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 A346338 Numbers that are the sum of nine fifth powers in exactly three ways. 7
 52418, 52449, 52660, 53441, 54519, 54550, 54761, 55690, 57643, 60193, 62294, 69224, 69635, 69666, 69877, 70658, 70955, 70986, 71197, 71325, 71978, 72759, 73001, 74079, 76031, 77410, 78730, 84162, 84459, 84490, 84521, 84701, 84732, 84943, 85185, 85482, 85513 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Differs from A345620 at term 8 because 55542 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5. LINKS Sean A. Irvine, Table of n, a(n) for n = 1..10000 EXAMPLE 52418 is a term because 52418 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5. PROG (Python) from itertools import combinations_with_replacement as cwr from collections import defaultdict keep = defaultdict(lambda: 0) power_terms = [x**5 for x in range(1, 1000)] for pos in cwr(power_terms, 9): tot = sum(pos) keep[tot] += 1 rets = sorted([k for k, v in keep.items() if v == 3]) for x in range(len(rets)): print(rets[x]) CROSSREFS Cf. A345620, A345845, A346328, A346337, A346339, A346348. Sequence in context: A345611 A346328 A345620 * A015359 A263123 A235237 Adjacent sequences: A346335 A346336 A346337 * A346339 A346340 A346341 KEYWORD nonn AUTHOR David Consiglio, Jr., Jul 13 2021 STATUS approved

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Last modified February 21 05:17 EST 2024. Contains 370219 sequences. (Running on oeis4.)