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.

A345544 Numbers that are the sum of nine cubes in five or more ways.

Original entry on oeis.org

409, 413, 428, 435, 439, 446, 465, 472, 479, 491, 498, 502, 505, 507, 512, 517, 524, 526, 531, 533, 535, 538, 540, 559, 561, 563, 566, 568, 570, 576, 579, 580, 587, 594, 596, 598, 600, 601, 603, 605, 613, 615, 617, 620, 622, 624, 627, 629, 631, 632, 633, 635
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			413 is a term because 413 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3.

Links

Crossrefs

Cf. A345502, A345535, A345543, A345545, A345553, A345589, A345797.

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 >= 5])
    for x in range(len(rets)):
        print(rets[x])

A345492 Numbers that are the sum of eight squares in five or more ways.

Original entry on oeis.org

46, 47, 49, 53, 54, 55, 56, 58, 59, 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, 104, 105, 106, 107, 108, 109, 110, 111, 112
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			47 is a term because 47 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 4^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2.

Links

Crossrefs

Cf. A345482, A345491, A345493, A345502, A345535.

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 >= 5])
    for x in range(len(rets)):
        print(rets[x])

A345503 Numbers that are the sum of nine squares in six or more ways.

Original entry on oeis.org

48, 56, 57, 59, 60, 62, 63, 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, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Comments

Numbers n such that A025433(n) >= 6. - David A. Corneth, Apr 26 2024

Examples

			56 is a term because 56 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 4^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2 = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2.

Links

Crossrefs

Cf. A025433, A345493, A345502, A345504, A345545, A346805.

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 >= 6])
    for x in range(len(rets)):
        print(rets[x])

Formula

Conjectures from Chai Wah Wu, Apr 25 2024: (Start)
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 - 7*x^2 - 40*x + 48)/(x - 1)^2. (End)

A345501 Numbers that are the sum of nine squares in four or more ways.

Original entry on oeis.org

33, 36, 39, 41, 42, 44, 45, 47, 48, 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, 104
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. A345491, A345500, A345502, A345543.

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 >= 4])
    for x in range(len(rets)):
        print(rets[x])

A346804 Numbers that are the sum of ten squares in five or more ways.

Original entry on oeis.org

40, 45, 48, 49, 51, 52, 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, 104, 105, 106, 107, 108, 109
Offset: 1

Views

Author

David Consiglio, Jr., Aug 04 2021

Keywords

Examples

			45 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 6^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2
so 45 is a term.

Links

Crossrefs

Cf. A345502, A345553, A346805.

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 >= 5])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-5 of 5 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.