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 1-6 of 6 results.

A345185 Numbers that are the sum of five third powers in nine or more ways.

Original entry on oeis.org

5860, 6112, 6138, 6462, 6497, 6588, 6651, 6859, 6947, 7001, 7038, 7057, 7064, 7099, 7190, 7316, 7328, 7372, 7433, 7561, 7587, 7703, 7759, 7841, 7902, 8056, 8163, 8289, 8352, 8371, 8443, 8506, 8560, 8569, 8630, 8632, 8758, 8928, 8991, 9017, 9045, 9080, 9099
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Examples

			6112 is a term because 6112 = 1^3 + 2^3 + 9^3 + 11^3 + 14^3  = 1^3 + 3^3 + 7^3 + 12^3 + 14^3  = 1^3 + 6^3 + 6^3 + 7^3 + 16^3  = 2^3 + 2^3 + 9^3 + 9^3 + 15^3  = 2^3 + 3^3 + 5^3 + 11^3 + 15^3  = 2^3 + 8^3 + 9^3 + 9^3 + 14^3  = 3^3 + 3^3 + 3^3 + 4^3 + 17^3  = 3^3 + 5^3 + 8^3 + 11^3 + 14^3  = 8^3 + 8^3 + 8^3 + 11^3 + 12^3.

Links

Crossrefs

Cf. A341891, A344802, A345146, A345183, A345186, A345187, A345518.

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

A341892 Numbers that are the sum of five fourth powers in exactly nine ways.

Original entry on oeis.org

619090, 775714, 1100579, 1179379, 1186834, 1243699, 1357315, 1367539, 1373859, 1422595, 1431234, 1436419, 1511299, 1536019, 1699234, 1734899, 1839874, 1858594, 1880850, 1950355, 1951650, 1978915, 2044819, 2052899, 2069955, 2085139, 2101779, 2119459, 2133234
Offset: 1

Views

Author

David Consiglio, Jr., Jun 04 2021

Keywords

Comments

Differs from A341781 at term 3 because
954979 = 1^4 + 2^4 + 11^4 + 19^4 + 30^4
= 1^4 + 7^4 + 18^4 + 25^4 + 26^4
= 3^4 + 8^4 + 17^4 + 20^4 + 29^4
= 4^4 + 8^4 + 13^4 + 25^4 + 27^4
= 4^4 + 9^4 + 10^4 + 11^4 + 31^4
= 6^4 + 6^4 + 15^4 + 21^4 + 29^4
= 7^4 + 10^4 + 18^4 + 19^4 + 29^4
= 11^4 + 11^4 + 20^4 + 22^4 + 27^4
= 16^4 + 17^4 + 17^4 + 24^4 + 25^4
= 18^4 + 19^4 + 20^4 + 23^4 + 23^4.

Examples

			619090 =  1^4 +  2^4 + 18^4 + 22^4 + 23^4
       =  1^4 +  3^4 +  4^4 +  8^4 + 28^4
       =  1^4 + 11^4 + 14^4 + 22^4 + 24^4
       =  2^4 +  2^4 +  8^4 + 17^4 + 27^4
       =  2^4 + 13^4 + 13^4 + 18^4 + 26^4
       =  3^4 +  6^4 + 12^4 + 16^4 + 27^4
       =  4^4 + 12^4 + 14^4 + 23^4 + 23^4
       =  9^4 + 12^4 + 16^4 + 21^4 + 24^4
       = 14^4 + 16^4 + 18^4 + 19^4 + 23^4
so 619090 is a term.

Links

Crossrefs

Cf. A341891, A341898, A344927, A344945, A345186, A345821.

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, 5):
    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])

A345154 Numbers that are the sum of four third powers in exactly nine ways.

Original entry on oeis.org

42120, 46683, 50806, 50904, 51408, 51480, 51688, 52208, 53865, 54971, 56385, 57113, 60515, 60984, 62433, 65303, 66276, 66339, 66430, 67158, 69048, 69832, 69930, 71162, 72072, 72520, 72576, 72800, 73017, 77714, 77903, 79345, 79667, 79849, 80066, 80073, 81207
Offset: 1

Views

Author

David Consiglio, Jr., Jun 09 2021

Keywords

Comments

Differs from A345146 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

			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. A025365, A344927, A345120, A345146, A345153, A345156, A345186.

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])

A345184 Numbers that are the sum of five third powers in exactly eight ways.

Original entry on oeis.org

4392, 4915, 5139, 5256, 5321, 5624, 5643, 5678, 5741, 5769, 5797, 5832, 5914, 6075, 6202, 6499, 6560, 6616, 6642, 6677, 6833, 6884, 7008, 7111, 7128, 7155, 7218, 7344, 7395, 7641, 7696, 7729, 7785, 7813, 7820, 7849, 7883, 8037, 8100, 8243, 8282, 8308, 8315
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Comments

Differs from A345183 at term 13 because 5860 = 1^3 + 1^3 + 5^3 + 8^3 + 16^3 = 1^3 + 2^3 + 3^3 + 11^3 + 15^3 = 1^3 + 3^3 + 8^3 + 11^3 + 14^3 = 1^3 + 5^3 + 5^3 + 10^3 + 15^3 = 1^3 + 9^3 + 10^3 + 10^3 + 12^3 = 2^3 + 3^3 + 8^3 + 9^3 + 15^3 = 2^3 + 3^3 + 5^3 + 12^3 + 14^3 = 2^3 + 8^3 + 8^3 + 12^3 + 12^3 = 3^3 + 8^3 + 8^3 + 9^3 + 14^3 = 3^3 + 6^3 + 7^3 + 12^3 + 13^3.

