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.

A345540 Numbers that are the sum of eight cubes in ten or more ways.

Original entry on oeis.org

1185, 1243, 1262, 1288, 1295, 1299, 1386, 1393, 1397, 1400, 1412, 1419, 1423, 1448, 1449, 1451, 1458, 1460, 1464, 1467, 1475, 1477, 1497, 1501, 1503, 1504, 1505, 1512, 1513, 1514, 1516, 1521, 1523, 1539, 1540, 1541, 1542, 1553, 1558, 1559, 1560, 1565, 1566
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345497, A345506, A345539, A345549, A345585, A345792.

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])

A345487 Numbers that are the sum of seven squares in ten or more ways.

Original entry on oeis.org

70, 79, 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, 137, 138, 139
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			79 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 5^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 5^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 8^2
   = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 4^2 + 4^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 5^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 3^2 + 3^2 + 4^2 + 6^2
   = 2^2 + 3^2 + 3^2 + 3^2 + 4^2 + 4^2 + 4^2
   = 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2
so 79 is a term.

Links

Crossrefs

Cf. A345477, A345486, A345497, A345506.

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 >= 10])
    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 > 10.
G.f.: x*(-x^9 + x^8 - x^7 + x^6 - x^5 - x^4 - 6*x^2 - 61*x + 70)/(x - 1)^2. (End)

A345496 Numbers that are the sum of eight squares in nine or more ways.

Original entry on oeis.org

62, 64, 67, 70, 71, 73, 74, 76, 77, 78, 79, 80, 81, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			64 is a term because 64 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 6^2 = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2 = 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2.

Links

Crossrefs

Cf. A345486, A345495, A345497, A345539.

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

A346803 Numbers that are the sum of nine squares in ten or more ways.

Original entry on oeis.org

63, 65, 68, 71, 72, 74, 75, 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, 123, 124, 125, 126, 127
Offset: 1

Views

Author

David Consiglio, Jr., Aug 04 2021

Keywords

Examples

			65 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 6^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2
   = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2
so 65 is a term.

Links

Crossrefs

Cf. A025428, A025429, A345497, A345549, A346808.

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, 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])
  • Python
    def A346803(n): return (63, 65, 68, 71, 72, 74, 75)[n-1] if n<8 else n+69 # Chai Wah Wu, May 09 2024

Formula

From Chai Wah Wu, May 09 2024: (Start)
All integers >= 77 are terms. Proof: since 246 can be written as the sum of 4 positive squares in 10 ways (see A025428) and any integer >= 34 can be written as a sum of 5 positive squares (see A025429), any integer >= 280 can be written as a sum of 9 positive squares in 10 or more ways. Integers from 77 to 279 are terms by inspection.
a(n) = 2*a(n-1) - a(n-2) for n > 9.
G.f.: x*(-x^8 + x^7 - x^6 + x^5 - 2*x^4 + x^2 - 61*x + 63)/(x - 1)^2. (End)
Showing 1-4 of 4 results.

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