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A346354
Numbers that are the sum of ten fifth powers in exactly nine ways.
6
1192180, 1226654, 1242437, 1431399, 1431430, 1431672, 1431883, 1432453, 1432664, 1434765, 1439174, 1441695, 1442718, 1447602, 1448447, 1455346, 1455377, 1464166, 1474431, 1474462, 1475485, 1491978, 1497619, 1531429, 1539173, 1614736, 1671199, 1671410, 1672937
OFFSET
1,1
COMMENTS
Differs from A345641 at term 6 because 1431641 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 6^5 + 7^5 + 10^5 + 12^5 + 16^5 = 1^5 + 1^5 + 4^5 + 6^5 + 7^5 + 7^5 + 8^5 + 9^5 + 12^5 + 16^5 = 1^5 + 3^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5 = 1^5 + 2^5 + 4^5 + 4^5 + 6^5 + 8^5 + 8^5 + 9^5 + 14^5 + 15^5 = 1^5 + 1^5 + 3^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 2^5 + 3^5 + 3^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 + 15^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 1^5 + 2^5 + 2^5 + 3^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 1^5 + 1^5 + 2^5 + 3^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 1^5 + 2^5 + 3^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5.
LINKS
EXAMPLE
1192180 is a term because 1192180 = 5^5 + 5^5 + 5^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 10^5 + 16^5 = 2^5 + 5^5 + 5^5 + 5^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 16^5 = 3^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 6^5 + 8^5 + 13^5 + 15^5 = 3^5 + 4^5 + 4^5 + 4^5 + 6^5 + 7^5 + 7^5 + 7^5 + 13^5 + 15^5 = 2^5 + 2^5 + 2^5 + 3^5 + 8^5 + 8^5 + 9^5 + 9^5 + 12^5 + 15^5 = 1^5 + 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 12^5 + 13^5 + 14^5 = 1^5 + 2^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 12^5 + 13^5 + 13^5 = 1^5 + 2^5 + 2^5 + 2^5 + 4^5 + 11^5 + 11^5 + 12^5 + 12^5 + 13^5 = 6^5 + 9^5 + 9^5 + 10^5 + 11^5 + 11^5 + 11^5 + 11^5 + 11^5 + 11^5.
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, 10):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 9])
for x in range(len(rets)):
print(rets[x])
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved