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Revision History for A346341 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing all changes.
Numbers that are the sum of nine fifth powers in exactly six ways.
(history; published version)
#6 by Sean A. Irvine at Sat Jul 31 19:01:00 EDT 2021
STATUS

editing

approved

#5 by Sean A. Irvine at Sat Jul 31 18:59:14 EDT 2021
STATUS

approved

editing

#4 by Sean A. Irvine at Tue Jul 13 21:45:49 EDT 2021
STATUS

editing

approved

#3 by Sean A. Irvine at Tue Jul 13 21:45:47 EDT 2021
LINKS

Sean A. Irvine, <a href="/A346341/b346341.txt">Table of n, a(n) for n = 1..10000</a>

#2 by Sean A. Irvine at Tue Jul 13 21:40:38 EDT 2021
NAME

allocated for Sean A. Irvine

Numbers that are the sum of nine fifth powers in exactly six ways.

DATA

926404, 936607, 952896, 985421, 993574, 993605, 993816, 1075779, 1123321, 1133344, 1134367, 1151406, 1160105, 1166111, 1177144, 1206514, 1209669, 1209847, 1215545, 1225630, 1251130, 1264929, 1265320, 1278611, 1414834, 1422367, 1422609, 1430384, 1431367

OFFSET

1,1

COMMENTS

Differs from A345623 at term 30 because 1431398 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 7^5 + 10^5 + 12^5 + 16^5 = 1^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5 = 1^5 + 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 2^5 + 3^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 + 15^5 = 1^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 1^5 + 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 1^5 + 1^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 1^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5.

EXAMPLE

926404 is a term because 926404 = 2^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 8^5 + 10^5 + 15^5 = 2^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 10^5 + 15^5 = 2^5 + 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 2^5 + 7^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 2^5 + 6^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5 = 1^5 + 1^5 + 4^5 + 4^5 + 7^5 + 11^5 + 12^5 + 12^5 + 12^5.

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, 9):

tot = sum(pos)

keep[tot] += 1

rets = sorted([k for k, v in keep.items() if v == 6])

for x in range(len(rets)):

print(rets[x])

KEYWORD

allocated

nonn,new

AUTHOR

David Consiglio, Jr., Jul 13 2021

STATUS

approved

editing

#1 by Sean A. Irvine at Tue Jul 13 21:35:43 EDT 2021
NAME

allocated for Sean A. Irvine

KEYWORD

allocated

STATUS

approved