A345816 - OEIS

%I #8 Mar 11 2023 14:04:28

%S 6626,6691,6866,9251,9491,10115,10706,10786,11555,12595,14225,14691,

%T 14771,15315,15330,15570,16051,16595,16660,16675,16850,17090,17091,

%U 17236,17316,17331,17346,17860,17875,17940,17955,18195,18786,18851,19155,19170,19475,19490

%N Numbers that are the sum of six fourth powers in exactly four ways.

%C Differs from A345561 at term 16 because 15395 = 1^4 + 1^4 + 1^4 + 6^4 + 8^4 + 10^4  = 1^4 + 2^4 + 5^4 + 8^4 + 8^4 + 9^4  = 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 10^4  = 3^4 + 5^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 11^4.

%H Sean A. Irvine, <a href="/A345816/b345816.txt">Table of n, a(n) for n = 1..10000</a>

%e 6691 is a term because 6691 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4.

%t Select[Range[20000],Count[PowersRepresentations[#,6,4],_?(#[[1]]>0&)]==4&] (* _Harvey P. Dale_, Mar 11 2023 *)

%o (Python)

%o from itertools import combinations_with_replacement as cwr

%o from collections import defaultdict

%o keep = defaultdict(lambda: 0)

%o power_terms = [x**4 for x in range(1, 1000)]

%o for pos in cwr(power_terms, 6):

%o     tot = sum(pos)

%o     keep[tot] += 1

%o     rets = sorted([k for k, v in keep.items() if v == 4])

%o     for x in range(len(rets)):

%o         print(rets[x])

%Y Cf. A344355, A345561, A345766, A345815, A345817, A345826, A346359.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 26 2021