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.

A344923 Numbers that are the sum of four fourth powers in exactly seven ways.

Original entry on oeis.org

6576339, 16020018, 16408434, 22673634, 23056803, 33734834, 39786098, 43583138, 51071619, 52652754, 53731458, 57976083, 63985314, 64365939, 67655779, 68846274, 73744563, 75951138, 77495778, 87038883, 88648914, 89148114, 90665058, 90818898, 92800178, 93830803
Offset: 1

Views

Author

David Consiglio, Jr., Jun 02 2021

Keywords

Comments

Differs from A344922 at term 2 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.

Examples

			6576339 is a term because 6576339 = 1^4 + 24^4 + 41^4 + 43^4  = 3^4 + 7^4 + 41^4 + 44^4  = 4^4 + 23^4 + 27^4 + 49^4  = 6^4 + 31^4 + 41^4 + 41^4  = 7^4 + 11^4 + 36^4 + 47^4  = 7^4 + 21^4 + 28^4 + 49^4  = 12^4 + 17^4 + 29^4 + 49^4.
		

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, 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])

A344648 Numbers that are the sum of three fourth powers in exactly six ways.

Original entry on oeis.org

292965218, 1010431058, 1110995522, 1500533762, 1665914642, 2158376402, 2373191618, 2636686962, 2689817858, 3019732898, 3205282178, 3642994082, 3831800882, 4324686002, 4687443488, 5064808658, 5175310322, 6317554418, 6450435362, 6720346178, 7018992162, 7635761042, 7781780258
Offset: 1

Views

Author

David Consiglio, Jr., May 25 2021

Keywords

Comments

Differs from A344647 at term 2 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.

Examples

			1010431058 is a term because 1010431058 = 13^4 + 143^4 + 156^4 = 31^4 + 132^4 + 163^4 = 44^4 + 123^4 + 167^4 = 52^4 + 117^4 + 169^4 = 69^4 + 103^4 + 172^4 = 81^4 + 92^4 + 173^4.
		

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, 500)]
    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])

A344738 Numbers that are the sum of three fourth powers in exactly eight ways.

Original entry on oeis.org

5745705602, 8185089458, 11054952818, 14355295682, 21789116258, 22247419922, 26839201298, 29428835618, 31861462178, 37314202562, 38214512882, 41923075922, 46543615202, 51711350418, 54438780578, 56255300738, 59223741122, 62862779042, 63429959138, 71035097042
Offset: 1

Views

Author

David Consiglio, Jr., May 27 2021

Keywords

Comments

Differs at term 14 because 49511121842 = 13^4 + 390^4 + 403^4 = 35^4 + 378^4 + 413^4 = 70^4 + 357^4 + 427^4 = 103^4 + 335^4 + 438^4 = 117^4 + 325^4 + 442^4 = 137^4 + 310^4 + 447^4 = 175^4 + 322^4 + 441^4 = 182^4 + 273^4 + 455^4 = 202^4 + 255^4 + 457^4 = 225^4 + 233^4 + 458^4.

Examples

			5745705602 is a term because 5745705602 = 3^4 + 230^4 + 233^4 = 25^4 + 218^4 + 243^4 = 43^4 + 207^4 + 250^4 = 58^4 + 197^4 + 255^4 = 85^4 + 177^4 + 262^4 = 90^4 + 173^4 + 263^4 = 102^4 + 163^4 + 265^4 = 122^4 + 145^4 + 267^4.
		

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 == 8])
    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 1-5 of 5 results.