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 1-8 of 8 results.

A345087 Numbers that are the sum of three third powers in eight or more ways.

Original entry on oeis.org

2562624, 5618250, 6525000, 6755328, 7374375, 12742920, 13581352, 14027112, 14288373, 14926248, 16819704, 18443160, 20168784, 20500992, 22783032, 23113728, 25305048, 26936064, 27131840, 29515968, 30205440, 32835375, 34012224, 38269440, 39317832, 39339000
Offset: 1

Views

Author

David Consiglio, Jr., Jun 07 2021

Keywords

Examples

			2562624 is a term because 2562624 = 7^3 + 35^3 + 135^3  = 7^3 + 63^3 + 131^3  = 11^3 + 99^3 + 115^3  = 16^3 + 45^3 + 134^3  = 29^3 + 102^3 + 112^3  = 35^3 + 59^3 + 131^3  = 50^3 + 84^3 + 121^3  = 68^3 + 71^3 + 122^3.

Links

Crossrefs

Cf. A025336, A344737, A345086, A345088, A345119, A345152.

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, 3):
    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])

A345146 Numbers that are the sum of four third powers in nine or more ways.

Original entry on oeis.org

21896, 36225, 42120, 46683, 46872, 48321, 48825, 50806, 50904, 51408, 51480, 51506, 51688, 52208, 52416, 53200, 53865, 54971, 55575, 56385, 57113, 58338, 58968, 59059, 60480, 60515, 60984, 62244, 62433, 65303, 66024, 66276, 66339, 66430, 67158, 67536, 67851
Offset: 1

Views

Author

David Consiglio, Jr., Jun 09 2021

Keywords

Examples

			42120 is a term because 42120 = 1^3 + 19^3 + 22^3 + 27^3  = 2^3 + 3^3 + 13^3 + 33^3  = 2^3 + 6^3 + 17^3 + 32^3  = 3^3 + 3^3 + 20^3 + 31^3  = 3^3 + 17^3 + 20^3 + 29^3  = 3^3 + 13^3 + 14^3 + 32^3  = 6^3 + 15^3 + 16^3 + 31^3  = 7^3 + 17^3 + 23^3 + 27^3  = 11^3 + 13^3 + 21^3 + 29^3.

Links

Crossrefs

Cf. A025374, A344926, A345119, A345152, A345154, A345155, A345185.

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, 4):
    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])

A345150 Numbers that are the sum of four third powers in seven or more ways.

Original entry on oeis.org

13104, 18928, 19376, 20755, 21203, 21896, 22743, 24544, 24570, 24787, 25172, 25928, 27720, 27755, 27846, 28917, 29582, 30429, 31031, 31248, 31339, 31402, 31528, 32858, 33579, 34056, 34624, 34713, 34776, 35289, 35317, 35441, 35497, 35712, 36162, 36190, 36225
Offset: 1

Views

Author

David Consiglio, Jr., Jun 09 2021

Keywords

Examples

			13104 is a term because 13104 = 1^3 + 10^3 + 16^3 + 18^3  = 1^3 + 11^3 + 14^3 + 19^3  = 2^3 + 9^3 + 15^3 + 19^3  = 4^3 + 6^3 + 14^3 + 20^3  = 4^3 + 9^3 + 10^3 + 21^3  = 5^3 + 7^3 + 11^3 + 21^3  = 8^3 + 9^3 + 14^3 + 19^3.

Links

Crossrefs

Cf. A025372, A344922, A345086, A345148, A345151, A345152, A345180.

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, 4):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 7])
for x in range(len(rets)):
    print(rets[x])

A345183 Numbers that are the sum of five third powers in eight or more ways.

Original entry on oeis.org

4392, 4915, 5139, 5256, 5321, 5624, 5643, 5678, 5741, 5769, 5797, 5832, 5860, 5914, 6075, 6112, 6138, 6202, 6462, 6497, 6499, 6560, 6588, 6616, 6642, 6651, 6677, 6833, 6859, 6884, 6947, 7001, 7008, 7038, 7057, 7064, 7099, 7111, 7128, 7155, 7190, 7218, 7316
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Examples

			4915 is a term because 4915 = 1^3 + 2^3 + 7^3 + 12^3 + 12^3  = 1^3 + 3^3 + 7^3 + 9^3 + 14^3  = 1^3 + 8^3 + 8^3 + 11^3 + 11^3  = 2^3 + 4^3 + 6^3 + 6^3 + 15^3  = 3^3 + 3^3 + 5^3 + 7^3 + 15^3  = 3^3 + 3^3 + 10^3 + 11^3 + 11^3  = 4^3 + 6^3 + 6^3 + 8^3 + 14^3  = 8^3 + 8^3 + 8^3 + 9^3 + 11^3.

Links

Crossrefs

Cf. A344801, A344944, A345152, A345180, A345184, A345185, A345517.

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, 5):
    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])

A344924 Numbers that are the sum of four fourth powers in eight or more ways.

