A003341 Numbers that are the sum of 7 positive 4th powers.
7, 22, 37, 52, 67, 82, 87, 97, 102, 112, 117, 132, 147, 162, 167, 177, 182, 197, 212, 227, 242, 247, 262, 277, 292, 307, 322, 327, 337, 342, 352, 357, 372, 387, 402, 407, 417, 422, 437, 452, 467, 482, 487, 502, 517, 532, 547, 562, 567, 577, 582, 592, 597, 612, 627
Offset: 1
Examples
From _David A. Corneth_, Aug 04 2020: (Start) 5971 is in the sequence as 5971 = 3^4 + 3^4 + 5^4 + 6^4 + 6^4 + 6^4 + 6^4. 12022 is in the sequence as 12022 = 1^4 + 2^4 + 7^4 + 7^4 + 7^4 + 7^4 + 7^4. 16902 is in the sequence as 16902 = 1^4 + 1^4 + 3^4 + 6^4 + 7^4 + 9^4 + 9^4. (End)
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Biquadratic Number.
Programs
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Maple
N:= 1000: S1:= {seq(i^4,i=1..floor(N^(1/4)))}: S2:= select(`<=`,{seq(seq(i+j,i=S1),j=S1)},N): S4:= select(`<=`,{seq(seq(i+j,i=S2),j=S2)},N): S6:= select(`<=`,{seq(seq(i+j,i=S2),j=S4)},N): sort(convert(select(`<=`,{seq(seq(i+j,i=S1),j=S6)},N),list)); # Robert Israel, Jul 21 2019 -
Python
from itertools import combinations_with_replacement as mc def aupto(limit): qd = [k**4 for k in range(1, int(limit**.25)+2) if k**4 + 6 <= limit] ss = set(sum(c) for c in mc(qd, 7)) return sorted(s for s in ss if s <= limit) print(aupto(630)) # Michael S. Branicky, Jul 22 2021