cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A345799 Numbers that are the sum of nine cubes in exactly seven ways.

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%I A345799 #6 Jul 31 2021 22:32:54
%S A345799 624,629,631,650,657,687,694,707,713,720,727,746,753,755,763,768,777,
%T A345799 779,781,784,786,789,792,796,798,803,807,820,822,824,831,833,848,849,
%U A345799 854,870,873,875,876,879,884,885,889,890,892,898,899,901,902,904,905,906
%N A345799 Numbers that are the sum of nine cubes in exactly seven ways.
%C A345799 Differs from A345546 at term 12 because 744 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 9^3  = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 7^3  = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3  = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 7^3  = 1^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3  = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 8^3  = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3  = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3.
%C A345799 Likely finite.
%H A345799 Sean A. Irvine, <a href="/A345799/b345799.txt">Table of n, a(n) for n = 1..102</a>
%e A345799 629 is a term because 629 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3.
%o A345799 (Python)
%o A345799 from itertools import combinations_with_replacement as cwr
%o A345799 from collections import defaultdict
%o A345799 keep = defaultdict(lambda: 0)
%o A345799 power_terms = [x**3 for x in range(1, 1000)]
%o A345799 for pos in cwr(power_terms, 9):
%o A345799     tot = sum(pos)
%o A345799     keep[tot] += 1
%o A345799     rets = sorted([k for k, v in keep.items() if v == 7])
%o A345799     for x in range(len(rets)):
%o A345799         print(rets[x])
%Y A345799 Cf. A345546, A345789, A345798, A345800, A345809, A345849.
%K A345799 nonn
%O A345799 1,1
%A A345799 _David Consiglio, Jr._, Jun 26 2021