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.
30429 is a term because 30429 = 1^3 + 4^3 + 7^3 + 30^3 = 1^3 + 16^3 + 17^3 + 26^3 = 2^3 + 12^3 + 21^3 + 25^3 = 3^3 + 3^3 + 14^3 + 29^3 = 4^3 + 17^3 + 21^3 + 23^3 = 5^3 + 11^3 + 15^3 + 28^3 = 6^3 + 6^3 + 22^3 + 25^3 = 7^3 + 14^3 + 18^3 + 26^3.
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 >= 8])
for x in range(len(rets)):
print(rets[x])
14926248 is a term because 14926248 = 2^3 + 33^3 + 245^3 = 11^3 + 185^3 + 203^3 = 14^3 + 32^3 + 245^3 = 50^3 + 113^3 + 236^3 = 71^3 + 89^3 + 239^3 = 74^3 + 189^3 + 196^3 = 89^3 + 185^3 + 197^3 = 98^3 + 148^3 + 219^3 = 105^3 + 149^3 + 217^3.
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 >= 9])
for x in range(len(rets)):
print(rets[x])
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.
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])
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.
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])
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.
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])
a(1) = 36 = 1^3 + 2^3 + 3^3. a(2) = 1009 = 1^3 + 2^3 + 10^3 = 4^3 + 6^3 + 9^3. a(3) = 12384 = 1^3 + 6^3 + 23^3 = 2^3 + 12^3 + 22^3 = 15^3 + 16^3 + 17^3.
Monitor[Do[k=1;While[Length@Union@Flatten[p=PowersRepresentations[k,3,3]]!=n*3||Length@p!=n||MemberQ[Flatten@p,0],k++];Print@k,{n,10}],k] (* Giorgos Kalogeropoulos, Sep 03 2021 *)
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