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. A294742, A344945, A345153, A345181, A345183, A345186, A345770.

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])

A345771 Numbers that are the sum of six cubes in exactly nine ways.

Original entry on oeis.org

2438, 2457, 2494, 2555, 2593, 2709, 2772, 2889, 2942, 2980, 3033, 3043, 3096, 3160, 3195, 3241, 3250, 3257, 3276, 3402, 3427, 3437, 3467, 3556, 3582, 3592, 3608, 3609, 3617, 3672, 3735, 3825, 3850, 3852, 3871, 3924, 3934, 3962, 3976, 3979, 3996, 3997, 4006
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345518 at term 14 because 3104 = 1^3 + 2^3 + 7^3 + 8^3 + 8^3 + 12^3 = 1^3 + 5^3 + 5^3 + 5^3 + 10^3 + 12^3 = 2^3 + 2^3 + 4^3 + 4^3 + 6^3 + 14^3 = 2^3 + 3^3 + 4^3 + 7^3 + 11^3 + 11^3 = 2^3 + 3^3 + 5^3 + 6^3 + 10^3 + 12^3 = 2^3 + 7^3 + 8^3 + 8^3 + 9^3 + 10^3 = 3^3 + 3^3 + 5^3 + 6^3 + 8^3 + 13^3 = 4^3 + 5^3 + 7^3 + 8^3 + 9^3 + 11^3 = 5^3 + 5^3 + 5^3 + 9^3 + 10^3 + 10^3 = 5^3 + 6^3 + 6^3 + 6^3 + 10^3 + 11^3 = 6^3 + 6^3 + 6^3 + 6^3 + 8^3 + 12^3.

Examples

			2457 is a term because 2457 = 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 12^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 12^3 = 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 11^3 = 1^3 + 5^3 + 5^3 + 7^3 + 7^3 + 9^3 = 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 11^3 = 2^3 + 3^3 + 3^3 + 3^3 + 9^3 + 10^3 = 2^3 + 5^3 + 5^3 + 6^3 + 6^3 + 10^3 = 3^3 + 3^3 + 5^3 + 8^3 + 8^3 + 8^3 = 3^3 + 3^3 + 4^3 + 7^3 + 8^3 + 9^3.

Links

Crossrefs

Cf. A345186, A345518, A345770, A345772, A345781, A345821.

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, 6):
    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])

A345188 Numbers that are the sum of five third powers in exactly ten ways.

Original entry on oeis.org

5860, 6588, 6651, 6859, 6947, 8056, 8289, 8569, 8758, 9045, 9099, 9227, 9414, 9612, 9829, 10009, 10277, 10485, 10522, 10529, 10800, 10963, 10970, 11008, 11061, 11089, 11241, 11385, 11458, 11656, 11719, 11782, 11817, 11845, 11934, 11990, 12016, 12060, 12088
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Comments

Differs from A345187 at term 8 because 8371 = 1^3 + 1^3 + 11^3 + 11^3 + 16^3 = 1^3 + 4^3 + 5^3 + 12^3 + 17^3 = 1^3 + 8^3 + 9^3 + 11^3 + 16^3 = 3^3 + 3^3 + 4^3 + 15^3 + 15^3 = 3^3 + 3^3 + 8^3 + 8^3 + 18^3 = 3^3 + 3^3 + 3^3 + 5^3 + 19^3 = 3^3 + 7^3 + 9^3 + 9^3 + 17^3 = 4^3 + 6^3 + 6^3 + 11^3 + 17^3 = 5^3 + 9^3 + 10^3 + 11^3 + 15^3 = 6^3 + 6^3 + 12^3 + 13^3 + 13^3 = 8^3 + 8^3 + 9^3 + 9^3 + 16^3.

Examples

			6588 is a term because 6588 = 1^3 + 3^3 + 5^3 + 7^3 + 17^3  = 1^3 + 4^3 + 6^3 + 13^3 + 14^3  = 1^3 + 5^3 + 8^3 + 8^3 + 16^3  = 1^3 + 10^3 + 10^3 + 11^3 + 12^3  = 2^3 + 2^3 + 9^3 + 12^3 + 14^3  = 2^3 + 3^3 + 8^3 + 11^3 + 15^3  = 3^3 + 8^3 + 8^3 + 11^3 + 14^3  = 3^3 + 3^3 + 5^3 + 10^3 + 16^3  = 5^3 + 5^3 + 8^3 + 10^3 + 15^3  = 8^3 + 9^3 + 10^3 + 10^3 + 12^3.

Links

Crossrefs

Cf. A294744, A341898, A345156, A345186, A345187, A345772.

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 == 10])
for x in range(len(rets)):
    print(rets[x])
Showing 1-6 of 6 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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