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.

A345573 Numbers that are the sum of seven fourth powers in seven or more ways.

Original entry on oeis.org

16691, 17347, 17971, 19491, 20706, 21252, 21267, 21332, 21507, 21636, 21876, 21956, 22547, 22612, 23156, 23587, 23652, 23827, 23892, 24436, 25107, 25347, 25427, 25716, 25971, 26051, 27812, 29092, 29187, 29332, 29427, 29442, 29636, 29701, 29716, 29956, 29971
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			17347 is a term because 17347 = 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 11^4 = 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 11^4 = 1^4 + 4^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 8^4 + 9^4 + 9^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 9^4 + 9^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 9^4 + 9^4.

Links

Crossrefs

Cf. A345525, A345564, A345572, A345574, A345582, A345629, A345829.

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

A345609 Numbers that are the sum of seven fifth powers in six or more ways.

Original entry on oeis.org

13562501, 14583968, 21555313, 22057487, 22066065, 23089782, 23345024, 24217918, 24401574, 24855016, 24952718, 24993517, 25052501, 25385064, 28608832, 29558618, 30653536, 31613713, 32559143, 33005785, 33533765, 33635825, 33828631, 34267551, 34268332, 35431351
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			14583968 is a term because 14583968 = 1^5 + 4^5 + 14^5 + 16^5 + 19^5 + 21^5 + 23^5 = 2^5 + 4^5 + 14^5 + 14^5 + 20^5 + 22^5 + 22^5 = 4^5 + 5^5 + 10^5 + 15^5 + 20^5 + 21^5 + 23^5 = 6^5 + 8^5 + 9^5 + 15^5 + 15^5 + 20^5 + 25^5 = 6^5 + 8^5 + 14^5 + 14^5 + 14^5 + 16^5 + 26^5 = 6^5 + 10^5 + 12^5 + 12^5 + 16^5 + 16^5 + 26^5.

Links

Crossrefs

Cf. A345572, A345608, A345614, A345629, A345720, A346283.

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 >= 6])
    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])

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])

A345721 Numbers that are the sum of six fifth powers in seven or more ways.

Original entry on oeis.org

1184966816, 1700336000, 1717860100, 1972000800, 2229475325, 2295937600, 2396275200, 2548597632, 2625460992, 2886251808, 3217068800, 3697267200, 3729261536, 3765398725, 4046532448, 4165116967, 4246566632, 4286704224, 4335900525, 4489548050
Offset: 1

Views

Author

David Consiglio, Jr., Jun 24 2021

Keywords

Examples

			1700336000 is a term because 1700336000 = 4^5 + 17^5 + 31^5 + 37^5 + 43^5 + 68^5 = 6^5 + 9^5 + 10^5 + 23^5 + 60^5 + 62^5 = 6^5 + 14^5 + 16^5 + 50^5 + 50^5 + 64^5 = 7^5 + 25^5 + 30^5 + 54^5 + 56^5 + 58^5 = 8^5 + 21^5 + 23^5 + 27^5 + 57^5 + 64^5 = 9^5 + 21^5 + 22^5 + 29^5 + 53^5 + 66^5 = 13^5 + 32^5 + 35^5 + 38^5 + 45^5 + 67^5.

Links

Crossrefs

Cf. A345564, A345629, A345720, A345722, 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, 6):
    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.

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

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