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.

A025335 Numbers that are the sum of 3 nonzero squares in 7 or more ways.

Original entry on oeis.org

341, 369, 374, 446, 461, 486, 494, 506, 509, 521, 545, 549, 566, 569, 581, 594, 614, 621, 626, 629, 641, 654, 666, 677, 686, 689, 701, 710, 726, 729, 731, 734, 749, 761, 770, 774, 789, 794, 797, 801, 806, 809, 810, 818, 821, 825, 833, 846, 849, 854, 857, 866, 869, 881
Offset: 1

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Keywords

Crossrefs

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.
		

Crossrefs

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])

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.
		

Crossrefs

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])

A345083 Numbers that are the sum of three third powers in six or more ways.

Original entry on oeis.org

1296378, 1371735, 1409400, 1614185, 1824040, 1885248, 2016496, 2101464, 2302028, 2305395, 2542968, 2562624, 2851848, 2889216, 2974392, 2988441, 3185792, 3380833, 3681280, 3689496, 3706984, 3775680, 3906657, 4109832, 4123008, 4142683, 4422592, 4525632, 4783680
Offset: 1

Views

Author

David Consiglio, Jr., Jun 07 2021

Keywords

Examples

			1296378 is a term because 1296378 = 3^3 + 75^3 + 94^3  = 8^3 + 32^3 + 107^3  = 20^3 + 76^3 + 93^3  = 30^3 + 58^3 + 101^3  = 32^3 + 80^3 + 89^3  = 59^3 + 74^3 + 86^3.
		

Crossrefs

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

A344729 Numbers that are the sum of three fourth powers in seven or more ways.

Original entry on oeis.org

779888018, 5745705602, 8185089458, 11054952818, 12478208288, 14355295682, 21789116258, 22247419922, 26839201298, 29428835618, 31861462178, 33038379458, 37314202562, 38214512882, 41923075922, 46543615202, 49511121842, 51711350418, 54438780578, 56255300738, 59223741122, 62862779042, 63170929458, 63429959138, 71035097042, 71447292098, 73526154338, 73665805122, 81629817458
Offset: 1

Views

Author

David Consiglio, Jr., May 27 2021

Keywords

Examples

			779888018 is a term because 779888018 = 3^4+ 139^4+ 142^4 = 9^4+ 38^4+ 167^4 = 14^4+ 133^4+ 147^4 = 43^4+ 114^4+ 157^4 = 47^4+ 111^4+ 158^4 = 63^4+ 98^4+ 161^4 = 73^4+ 89^4+ 162^4
		

Crossrefs

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, 3):
        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])

A345085 Numbers that are the sum of three third powers in exactly seven ways.

Original entry on oeis.org

2016496, 4525632, 4783680, 5268024, 6366816, 7451352, 7457120, 8275392, 9063144, 9086104, 9931167, 10036872, 10266138, 10371024, 10973880, 12002472, 12452049, 12983517, 13639816, 13641480, 13818384, 13832729, 14090112, 15081984, 15212016, 15685704, 16131968
Offset: 1

Views

Author

David Consiglio, Jr., Jun 07 2021

Keywords

Comments

Differs from A345086 at term 2 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.

Examples

			2016496 is a term because 2016496 = 5^3 + 71^3 + 117^3 = 9^3 + 65^3 + 119^3 = 18^3 + 20^3 + 125^3 = 46^3 + 96^3 + 99^3 = 53^3 + 59^3 + 117^3 = 65^3 + 89^3 + 99^3 = 82^3 + 84^3 + 93^3.
		

Crossrefs

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 == 7])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-6 of 6 results.