A346353 -id:A346353 - cp's OEIS frontend

A345860 Numbers that are the sum of ten fourth powers in exactly eight ways.

6675, 6740, 6755, 6805, 6995, 7015, 7030, 7045, 7095, 7270, 7300, 7365, 7429, 7494, 7525, 7540, 7590, 7605, 7750, 7780, 7845, 7955, 8005, 8085, 8150, 8195, 8215, 8310, 8450, 8470, 8500, 8630, 8644, 8709, 8710, 8790, 8885, 8949, 9124, 9189, 9190, 9250, 9255

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345601 at term 5 because 6820 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

			6740 is a term because 6740 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 8^4.

Sean A. Irvine, Table of n, a(n) for n = 1..8900

Cf. A345601, A345810, A345850, A345859, A345861, A346353.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 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 == 8])
    for x in range(len(rets)):
        print(rets[x])

A346343 Numbers that are the sum of nine fifth powers in exactly eight ways.

Original entry on oeis.org

1431398, 1431640, 1531397, 1952415, 2247917, 2530399, 2652563, 2652860, 2736790, 2851254, 2965588, 3088909, 3148674, 3273590, 3297416, 3329120, 3329362, 3332244, 3336895, 3345442, 3345653, 3361614, 3362217, 3364738, 3553641, 3571549, 3577951, 3609926, 3610155

Offset: 1

David Consiglio, Jr., Jul 13 2021

nonn

Differs from A345625 at term 5 because 1969221 = 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 + 16^5 = 3^5 + 5^5 + 6^5 + 6^5 + 8^5 + 12^5 + 12^5 + 13^5 + 16^5 = 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 12^5 + 12^5 + 13^5 + 16^5 = 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 15^5 = 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 14^5 + 15^5 = 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 14^5 + 15^5 = 1^5 + 4^5 + 5^5 + 8^5 + 9^5 + 13^5 + 13^5 + 13^5 + 15^5 = 1^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 + 14^5 = 1^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5 + 14^5.

			1431398 is a term 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.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345625, A345850, A346333, A346342, A346344, A346353.

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 == 8])
    for x in range(len(rets)):
        print(rets[x])

A345640 Numbers that are the sum of ten fifth powers in eight or more ways.

Original entry on oeis.org

944383, 953139, 953414, 985453, 1118585, 1151438, 1185375, 1192180, 1198879, 1206546, 1209912, 1216569, 1217172, 1218912, 1223321, 1225398, 1226654, 1234631, 1241834, 1242437, 1251195, 1251406, 1252123, 1259685, 1265563, 1265594, 1267937, 1275375, 1281736

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			953139 is a term because 953139 = 1^5 + 1^5 + 1^5 + 3^5 + 8^5 + 10^5 + 10^5 + 10^5 + 12^5 + 13^5 = 1^5 + 2^5 + 2^5 + 6^5 + 6^5 + 8^5 + 9^5 + 9^5 + 12^5 + 14^5 = 2^5 + 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 12^5 + 12^5 + 13^5 = 2^5 + 2^5 + 3^5 + 3^5 + 7^5 + 7^5 + 9^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 11^5 + 11^5 + 12^5 + 13^5 = 2^5 + 3^5 + 3^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5 = 3^5 + 3^5 + 3^5 + 4^5 + 4^5 + 6^5 + 10^5 + 10^5 + 13^5 + 13^5 = 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 9^5 + 10^5 + 15^5.

Sean A. Irvine, Table of n, a(n) for n = 1..1000

Cf. A345601, A345625, A345639, A345641, A346353.

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 >= 8])
    for x in range(len(rets)):
        print(rets[x])

A346352 Numbers that are the sum of ten fifth powers in exactly seven ways.

Original entry on oeis.org

555098, 674040, 683166, 707315, 763631, 777852, 778844, 780945, 783224, 893654, 896500, 897668, 920887, 926616, 927819, 928802, 936850, 937631, 945017, 952897, 953077, 953350, 955178, 963131, 975133, 979482, 984133, 985664, 987257, 991908, 993575, 993606

Offset: 1

David Consiglio, Jr., Jul 13 2021

nonn

Differs from A345639 at term 19 because 944383 = 4^5 + 4^5 + 4^5 + 6^5 + 7^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 + 10^5 + 12^5 + 14^5 = 2^5 + 4^5 + 5^5 + 5^5 + 7^5 + 7^5 + 7^5 + 10^5 + 12^5 + 14^5 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 6^5 + 9^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 4^5 + 6^5 + 7^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5 = 1^5 + 1^5 + 2^5 + 6^5 + 7^5 + 10^5 + 11^5 + 11^5 + 12^5 + 12^5.

			555098 is a term because 555098 = 2^5 + 2^5 + 2^5 + 2^5 + 2^5 + 2^5 + 2^5 + 3^5 + 7^5 + 14^5 = 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 7^5 + 8^5 + 10^5 + 13^5 = 1^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 7^5 + 10^5 + 13^5 = 1^5 + 2^5 + 5^5 + 7^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 13^5 = 4^5 + 4^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 11^5 + 12^5 = 3^5 + 3^5 + 4^5 + 4^5 + 6^5 + 7^5 + 9^5 + 9^5 + 11^5 + 12^5 = 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 + 11^5.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345639, A345859, A346342, A346351, A346353.

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 == 7])
    for x in range(len(rets)):
        print(rets[x])

A346354 Numbers that are the sum of ten fifth powers in exactly nine ways.

Original entry on oeis.org

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

David Consiglio, Jr., Jul 13 2021

nonn

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.

			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.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345641, A345861, A346344, A346353, A346355.

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])

cp's OEIS Frontend

A345860 Numbers that are the sum of ten fourth powers in exactly eight ways.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

A346343 Numbers that are the sum of nine fifth powers in exactly eight ways.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

A345640 Numbers that are the sum of ten fifth powers in eight or more ways.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Python

A346352 Numbers that are the sum of ten fifth powers in exactly seven ways.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

A346354 Numbers that are the sum of ten fifth powers in exactly nine ways.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python