A345771 - OEIS

%I #6 Jul 31 2021 22:50:04

%S 2438,2457,2494,2555,2593,2709,2772,2889,2942,2980,3033,3043,3096,

%T 3160,3195,3241,3250,3257,3276,3402,3427,3437,3467,3556,3582,3592,

%U 3608,3609,3617,3672,3735,3825,3850,3852,3871,3924,3934,3962,3976,3979,3996,3997,4006

%N Numbers that are the sum of six cubes in exactly nine ways.

%C Differs from A345518 at term 14 because 3104 = 1^3 + 2^3 + 7^3 + 8^3 + 8^3 + 12^3  = 1^3 + 5^3 + 5^3 + 5^3 + 10^3 + 12^3  = 2^3 + 2^3 + 4^3 + 4^3 + 6^3 + 14^3  = 2^3 + 3^3 + 4^3 + 7^3 + 11^3 + 11^3  = 2^3 + 3^3 + 5^3 + 6^3 + 10^3 + 12^3  = 2^3 + 7^3 + 8^3 + 8^3 + 9^3 + 10^3  = 3^3 + 3^3 + 5^3 + 6^3 + 8^3 + 13^3  = 4^3 + 5^3 + 7^3 + 8^3 + 9^3 + 11^3  = 5^3 + 5^3 + 5^3 + 9^3 + 10^3 + 10^3  = 5^3 + 6^3 + 6^3 + 6^3 + 10^3 + 11^3  = 6^3 + 6^3 + 6^3 + 6^3 + 8^3 + 12^3.

%H Sean A. Irvine, <a href="/A345771/b345771.txt">Table of n, a(n) for n = 1..1326</a>

%e 2457 is a term because 2457 = 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 12^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 12^3 = 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 11^3 = 1^3 + 5^3 + 5^3 + 7^3 + 7^3 + 9^3 = 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 11^3 = 2^3 + 3^3 + 3^3 + 3^3 + 9^3 + 10^3 = 2^3 + 5^3 + 5^3 + 6^3 + 6^3 + 10^3 = 3^3 + 3^3 + 5^3 + 8^3 + 8^3 + 8^3 = 3^3 + 3^3 + 4^3 + 7^3 + 8^3 + 9^3.

%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**3 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 == 9])

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

%o         print(rets[x])

%Y Cf. A345186, A345518, A345770, A345772, A345781, A345821.

%K nonn

%O 1,1

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