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

1,1

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Eric Weisstein's World of Mathematics, Biquadratic Number.

Example

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)

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

Prog

(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

Crossrefs

Cf. A000583, A003330, A003340, A003342, A003352, A047718, A345568, A345823.

Sequence in context: A316840 A063240 A063157 * A345823 A161447 A047718

Adjacent sequences: A003338 A003339 A003340 * A003342 A003343 A003344

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved