OFFSET
1,1
COMMENTS
a(88) = 636 = 5^4 + 11 and a(91) = 651 = 5^4 + 2^4 + 10 are the first two terms not congruent to 2 or 7 (mod 10). - M. F. Hasler, Aug 03 2020
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Eric Weisstein's World of Mathematics, Biquadratic Number.
EXAMPLE
From David A. Corneth, Aug 03 2020: (Start)
3740 is in the sequence as 3740 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 5^4 + 5^4 + 7^4.
4690 is in the sequence as 4690 = 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 6^4 + 6^4.
7193 is in the sequence as 7193 = 2^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4. (End)
PROG
(PARI) (A003346_upto(N, k=12, m=4)=[i|i<-[1..#N=sum(n=1, sqrtnint(N, m), 'x^n^m, O('x^N))^k], polcoef(N, i)])(500) \\ 2nd & 3rd optional arg allow to get other sequences of this group. See A003333 for alternate code. - M. F. Hasler, Aug 03 2020
(Python)
from itertools import count, takewhile, combinations_with_replacement as mc
def aupto(limit):
qd = takewhile(lambda x: x <= limit, (k**4 for k in count(1)))
ss = set(sum(c) for c in mc(qd, 12))
return sorted(s for s in ss if s <= limit)
print(aupto(417)) # Michael S. Branicky, Dec 27 2021
CROSSREFS
Cf. A000583 (4th powers).
Other numbers that are the sum of k positive m-th powers:
KEYWORD
nonn,easy
AUTHOR
STATUS
approved