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 A345537 Numbers that are the sum of eight cubes in seven or more ways. 8
 902, 908, 921, 938, 958, 963, 970, 977, 982, 984, 991, 996, 1003, 1008, 1010, 1017, 1019, 1028, 1029, 1033, 1047, 1054, 1055, 1058, 1061, 1062, 1070, 1073, 1075, 1080, 1087, 1090, 1091, 1094, 1096, 1097, 1099, 1104, 1106, 1108, 1110, 1111, 1113, 1115, 1116 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 LINKS Sean A. Irvine, Table of n, a(n) for n = 1..10000 EXAMPLE 908 is a term because 908 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3. PROG (Python) from itertools import combinations_with_replacement as cwr from collections import defaultdict keep = defaultdict(lambda: 0) power_terms = [x**3 for x in range(1, 1000)] for pos in cwr(power_terms, 8): tot = sum(pos) keep[tot] += 1 rets = sorted([k for k, v in keep.items() if v >= 7]) for x in range(len(rets)): print(rets[x]) CROSSREFS Cf. A345494, A345525, A345536, A345538, A345546, A345582, A345789. Sequence in context: A031738 A268585 A224611 * A345789 A031528 A158406 Adjacent sequences: A345534 A345535 A345536 * A345538 A345539 A345540 KEYWORD nonn AUTHOR David Consiglio, Jr., Jun 20 2021 STATUS approved

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Last modified October 2 13:09 EDT 2023. Contains 365833 sequences. (Running on oeis4.)