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 1-3 of 3 results.

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])

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)

A346808 Numbers that are the sum of ten squares in ten or more ways.

Original entry on oeis.org

61, 64, 66, 67, 69, 70, 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, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124
Offset: 1

Views

Author

David Consiglio, Jr., Aug 04 2021

Keywords

Examples

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

Links

Crossrefs

Cf. A345558, A346803. Subsequence of A346807.

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 >= 10])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-3 of 3 results.

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