cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A345830 Numbers that are the sum of seven fourth powers in exactly eight ways.

Original entry on oeis.org

21252, 21507, 21636, 21876, 25347, 30372, 31412, 31652, 32116, 32356, 33811, 33907, 35637, 35652, 35892, 36261, 37827, 38052, 38596, 38676, 39267, 39347, 39971, 39972, 40212, 40452, 41506, 41731, 41987, 42147, 42227, 42357, 42532, 42771, 42852, 43027, 43282
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345574 at term 1 because 19491 = 1^4 + 1^4 + 1^4 + 6^4 + 8^4 + 8^4 + 10^4 = 1^4 + 2^4 + 4^4 + 4^4 + 7^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 8^4 + 11^4 = 2^4 + 4^4 + 4^4 + 5^4 + 6^4 + 7^4 + 11^4 = 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 8^4 + 10^4 = 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 11^4 = 3^4 + 5^4 + 7^4 + 8^4 + 8^4 + 8^4 + 8^4.

Examples

			21252 is a term because 21252 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 12^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 1^4 + 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 1^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4 = 3^4 + 4^4 + 6^4 + 7^4 + 8^4 + 9^4 + 9^4.

Links

Crossrefs

Cf. A345574, A345780, A345820, A345829, A345831, A345840, A346285.

Programs

  • Python
    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, 7):
    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])

A345630 Numbers that are the sum of seven fifth powers in eight or more ways.

Original entry on oeis.org

36620574, 80552143, 81401376, 82078424, 92347417, 93653176, 94486699, 94626949, 98873875, 105674625, 110276376, 121050874, 124732805, 125959393, 127808693, 129228307, 130298618, 134581976, 144209018, 145340799, 147245218, 147898763, 151727082
Offset: 1

Views

Author

David Consiglio, Jr., Jun 22 2021

Keywords

Examples

			80552143 is a term because 80552143 = 1^5 + 4^5 + 21^5 + 21^5 + 23^5 + 29^5 + 34^5 = 1^5 + 8^5 + 14^5 + 23^5 + 23^5 + 32^5 + 32^5 = 1^5 + 8^5 + 16^5 + 19^5 + 27^5 + 28^5 + 34^5 = 3^5 + 12^5 + 13^5 + 14^5 + 28^5 + 31^5 + 32^5 = 3^5 + 14^5 + 17^5 + 18^5 + 18^5 + 27^5 + 36^5 = 4^5 + 11^5 + 13^5 + 22^5 + 23^5 + 24^5 + 36^5 = 5^5 + 6^5 + 19^5 + 20^5 + 23^5 + 24^5 + 36^5 = 6^5 + 23^5 + 25^5 + 25^5 + 25^5 + 29^5 + 30^5.

Links

Crossrefs

Cf. A345574, A345616, A345629, A345631, A345722, A346285.

Programs

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

A346284 Numbers that are the sum of seven fifth powers in exactly seven ways.

Original entry on oeis.org

28608832, 35663099, 36090526, 46998599, 51095638, 52541851, 54233651, 54827543, 54886349, 61263643, 61634374, 63514593, 64810976, 65198607, 66708676, 67887843, 70979107, 72970305, 74002457, 74115801, 74132607, 74487093, 75044651, 75378359, 75612250, 75997624
Offset: 1

Views

Author

David Consiglio, Jr., Jul 12 2021

Keywords

Comments

Differs from A345629 at term 4 because 36620574 = 4^5 + 9^5 + 14^5 + 17^5 + 18^5 + 21^5 + 31^5 = 1^5 + 12^5 + 13^5 + 14^5 + 20^5 + 24^5 + 30^5 = 8^5 + 9^5 + 12^5 + 13^5 + 16^5 + 27^5 + 29^5 = 5^5 + 7^5 + 7^5 + 20^5 + 23^5 + 23^5 + 29^5 = 17^5 + 18^5 + 20^5 + 20^5 + 20^5 + 20^5 + 29^5 = 2^5 + 7^5 + 14^5 + 14^5 + 23^5 + 26^5 + 28^5 = 4^5 + 8^5 + 8^5 + 17^5 + 23^5 + 27^5 + 27^5 = 2^5 + 3^5 + 14^5 + 18^5 + 24^5 + 26^5 + 27^5.

