A344196 - OEIS

%I #37 May 10 2024 02:20:37

%S 55302546200,89999127392,96110537743,104484239200,120492759200,

%T 121258798144,127794946400,133364991375,135030535200,136156575744,

%U 151305014432,155434423925,174388570400,177099008000,179272687000,180336745600,182844944832,184948721056,187873845500

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

%H Sean A. Irvine, <a href="/A344196/b344196.txt">Table of n, a(n) for n = 1..61</a>

%e 89999127392 =  4^5 + 36^5 + 39^5 +  40^5 +  90^5 + 153^5

%e             =  8^5 + 21^5 + 90^5 + 109^5 + 119^5 + 135^5

%e             =  8^5 + 28^5 + 98^5 + 102^5 + 104^5 + 142^5

%e             = 10^5 + 38^5 + 74^5 + 102^5 + 118^5 + 140^5

%e             = 13^5 + 51^5 + 64^5 +  98^5 + 112^5 + 144^5

%e             = 18^5 + 44^5 + 66^5 +  98^5 + 112^5 + 144^5

%e             = 18^5 + 52^5 + 72^5 +  78^5 + 118^5 + 144^5

%e             = 28^5 + 60^5 + 63^5 +  65^5 + 124^5 + 142^5

%e             = 36^5 + 53^5 + 62^5 +  63^5 + 129^5 + 139^5

%e             = 39^5 + 41^5 + 64^5 +  91^5 +  98^5 + 149^5

%e so 89999127392 is a term.

%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, 6):

%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. A345567, A345643, A345723, A346365.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 25 2021