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 1-5 of 5 results.

A345541 Numbers that are the sum of nine cubes in two or more ways.

Original entry on oeis.org

72, 133, 140, 147, 159, 161, 166, 168, 175, 182, 185, 187, 189, 194, 196, 198, 201, 203, 205, 208, 213, 217, 220, 222, 224, 227, 231, 238, 239, 243, 245, 246, 250, 252, 257, 259, 261, 264, 265, 266, 271, 273, 276, 278, 280, 283, 285, 287, 289, 290, 292, 294
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			133 is a term because 133 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3.

Links

Crossrefs

Cf. A003332, A345499, A345532, A345542, A345550, A345586, A345794.

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 >= 2])
    for x in range(len(rets)):
        print(rets[x])

A345489 Numbers that are the sum of eight squares in two or more ways.

Original entry on oeis.org

23, 26, 29, 31, 32, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			26 is a term because 26 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 4^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2.

Links

Crossrefs

Cf. A345479, A345488, A345490, A345499, A345532.

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 >= 2])
    for x in range(len(rets)):
        print(rets[x])

A345500 Numbers that are the sum of nine squares in three or more ways.

Original entry on oeis.org

33, 36, 39, 41, 42, 44, 45, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			36 is a term because 36 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2.

Links

Crossrefs

Cf. A345490, A345499, A345501, A345510, A345542.

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 >= 3])
    for x in range(len(rets)):
        print(rets[x])

A345509 Numbers that are the sum of ten squares in two or more ways.

Original entry on oeis.org

25, 28, 31, 33, 34, 36, 37, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			28 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 4^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2
so 28 is a term.

Links

Crossrefs

Cf. A025429, A345499, A345508, A345510, A345550.

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, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 2])
    for x in range(len(rets)):
        print(rets[x])
  • Python
    def A345509(n): return (25, 28, 31, 33, 34, 36, 37)[n-1] if n<8 else n+31 # Chai Wah Wu, May 09 2024

Formula

From Chai Wah Wu, May 09 2024: (Start)
All integers >= 39 are terms. Proof: since 20 can be written as the sum of 5 positive squares in 2 ways and any integer >= 34 can be written as a sum of 5 positive squares (see A025429), any integer >= 54 can be written as a sum of 10 positive squares in 2 or more ways. Integers from 39 to 53 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 - x^4 - x^3 - 22*x + 25)/(x - 1)^2. (End)

A345498 Numbers that are the sum of nine squares in one or more ways.

Original entry on oeis.org

9, 12, 15, 17, 18, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80
Offset: 1

Views

Author

David Consiglio, Jr., Jun 19 2021

Keywords

Examples

			12 is a term because 12 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2.

Links

Crossrefs

Cf. A003332, A345488, A345499, A345508.

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 >= 1])
    for x in range(len(rets)):
        print(rets[x])

Formula

From Chai Wah Wu, Jun 12 2025: (Start)
All integers >= 23 are terms. See A345508 for similar proof.
a(n) = 2*a(n-1) - a(n-2) for n > 9.
G.f.: x*(-x^8 + x^7 - x^6 + x^5 - x^4 - x^3 - 6*x + 9)/(x - 1)^2. (End)
Showing 1-5 of 5 results.

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