Examples

			28608832 is a term because 28608832 = 3^5 + 4^5 + 4^5 + 8^5 + 10^5 + 24^5 + 29^5 = 2^5 + 12^5 + 12^5 + 16^5 + 18^5 + 24^5 + 28^5 = 6^5 + 6^5 + 14^5 + 14^5 + 22^5 + 22^5 + 28^5 = 7^5 + 8^5 + 13^5 + 14^5 + 17^5 + 26^5 + 27^5 = 2^5 + 8^5 + 11^5 + 19^5 + 22^5 + 23^5 + 27^5 = 6^5 + 6^5 + 12^5 + 14^5 + 24^5 + 24^5 + 26^5 = 7^5 + 7^5 + 8^5 + 16^5 + 24^5 + 25^5 + 25^5.

Links

Crossrefs

Cf. A345629, A345829, A346283, A346285, A346332, A346362.

Programs

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

A346286 Numbers that are the sum of seven fifth powers in exactly nine ways.

Original entry on oeis.org

110276376, 124732805, 127808693, 130298618, 188116743, 202274051, 202686274, 203343582, 230909843, 233137574, 233549568, 234250752, 244250335, 251138524, 253480833, 254017026, 254380543, 265006057, 265072501, 273628068, 279536432, 279770326, 280361082
Offset: 1

Views

Author

David Consiglio, Jr., Jul 12 2021

Keywords

Comments

Differs from A345631 at term 5 because 134581976 = 1^5 + 14^5 + 17^5 + 18^5 + 26^5 + 31^5 + 39^5 = 1^5 + 1^5 + 10^5 + 12^5 + 19^5 + 35^5 + 38^5 = 8^5 + 11^5 + 12^5 + 17^5 + 27^5 + 33^5 + 38^5 = 3^5 + 12^5 + 12^5 + 21^5 + 28^5 + 32^5 + 38^5 = 4^5 + 11^5 + 13^5 + 22^5 + 24^5 + 36^5 + 36^5 = 5^5 + 6^5 + 19^5 + 20^5 + 24^5 + 36^5 + 36^5 = 1^5 + 4^5 + 21^5 + 21^5 + 29^5 + 34^5 + 36^5 = 1^5 + 8^5 + 14^5 + 23^5 + 32^5 + 32^5 + 36^5 = 6^5 + 25^5 + 25^5 + 25^5 + 29^5 + 30^5 + 36^5 = 12^5 + 20^5 + 21^5 + 26^5 + 28^5 + 34^5 + 35^5.

Examples

			110276376 is a term because 110276376 = 1^5 + 3^5 + 5^5 + 7^5 + 17^5 + 23^5 + 40^5 = 5^5 + 10^5 + 16^5 + 16^5 + 19^5 + 20^5 + 40^5 = 1^5 + 8^5 + 14^5 + 16^5 + 21^5 + 27^5 + 39^5 = 7^5 + 8^5 + 11^5 + 14^5 + 16^5 + 33^5 + 37^5 = 4^5 + 7^5 + 8^5 + 13^5 + 26^5 + 31^5 + 37^5 = 1^5 + 5^5 + 6^5 + 20^5 + 28^5 + 29^5 + 37^5 = 3^5 + 3^5 + 7^5 + 18^5 + 27^5 + 32^5 + 36^5 = 6^5 + 12^5 + 18^5 + 25^5 + 30^5 + 31^5 + 34^5 = 6^5 + 10^5 + 20^5 + 27^5 + 27^5 + 33^5 + 33^5.

Links

Crossrefs

Cf. A345631, A345831, A346259, A346285, A346334, A346364.

Programs

  • 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 == 9])
    for x in range(len(rets)):
        print(rets[x])

A346333 Numbers that are the sum of eight fifth powers in exactly eight ways.

