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 1-7 of 7 results.

A343704 Numbers that are the sum of five positive cubes in three or more ways.

Original entry on oeis.org

766, 810, 827, 829, 865, 883, 981, 1018, 1025, 1044, 1070, 1105, 1108, 1142, 1145, 1161, 1168, 1226, 1233, 1252, 1259, 1289, 1350, 1368, 1376, 1424, 1431, 1439, 1441, 1457, 1461, 1487, 1492, 1494, 1522, 1529, 1531, 1538, 1548, 1550, 1555, 1568, 1583, 1585, 1587, 1590, 1592, 1593, 1594, 1609, 1611, 1613, 1639
Offset: 1

Views

Author

David Consiglio, Jr., Apr 26 2021

Keywords

Comments

This sequence differs from A343705 at term 20 because 1252 = 1^3+1^3+5^3+5^3+10^3= 1^3+2^3+3^3+6^3+10^3 = 3^3+3^3+7^3+7^3+8^3 = 3^3+4^3+6^3+6^3+9^3. Thus this term is in this sequence but not A343705.

Examples

			827 is a member of this sequence because 827 = 1^3 + 4^3 + 5^3 + 5^3 + 8^3 = 2^3 + 2^3 + 5^3 + 7^3 + 7^3 = 2^3 + 3^3 + 4^3 + 6^3 + 8^3.
		

Crossrefs

Programs

  • Mathematica
    Select[Range@2000,Length@Select[PowersRepresentations[#,5,3],FreeQ[#,0]&]>2&] (* Giorgos Kalogeropoulos, Apr 26 2021 *)
  • 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)]#n
    for pos in cwr(power_terms,5):#m
        tot = sum(pos)
        keep[tot] += 1
    rets = sorted([k for k,v in keep.items() if v >= 3])#s
    for x in range(len(rets)):
        print(rets[x])

A343971 Numbers that are the sum of four positive cubes in four or more ways.

Original entry on oeis.org

1979, 2737, 3663, 4384, 4445, 4474, 4949, 5105, 5131, 5257, 5320, 5473, 5499, 5553, 5616, 5733, 5768, 5833, 5852, 5859, 6064, 6104, 6328, 6372, 6435, 6587, 6643, 6832, 6883, 6912, 6974, 7000, 7030, 7120, 7217, 7371, 7560, 7686, 7777, 7840, 8099, 8108, 8281, 8316, 8344, 8379, 8414, 8505, 8568, 8927, 9016, 9018
Offset: 1

Views

Author

David Consiglio, Jr., May 05 2021

Keywords

Examples

			3663 = 1^3 + 10^3 + 11^3 + 11^3
     = 2^3 +  4^3 +  6^3 + 15^3
     = 2^3 +  9^3 +  9^3 + 13^3
     = 4^3 +  7^3 +  8^3 + 14^3
so 3663 is a term.
		

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,4):
        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])

A343989 Numbers that are the sum of five positive cubes in five or more ways.

Original entry on oeis.org

1765, 1980, 2043, 2104, 2195, 2250, 2430, 2449, 2486, 2491, 2493, 2547, 2584, 2592, 2738, 2745, 2764, 2817, 2888, 2915, 2953, 2969, 2979, 3095, 3096, 3133, 3142, 3186, 3188, 3214, 3240, 3249, 3275, 3277, 3310, 3312, 3366, 3403, 3422, 3459, 3464, 3466, 3483, 3492, 3520, 3529, 3583, 3608, 3627, 3653, 3664, 3671
Offset: 1

Views

Author

David Consiglio, Jr., May 06 2021

Keywords

Examples

			2043 = 1^3 + 4^3 + 5^3 +  5^3 + 12^3
     = 2^3 + 2^3 + 3^3 + 10^3 + 10^3
     = 2^3 + 3^3 + 4^3 +  6^3 + 12^3
     = 4^3 + 5^3 + 5^3 +  9^3 + 10^3
     = 4^3 + 6^3 + 6^3 +  6^3 + 11^3
so 2043 is a term.
		

