login
A345643
Numbers that are the sum of seven fifth powers in ten or more ways.
6
134581976, 189642309, 219063107, 235438301, 252277376, 275782407, 281935070, 290928076, 300919884, 308188849, 309631268, 315635200, 322947868, 327287951, 335530174, 342030094, 358852218, 361946949, 379913293, 384699424, 387538625, 391133568
OFFSET
1,1
LINKS
EXAMPLE
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.
PROG
(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**5 for x in range(1, 1000)]
for pos in cwr(power_terms, 7):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 10])
for x in range(len(rets)):
print(rets[x])
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved