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.

A048931 Numbers that are the sum of 6 positive cubes in exactly 3 ways.

Original entry on oeis.org

221, 254, 369, 411, 443, 469, 495, 502, 576, 595, 600, 648, 658, 684, 704, 711, 720, 739, 746, 753, 760, 765, 767, 772, 774, 779, 786, 793, 811, 818, 828, 835, 844, 854, 863, 866, 874, 880, 884, 886, 892, 893, 899, 905, 910, 919, 928, 929, 935, 936, 937
Offset: 1

Views

Author

Eric W. Weisstein

Keywords

Comments

It appears that this sequence has 1141 terms, the last of which is 26132. - Donovan Johnson, Jan 09 2013

Examples

			221 is in the sequence since 221 = 216+1+1+1+1+1 = 125+64+8+8+8+8 = 64+64+64+27+1+1.

Links

Crossrefs

Cf. A003329, A048929, A048930, A343705, A345512, A345766, A345775, A345815.

Programs

  • Mathematica
    Reap[For[n = 1, n <= 1000, n++, pr = Select[ PowersRepresentations[n, 6, 3], Times @@ # != 0 &]; If[pr != {} && Length[pr] == 3, Print[n, pr]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jul 31 2013 *)
  • PARI
    mx=10^6; ct=vector(mx); cb=vector(99); for(i=1, 99, cb[i]=i^3); for(i1=1, 99, s1=cb[i1]; for(i2=i1, 99, s2=s1+cb[i2]; if(s2+4*cb[i2]>mx, next(2)); for(i3=i2, 99, s3=s2+cb[i3]; if(s3+3*cb[i3]>mx, next(2)); for(i4=i3, 99, s4=s3+cb[i4]; if(s4+2*cb[i4]>mx, next(2)); for(i5=i4, 99, s5=s4+cb[i5]; if(s5+cb[i5]>mx, next(2)); for(i6=i5, 99, s6=s5+cb[i6]; if(s6>mx, next(2)); ct[s6]++)))))); n=0; for(i=6, mx, if(ct[i]==3, n++; write("b048931.txt", n " " i))) /* Donovan Johnson, Jan 09 2013 */

Extensions

Corrected and extended by Larry Reeves (larryr(AT)acm.org), Oct 02 2000

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.

Links

Crossrefs

Cf. A025407, A343702, A343705, A344034, A344243, A344796, A345512.

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])

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.

Links

Crossrefs

Cf. A344034, A344808, A345512, A345514, A345522, A345561, A345766.

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])

A345521 Numbers that are the sum of seven cubes in three or more ways.

Original entry on oeis.org

222, 229, 248, 255, 262, 281, 283, 285, 318, 346, 370, 374, 377, 379, 381, 396, 400, 407, 412, 419, 426, 433, 437, 438, 444, 451, 463, 470, 472, 475, 477, 489, 494, 496, 501, 503, 505, 507, 510, 522, 529, 533, 536, 559, 564, 566, 568, 570, 577, 578, 584, 585
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			229 is a term because 229 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.

Links

Crossrefs

Cf. A345480, A345512, A345520, A345522, A345533, A345569, A345775.

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, 7):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 3])
    for x in range(len(rets)):
        print(rets[x])

A345560 Numbers that are the sum of six fourth powers in three or more ways.

Original entry on oeis.org

2676, 2851, 2916, 4131, 4226, 4241, 4306, 4371, 4481, 4850, 5346, 5411, 5521, 5586, 5651, 6561, 6611, 6626, 6691, 6756, 6771, 6801, 6821, 6836, 6851, 6866, 6931, 7106, 7235, 7475, 7491, 7666, 7841, 7906, 7971, 8146, 8211, 8321, 8386, 8451, 8531, 8706, 9011
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			2851 is a term because 2851 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

Links

Crossrefs

Cf. A344243, A345512, A345559, A345561, A345569, A345604, A345815.

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, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 3])
    for x in range(len(rets)):
        print(rets[x])

A345511 Numbers that are the sum of six cubes in two or more ways.

Original entry on oeis.org

158, 165, 184, 221, 228, 235, 247, 254, 256, 261, 268, 273, 275, 280, 282, 284, 287, 291, 294, 306, 310, 313, 317, 324, 331, 332, 343, 345, 347, 350, 352, 362, 369, 371, 373, 376, 378, 380, 385, 387, 388, 392, 395, 399, 404, 406, 408, 411, 418, 425, 430, 432
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Links

Crossrefs

Cf. A003329, A048930, A343702, A344806, A345512, A345520, A345559.

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 >= 2])
    for x in range(len(rets)):
        print(rets[x])

A344807 Numbers that are the sum of six squares in three or more ways.

Original entry on oeis.org

30, 33, 36, 38, 39, 41, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			33 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 5^2
   = 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2
so 33 is a term.

Links

Crossrefs

Cf. A344796, A344806, A344808, A345480, A345512.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 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 >= 3])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-7 of 7 results.

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