Original entry on oeis.org

8625619, 9773236, 10036233, 10071050, 12247994, 13180706, 13377868, 13662501, 14584992, 14591744, 14611077, 15251119, 16112362, 16374250, 16391025, 16472544, 16588000, 16667851, 17059075, 17216298, 17405300, 17917097, 18107564, 18392902, 18470839, 18541635
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345616 at term 2 because 8742208 = 1^5 + 1^5 + 2^5 + 3^5 + 5^5 + 7^5 + 15^5 + 24^5 = 4^5 + 4^5 + 8^5 + 8^5 + 9^5 + 15^5 + 17^5 + 23^5 = 1^5 + 3^5 + 7^5 + 12^5 + 12^5 + 13^5 + 17^5 + 23^5 = 2^5 + 5^5 + 6^5 + 7^5 + 15^5 + 15^5 + 15^5 + 23^5 = 1^5 + 1^5 + 9^5 + 9^5 + 11^5 + 17^5 + 18^5 + 22^5 = 3^5 + 3^5 + 7^5 + 9^5 + 12^5 + 12^5 + 21^5 + 21^5 = 4^5 + 4^5 + 4^5 + 11^5 + 11^5 + 12^5 + 21^5 + 21^5 = 10^5 + 12^5 + 12^5 + 13^5 + 16^5 + 16^5 + 19^5 + 20^5 = 8^5 + 13^5 + 14^5 + 14^5 + 14^5 + 16^5 + 19^5 + 20^5.

Examples

			8625619 is a term because 8625619 = 2^5 + 5^5 + 5^5 + 9^5 + 10^5 + 12^5 + 12^5 + 24^5 = 1^5 + 3^5 + 8^5 + 9^5 + 11^5 + 11^5 + 12^5 + 24^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 16^5 + 23^5 = 1^5 + 3^5 + 3^5 + 4^5 + 11^5 + 17^5 + 18^5 + 22^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 16^5 + 22^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 16^5 + 19^5 + 20^5 = 3^5 + 10^5 + 12^5 + 12^5 + 16^5 + 18^5 + 18^5 + 20^5 = 3^5 + 8^5 + 14^5 + 14^5 + 14^5 + 18^5 + 18^5 + 20^5.

Links

Crossrefs

Cf. A345616, A345840, A346285, A346332, A346334, A346343.

Programs

  • 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, 8):
    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])

A346363 Numbers that are the sum of six fifth powers in exactly eight ways.

Original entry on oeis.org

2295937600, 4335900525, 6251954544, 8986552608, 13413708308, 14539246326, 15277569450, 15728636000, 16770321920, 16873011232, 17572402769, 17713454592, 17960776999, 18190647200, 19621666592, 20570070125, 20827689300, 22322555200, 23461554774, 23613244800
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

This sequence differs from A345722:
9085584992 = 24^5 + 38^5 + 42^5 + 48^5 + 54^5 + 96^5
= 21^5 + 34^5 + 38^5 + 43^5 + 74^5 + 92^5
= 8^5 + 34^5 + 38^5 + 62^5 + 68^5 + 92^5
= 18^5 + 18^5 + 44^5 + 64^5 + 66^5 + 92^5
= 13^5 + 18^5 + 51^5 + 64^5 + 64^5 + 92^5
= 8^5 + 38^5 + 41^5 + 47^5 + 79^5 + 89^5
= 5^5 + 23^5 + 29^5 + 45^5 + 85^5 + 85^5
= 8^5 + 23^5 + 41^5 + 64^5 + 82^5 + 84^5
= 12^5 + 18^5 + 38^5 + 72^5 + 78^5 + 84^5,
so 9085584992 is in A345722, but is not in this sequence.

Examples

			2295937600 =  4^5 + 21^5 + 38^5 + 42^5 + 43^5 + 72^5
           =  8^5 + 16^5 + 30^5 + 42^5 + 54^5 + 70^5
           =  8^5 + 13^5 + 36^5 + 37^5 + 57^5 + 69^5
           = 14^5 + 16^5 + 16^5 + 52^5 + 54^5 + 68^5
           =  3^5 + 14^5 + 32^5 + 44^5 + 61^5 + 66^5
           =  4^5 + 18^5 + 22^5 + 52^5 + 58^5 + 66^5
           = 10^5 + 14^5 + 26^5 + 42^5 + 63^5 + 65^5
           =  1^5 +  7^5 + 34^5 + 57^5 + 58^5 + 63^5,
so 2295937600 is a term.

Links

Crossrefs

Cf. A345722, A345820, A346285, A346362, A346364.

Programs

  • 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, 6):
    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])
Showing 1-6 of 6 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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