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.

A346333 Numbers that are the sum of eight fifth powers in exactly eight ways.

This page as a plain text file.
%I A346333 #6 Jul 31 2021 19:04:02
%S A346333 8625619,9773236,10036233,10071050,12247994,13180706,13377868,
%T A346333 13662501,14584992,14591744,14611077,15251119,16112362,16374250,
%U A346333 16391025,16472544,16588000,16667851,17059075,17216298,17405300,17917097,18107564,18392902,18470839,18541635
%N A346333 Numbers that are the sum of eight fifth powers in exactly eight ways.
%C A346333 Differs from A345616 at term 2 because 8742208 = 1^5 + 1^5 + 2^5 + 3^5 + 5^5 + 7^5 + 15^5 + 24^5 = 4^5 + 4^5 + 8^5 + 8^5 + 9^5 + 15^5 + 17^5 + 23^5 = 1^5 + 3^5 + 7^5 + 12^5 + 12^5 + 13^5 + 17^5 + 23^5 = 2^5 + 5^5 + 6^5 + 7^5 + 15^5 + 15^5 + 15^5 + 23^5 = 1^5 + 1^5 + 9^5 + 9^5 + 11^5 + 17^5 + 18^5 + 22^5 = 3^5 + 3^5 + 7^5 + 9^5 + 12^5 + 12^5 + 21^5 + 21^5 = 4^5 + 4^5 + 4^5 + 11^5 + 11^5 + 12^5 + 21^5 + 21^5 = 10^5 + 12^5 + 12^5 + 13^5 + 16^5 + 16^5 + 19^5 + 20^5 = 8^5 + 13^5 + 14^5 + 14^5 + 14^5 + 16^5 + 19^5 + 20^5.
%H A346333 Sean A. Irvine, <a href="/A346333/b346333.txt">Table of n, a(n) for n = 1..10000</a>
%e A346333 8625619 is a term because 8625619 = 2^5 + 5^5 + 5^5 + 9^5 + 10^5 + 12^5 + 12^5 + 24^5 = 1^5 + 3^5 + 8^5 + 9^5 + 11^5 + 11^5 + 12^5 + 24^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 16^5 + 23^5 = 1^5 + 3^5 + 3^5 + 4^5 + 11^5 + 17^5 + 18^5 + 22^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 16^5 + 22^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 16^5 + 19^5 + 20^5 = 3^5 + 10^5 + 12^5 + 12^5 + 16^5 + 18^5 + 18^5 + 20^5 = 3^5 + 8^5 + 14^5 + 14^5 + 14^5 + 18^5 + 18^5 + 20^5.
%o A346333 (Python)
%o A346333 from itertools import combinations_with_replacement as cwr
%o A346333 from collections import defaultdict
%o A346333 keep = defaultdict(lambda: 0)
%o A346333 power_terms = [x**5 for x in range(1, 1000)]
%o A346333 for pos in cwr(power_terms, 8):
%o A346333     tot = sum(pos)
%o A346333     keep[tot] += 1
%o A346333     rets = sorted([k for k, v in keep.items() if v == 8])
%o A346333     for x in range(len(rets)):
%o A346333         print(rets[x])
%Y A346333 Cf. A345616, A345840, A346285, A346332, A346334, A346343.
%K A346333 nonn
%O A346333 1,1
%A A346333 _David Consiglio, Jr._, Jul 13 2021