A003337 Numbers n which are the sum of 3 nonzero 4th powers.

3, 18, 33, 48, 83, 98, 113, 163, 178, 243, 258, 273, 288, 338, 353, 418, 513, 528, 593, 627, 642, 657, 707, 722, 768, 787, 882, 897, 962, 1137, 1251, 1266, 1298, 1313, 1328, 1331, 1378, 1393, 1458, 1506, 1553, 1568, 1633, 1808, 1875, 1922, 1937, 2002, 2177

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Numbers which are in this sequence but not in A047714 must also be the sum of 2 biquadrates, or equal to a fourth power. Among the first 1000 terms of this sequence, this is the case for 4802 = 2*7^4, 57122 = 2*13^4 and 76832 = 2*14^4. - M. F. Hasler, Dec 31 2012

The union of A047714, A336536, and fourth powers of A003294. - Robert Israel, Jul 24 2020

As the order of addition doesn't matter we can assume terms are in nondecreasing order. - David A. Corneth, Aug 01 2020

Examples

			From _David A. Corneth_, Aug 01 2020: (Start)
194818 is in the sequence as 194818 = 3^4 + 4^4 + 21^4.
480113 is in the sequence as 480113 = 7^4 + 12^4 + 26^4.
693842 is in the sequence as 693842 = 13^4 + 15^4 + 28^4. (End)
		

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.

Crossrefs

Cf. A003294, A047714, A336536, A309762, A344188.

A###### (x, y): Numbers that are the form of x nonzero y-th powers.

Cf. A000404 (2, 2), A000408 (3, 2), A000414 (4, 2), A003072 (3, 3), A003325 (3, 2), A003327 (4, 3), A003328 (5, 3), A003329 (6, 3), A003330 (7, 3), A003331 (8, 3), A003332 (9, 3), A003333 (10, 3), A003334 (11, 3), A003335 (12, 3), A003336 (2, 4), A003337 (3, 4), A003338 (4, 4), A003339 (5, 4), A003340 (6, 4), A003341 (7, 4), A003342 (8, 4), A003343 (9, 4), A003344 (10, 4), A003345 (11, 4), A003346 (12, 4), A003347 (2, 5), A003348 (3, 5), A003349 (4, 5), A003350 (5, 5), A003351 (6, 5), A003352 (7, 5), A003353 (8, 5), A003354 (9, 5), A003355 (10, 5), A003356 (11, 5), A003357 (12, 5), A003358 (2, 6), A003359 (3, 6), A003360 (4, 6), A003361 (5, 6), A003362 (6, 6), A003363 (7, 6), A003364 (8, 6), A003365 (9, 6), A003366 (10, 6), A003367 (11, 6), A003368 (12, 6), A003369 (2, 7), A003370 (3, 7), A003371 (4, 7), A003372 (5, 7), A003373 (6, 7), A003374 (7, 7), A003375 (8, 7), A003376 (9, 7), A003377 (10, 7), A003378 (11, 7), A003379 (12, 7), A003380 (2, 8), A003381 (3, 8), A003382 (4, 8), A003383 (5, 8), A003384 (6, 8), A003385 (7, 8), A003387 (9, 8), A003388 (10, 8), A003389 (11, 8), A003390 (12, 8), A003391 (2, 9), A003392 (3, 9), A003393 (4, 9), A003394 (5, 9), A003395 (6, 9), A003396 (7, 9), A003397 (8, 9), A003398 (9, 9), A003399 (10, 9), A004800 (11, 9), A004801 (12, 9), A004802 (2, 10), A004803 (3, 10), A004804 (4, 10), A004805 (5, 10), A004806 (6, 10), A004807 (7, 10), A004808 (8, 10), A004809 (9, 10), A004810 (10, 10), A004811 (11, 10), A004812 (12, 10), A004813 (2, 11), A004814 (3, 11), A004815 (4, 11), A004816 (5, 11), A004817 (6, 11), A004818 (7, 11), A004819 (8, 11), A004820 (9, 11), A004821 (10, 11), A004822 (11, 11), A004823 (12, 11), A047700 (5, 2).

Programs

Python

def aupto(lim):
  p1 = set(i**4 for i in range(1, int(lim**.25)+2) if i**4 <= lim)
  p2 = set(a+b for a in p1 for b in p1 if a+b <= lim)
  p3 = set(apb+c for apb in p2 for c in p1 if apb+c <= lim)
  return sorted(p3)
print(aupto(2400)) # Michael S. Branicky, Mar 18 2021