A345643 - OEIS

%I #6 Jul 31 2021 16:25:40

%S 134581976,189642309,219063107,235438301,252277376,275782407,

%T 281935070,290928076,300919884,308188849,309631268,315635200,

%U 322947868,327287951,335530174,342030094,358852218,361946949,379913293,384699424,387538625,391133568

%N Numbers that are the sum of seven fifth powers in ten or more ways.

%H Sean A. Irvine, <a href="/A345643/b345643.txt">Table of n, a(n) for n = 1..5000</a>

%e 189642309 is a term because 189642309 = 1^5 + 1^5 + 2^5 + 19^5 + 30^5 + 36^5 + 40^5 = 1^5 + 2^5 + 6^5 + 7^5 + 18^5 + 20^5 + 45^5 = 1^5 + 6^5 + 21^5 + 27^5 + 29^5 + 36^5 + 39^5 = 2^5 + 9^5 + 19^5 + 23^5 + 33^5 + 33^5 + 40^5 = 3^5 + 4^5 + 21^5 + 28^5 + 29^5 + 34^5 + 40^5 = 6^5 + 7^5 + 11^5 + 29^5 + 33^5 + 36^5 + 37^5 = 7^5 + 12^5 + 17^5 + 20^5 + 29^5 + 32^5 + 42^5 = 8^5 + 11^5 + 21^5 + 21^5 + 22^5 + 34^5 + 42^5 = 13^5 + 14^5 + 14^5 + 19^5 + 21^5 + 38^5 + 40^5 = 20^5 + 21^5 + 24^5 + 24^5 + 24^5 + 38^5 + 38^5.

%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**5 for x in range(1, 1000)]

%o for pos in cwr(power_terms, 7):

%o     tot = sum(pos)

%o     keep[tot] += 1

%o     rets = sorted([k for k, v in keep.items() if v >= 10])

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

%o         print(rets[x])

%Y Cf. A344196, A345576, A345618, A345631, A346259.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 22 2021