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,5):
        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])

A345513 Numbers that are the sum of six cubes in four or more ways.

Original entry on oeis.org

626, 830, 837, 856, 873, 891, 947, 954, 982, 1008, 1026, 1045, 1052, 1053, 1071, 1094, 1097, 1106, 1109, 1134, 1143, 1150, 1153, 1169, 1172, 1195, 1208, 1227, 1234, 1241, 1253, 1260, 1267, 1278, 1279, 1283, 1286, 1290, 1297, 1316, 1323, 1324, 1358, 1361, 1368
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			830 is a term because 830 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 8^3 = 1^3 + 3^3 + 3^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 6^3 + 6^3.
		

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, 6):
        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])

A344035 Numbers that are the sum of five positive cubes in exactly four ways.

Original entry on oeis.org

1252, 1376, 1461, 1522, 1548, 1585, 1590, 1646, 1702, 1709, 1737, 1739, 1772, 1798, 1802, 1810, 1864, 1889, 1954, 1987, 2006, 2033, 2081, 2096, 2152, 2160, 2225, 2241, 2251, 2276, 2313, 2322, 2339, 2341, 2367, 2374, 2377, 2416, 2423, 2456, 2458, 2465, 2467, 2512, 2521, 2528, 2530, 2537, 2540, 2549, 2556, 2582
Offset: 1

Views

Author

David Consiglio, Jr., May 07 2021

Keywords

Comments

Differs from A344034 at term 13 because 1765 = 1^3 + 1^3 + 2^3 + 3^3 + 12^3 = 1^3 + 1^3 + 6^3 + 6^3 + 11^3 = 1^3 + 2^3 + 3^3 + 9^3 + 10^3 = 3^3 + 4^3 + 6^3 + 9^3 + 9^3 = 4^3 + 4^3 + 5^3 + 8^3 + 10^3

Examples

			1461 is a member of this sequence because 1461 = 1^3 + 1^3 + 1^3 + 9^3 + 9^3 = 1^3 + 1^3 + 4^3 + 4^3 + 11^3 = 3^3 + 3^3 + 4^3 + 7^3 + 10^3 = 6^3 + 6^3 + 7^3 + 7^3 + 7^3
		

Crossrefs

Programs

  • Mathematica
    s5pcQ[n_]:=Length[Select[PowersRepresentations[n,5,3],FreeQ[#,0]&]]==4; Select[Range[ 3000],s5pcQ] (* Harvey P. Dale, Sep 15 2024 *)
  • 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,5):
        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])

A344354 Numbers that are the sum of five fourth powers in four or more ways.

Original entry on oeis.org

20995, 21235, 31250, 41474, 43235, 43250, 43315, 43490, 43859, 45139, 46290, 47570, 51939, 53234, 53299, 54994, 56274, 57379, 57410, 57779, 59329, 59779, 63970, 67010, 67859, 68035, 68290, 71795, 71954, 73730, 73954, 75714, 75794, 77890, 82099, 84499, 86275, 86450, 87730, 92500, 93394, 93474, 93859
Offset: 1

Views

Author

David Consiglio, Jr., May 15 2021

Keywords

Examples

			31250 is a term of this sequence because 31250 = 2^4 + 2^4 + 4^4 + 7^4 + 13^4 = 2^4 + 3^4 + 6^4 + 6^4 + 13^4 = 4^4 + 6^4 + 7^4 + 9^4 + 12^4 = 5^4 + 5^4 + 10^4 + 10^4 + 10^4.
		

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, 5):
        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])

A344797 Numbers that are the sum of five squares in four or more ways.

Original entry on oeis.org

53, 56, 59, 61, 62, 64, 67, 68, 70, 71, 72, 74, 75, 76, 77, 79, 80, 82, 83, 84, 85, 86, 88, 89, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126
Offset: 1

Views

Author

Sean A. Irvine, May 28 2021

Keywords

Crossrefs

Showing 1-7 of 7 results.