Original entry on oeis.org

13155858, 26421474, 35965458, 39803778, 98926434, 128198994, 143776179, 156279618, 210493728, 237073554, 248075538, 255831858, 257931378, 269965938, 270289698, 292967619, 293579874, 295880274, 300120003, 301080243, 302115843, 305670834, 309742434, 328118259
Offset: 1

Views

Author

David Consiglio, Jr., Jun 02 2021

Keywords

Examples

			13155858 is a term because 13155858 = 1^4 + 16^4 + 19^4 + 60^4  = 3^4 + 6^4 + 21^4 + 60^4  = 10^4 + 18^4 + 31^4 + 59^4  = 12^4 + 27^4 + 45^4 + 54^4  = 15^4 + 44^4 + 46^4 + 47^4  = 18^4 + 25^4 + 41^4 + 56^4  = 29^4 + 30^4 + 44^4 + 53^4  = 35^4 + 36^4 + 38^4 + 53^4.

Links

Crossrefs

Cf. A344737, A344922, A344925, A344926, A344944, A345152.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1, 1000)]
for pos in cwr(power_terms, 4):
    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])

A345153 Numbers that are the sum of four third powers in exactly eight ways.

Original entry on oeis.org

27720, 30429, 31339, 31402, 33579, 34624, 34776, 36162, 40105, 42695, 44037, 44163, 44226, 44947, 45162, 45675, 46277, 46900, 47600, 49042, 50112, 50689, 51058, 51597, 51805, 52227, 52264, 52507, 53144, 54271, 54873, 55692, 55790, 56240, 58032, 58221, 58312
Offset: 1

Views

Author

David Consiglio, Jr., Jun 09 2021

Keywords

Comments

Differs from A345152 at term 1 because 21896 = 1^3 + 11^3 + 19^3 + 22^3 = 2^3 + 2^3 + 12^3 + 26^3 = 2^3 + 3^3 + 19^3 + 23^3 = 2^3 + 5^3 + 15^3 + 25^3 = 3^3 + 10^3 + 16^3 + 24^3 = 3^3 + 17^3 + 19^3 + 19^3 = 4^3 + 6^3 + 20^3 + 22^3 = 5^3 + 8^3 + 14^3 + 25^3 = 7^3 + 11^3 + 17^3 + 23^3 = 8^3 + 9^3 + 19^3 + 22^3.

Examples

			30429 is a term because 30429 = 1^3 + 4^3 + 7^3 + 30^3  = 1^3 + 16^3 + 17^3 + 26^3  = 2^3 + 12^3 + 21^3 + 25^3  = 3^3 + 3^3 + 14^3 + 29^3  = 4^3 + 17^3 + 21^3 + 23^3  = 5^3 + 11^3 + 15^3 + 28^3  = 6^3 + 6^3 + 22^3 + 25^3  = 7^3 + 14^3 + 18^3 + 26^3.

Links

Crossrefs

Cf. A025364, A344925, A345088, A345151, A345152, A345154, A345184.

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, 4):
    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])

A025373 Numbers that are the sum of 4 nonzero squares in 8 or more ways.

Original entry on oeis.org

130, 138, 150, 154, 162, 175, 178, 180, 186, 195, 196, 198, 202, 207, 210, 213, 214, 217, 218, 220, 222, 223, 225, 226, 228, 230, 231, 234, 235, 237, 238, 242, 243, 244, 246, 247, 250, 252, 253, 255, 258, 259, 262, 265, 266, 267, 268, 270, 271, 273, 274, 275, 276, 277
Offset: 1

Views

Author

David W. Wilson

Keywords

Links

Crossrefs

Cf. A025336, A025364, A025372, A025374, A344801, A345152.

Programs

  • Mathematica
    selQ[n_] := Length[ Select[ PowersRepresentations[n, 4, 2], Times @@ # != 0 &]] >= 8; Select[ Range[300], selQ] (* Jean-François Alcover, Oct 03 2013 *)

Formula

{n: A025428(n) >= 8}. - R. J. Mathar, Jun 15 2018

A347362 Smallest number which can be decomposed into exactly n sums of three distinct positive cubes, but cannot be decomposed into more than one such sum containing the same cube.

Original entry on oeis.org

36, 1009, 12384, 82278, 746992, 5401404, 15685704, 26936064, 137763072, 251066304, 857520000, 618817536, 3032856000, 2050677000, 6100691904, 36013192704, 16405416000, 96569712000, 48805535232, 131243328000, 611996202000, 201153672000
Offset: 1

Views

Author

Gleb Ivanov, Aug 29 2021

Keywords

Comments

No cube should appear in two or more sums. 5104 = 15^3 + 10^3 + 9^3 = 15^3 + 12^3 + 1^3 = 16^3 + 10^3 + 2^3 is not a(3), because 15^3 appears in more than one sum.

Examples

			a(1) = 36 = 1^3 + 2^3 + 3^3.
a(2) = 1009 = 1^3 + 2^3 + 10^3 = 4^3 + 6^3 + 9^3.
a(3) = 12384 = 1^3 + 6^3 + 23^3 = 2^3 + 12^3 + 22^3 = 15^3 + 16^3 + 17^3.

Links

Crossrefs

Cf. A025333, A025419, A025469, A025398.
Cf. A343968, A343967, A345088, A345087, A345119, A345152.

Programs

  • Mathematica
    Monitor[Do[k=1;While[Length@Union@Flatten[p=PowersRepresentations[k,3,3]]!=n*3||Length@p!=n||MemberQ[Flatten@p,0],k++];Print@k,{n,10}],k] (* Giorgos Kalogeropoulos, Sep 03 2021 *)

Extensions

a(13)-a(15) from Jon E. Schoenfield, Sep 02 2021
a(16)-a(22) from Gleb Ivanov, Sep 12 2021
Showing 1-8 of 8 results.

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