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.

A344357 Numbers that are the sum of four fourth powers in exactly five ways.

Original entry on oeis.org

2147874, 2266338, 2690658, 3189603, 3464178, 3754674, 4030419, 4165794, 4457298, 4884114, 5229378, 5978883, 5980178, 5981283, 6014178, 6134994, 6258723, 6313953, 6400194, 6612354, 7088898, 7498323, 7811874, 7918498, 8064018, 8292323, 8630259, 9146034, 9269523, 9388978, 9397683, 9726978
Offset: 1

Views

Author

David Consiglio, Jr., May 15 2021

Keywords

Comments

Differs from A344356 at term 7 because 3847554 = 2^4 + 13^4 + 29^4 + 42^4 = 2^4 + 21^4 + 22^4 + 43^4 = 6^4 + 11^4 + 17^4 + 44^4 = 6^4 + 31^4 + 32^4 + 37^4 = 9^4 + 29^4 + 32^4 + 38^4 = 13^4 + 26^4 + 32^4 + 39^4.

Examples

			2690658 is a term of this sequence because 2690658 = 2^4 + 8^4 + 33^4 + 35^4 = 3^4 + 4^4 + 19^4 + 40^4 = 7^4 + 8^4 + 30^4 + 37^4 = 9^4 + 21^4 + 30^4 + 36^4 = 16^4 + 25^4 + 32^4 + 33^4.
		

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, 50)]
    for pos in cwr(power_terms, 4):
        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])

A343970 Numbers that are the sum of three positive cubes in exactly five ways.

Original entry on oeis.org

161568, 262683, 314712, 326808, 359568, 443197, 444536, 471960, 503208, 513729, 515376, 526023, 529199, 532683, 552824, 597960, 702729, 736371, 746992, 806688, 844416, 863379, 907479, 924048, 931419, 975213, 1011067, 1028663, 1062937, 1092853, 1152152, 1172016, 1211048, 1232496, 1258011
Offset: 1

Views

Author

David Consiglio, Jr., May 05 2021

Keywords

Comments

This sequence differs from A343967 at term 40 because 1296378 = 3^3 + 76^3 + 95^3 = 9^3 + 33^3 + 108^3 = 21^3 + 77^3 + 94^3 = 31^3 + 59^3 + 102^3 = 33^3 + 81^3 + 90^3 = 60^3 + 75^3 + 87^3.

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(1,50)]
    for pos in cwr(power_terms,3):
        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])

A344278 Numbers that are the sum of three fourth powers in exactly four ways.

Original entry on oeis.org

5978882, 15916082, 20621042, 22673378, 30623138, 33998258, 39765362, 48432482, 53809938, 61627202, 65413922, 74346818, 84942578, 88258898, 95662112, 103363442, 117259298, 128929682, 131641538, 137149922, 143244738, 155831858, 158811842, 167042642, 174135122, 175706258, 188529362
Offset: 1

Views

Author

David Consiglio, Jr., May 13 2021

Keywords

Comments

Differs from A344277 at term 37 because 292965218 = 2^4 + 109^4 + 111^4 = 21^4 + 98^4 + 119^4 = 27^4 + 94^4 + 121^4 = 34^4 + 89^4 + 123^4 = 49^4 + 77^4 + 126^4 = 61^4 + 66^4 + 127^4

Examples

			20621042 is a member of this sequence because 20621042 = 5^4 + 54^4 + 59^4 = 10^4 + 51^4 + 61^4 = 25^4 + 46^4 + 63^4 = 26^4 + 39^4 + 65^4
		

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(1,50)]
    for pos in cwr(power_terms,3):
        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])

A344364 Numbers that are the sum of three fourth powers in five or more ways.

Original entry on oeis.org

292965218, 779888018, 1010431058, 1110995522, 1234349298, 1289202642, 1500533762, 1665914642, 1948502738, 2158376402, 2373191618, 2636686962, 2689817858, 2935465442, 3019732898, 3205282178, 3642994082, 3831800882, 4162186322, 4324686002, 4687443488, 5064808658
Offset: 1

Views

Author

Sean A. Irvine, May 15 2021

Keywords

Examples

			292965218 is a member of this sequence because 292965218 = 2^4 + 109^4 + 111^4 = 21^4 + 98^4 + 119^4 = 27^4 + 94^4 + 121^4 = 34^4 + 89^4 + 123^4 = 49^4 + 77^4 + 126^4 = 61^4 + 66^4 + 127^4 (actually has 6 representations, so is a member of this sequence but not of A344365).
1234349298 is a member of this sequence because 1234349298 = 7^4 + 154^4 + 161^4 = 26^4 + 143^4 + 169^4 = 61^4 + 118^4 + 179^4 = 74^4 + 107^4 + 181^4 = 91^4 + 91^4 + 182^4.
		

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 >= 5])
    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])
Showing 1-5 of 5 results.