2.

Links

Crossrefs

Cf. A345485, A345494, A345496, A345505, A345538.

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 >= 8])
    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 1-4 of 4 results.

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