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.

A345839 Numbers that are the sum of eight fourth powers in exactly seven ways.

Original entry on oeis.org

8003, 8243, 9043, 9218, 9283, 9523, 10372, 10803, 10868, 10948, 11043, 11412, 11557, 11587, 12083, 12692, 12932, 13188, 13333, 13508, 13972, 14147, 14387, 14883, 14933, 14948, 14963, 15013, 15028, 15093, 15173, 15268, 15317, 15332, 15397, 15412, 15413, 15492
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345582 at term 19 because 13268 = 1^4 + 1^4 + 1^4 + 2^4 + 6^4 + 6^4 + 8^4 + 9^4 = 1^4 + 1^4 + 2^4 + 4^4 + 7^4 + 7^4 + 8^4 + 8^4 = 1^4 + 1^4 + 3^4 + 6^4 + 6^4 + 7^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 9^4 = 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 9^4 = 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 9^4 = 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 10^4.

Examples

			8243 is a term because 8243 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 6^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 8^4 = 2^4 + 4^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 3^4 + 4^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4.

Links

Crossrefs

Cf. A345582, A345789, A345829, A345838, A345840, A345849, A346332.

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, 8):
    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])

A345615 Numbers that are the sum of eight fifth powers in seven or more ways.

Original entry on oeis.org

4104553, 4915506, 6011150, 6027989, 6323394, 6563733, 6622231, 6776363, 6785394, 7982834, 8181481, 8288806, 8625619, 8658144, 8710484, 8742208, 8773477, 8932244, 8996669, 9252219, 9253706, 9311478, 9773236, 9904983, 9976120, 10036233, 10045233, 10053008
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			4915506 is a term because 4915506 = 1^5 + 3^5 + 5^5 + 5^5 + 8^5 + 8^5 + 15^5 + 21^5 = 1^5 + 8^5 + 12^5 + 12^5 + 14^5 + 14^5 + 17^5 + 18^5 = 1^5 + 9^5 + 9^5 + 13^5 + 14^5 + 16^5 + 17^5 + 17^5 = 2^5 + 4^5 + 4^5 + 5^5 + 6^5 + 9^5 + 15^5 + 21^5 = 4^5 + 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 15^5 + 19^5 = 4^5 + 8^5 + 10^5 + 12^5 + 12^5 + 15^5 + 16^5 + 19^5 = 9^5 + 9^5 + 10^5 + 10^5 + 10^5 + 12^5 + 16^5 + 20^5.

Links

Crossrefs

Cf. A345582, A345614, A345616, A345624, A345629, A346332.

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 >= 7])
    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])

A346331 Numbers that are the sum of eight fifth powers in exactly six ways.

Original entry on oeis.org

1431397, 2593811, 3329119, 3345410, 3609912, 3800722, 3932480, 4093604, 4096697, 4114187, 4129433, 4154031, 4169869, 4377714, 4451412, 4475603, 4484634, 4501409, 4730845, 4756642, 4882770, 4912477, 4970823, 5003645, 5112274, 5259111, 5449985, 5523925, 5722189
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345614 at term 10 because 4104553 = 1^5 + 1^5 + 2^5 + 3^5 + 3^5 + 5^5 + 7^5 + 21^5 = 3^5 + 3^5 + 4^5 + 6^5 + 8^5 + 14^5 + 16^5 + 19^5 = 3^5 + 3^5 + 3^5 + 7^5 + 9^5 + 12^5 + 18^5 + 18^5 = 3^5 + 4^5 + 4^5 + 4^5 + 11^5 + 11^5 + 18^5 + 18^5 = 1^5 + 1^5 + 4^5 + 7^5 + 10^5 + 16^5 + 16^5 + 18^5 = 7^5 + 11^5 + 11^5 + 13^5 + 14^5 + 15^5 + 16^5 + 16^5 = 6^5 + 12^5 + 12^5 + 13^5 + 13^5 + 15^5 + 16^5 + 16^5.

Examples

			1431397 is a term because 1431397 = 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5 = 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 1^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5.

Links

Crossrefs

Cf. A345614, A345838, A346283, A346330, A346332, A346341.

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 == 6])
    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])

A346342 Numbers that are the sum of nine fifth powers in exactly seven ways.

Original entry on oeis.org

1431429, 1439173, 1447570, 1504636, 1597929, 1671167, 1696159, 1697686, 1697928, 1778835, 1936454, 1975049, 2017344, 2092122, 2182161, 2198967, 2208680, 2280818, 2283911, 2289343, 2314335, 2329845, 2340319, 2345806, 2362370, 2388651, 2497771, 2529407, 2530672
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345624 at term 1 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.

Examples

			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.

Links

Crossrefs

Cf. A345624, A345849, A346332, A346341, A346343, A346352.

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, 9):
    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])
Showing 1-6 of 6 results.

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