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 1-5 of 5 results.

A345842 Numbers that are the sum of eight fourth powers in exactly ten ways.

Original entry on oeis.org

17972, 17987, 19492, 19507, 19747, 20116, 21283, 21333, 21413, 21508, 21588, 22067, 22563, 23237, 23252, 23587, 23588, 23603, 23653, 24277, 24452, 24802, 24948, 25603, 26228, 27347, 27683, 27813, 27893, 27973, 28532, 28852, 28853, 28933, 29108, 29173, 29491
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345585 at term 7 because 20787 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 8^4 + 9^4 + 10^4 = 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 6^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 8^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 8^4 + 10^4 = 2^4 + 4^4 + 4^4 + 5^4 + 6^4 + 6^4 + 7^4 + 11^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 8^4 + 10^4 = 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 6^4 + 11^4 = 3^4 + 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 + 8^4.

Examples

			17987 is a term because 17987 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 6^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 6^4 + 6^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 5^4 + 7^4 + 11^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 6^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 + 10^4 = 2^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 + 10^4 = 2^4 + 4^4 + 5^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 = 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 + 10^4 = 3^4 + 5^4 + 6^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4.

Links

Crossrefs

Cf. A345585, A345792, A345832, A345841, A345852, A346335.

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 == 10])
    for x in range(len(rets)):
        print(rets[x])

A346334 Numbers that are the sum of eight fifth powers in exactly nine ways.

Original entry on oeis.org

8742208, 18913169, 19987308, 20135313, 21505583, 21512966, 21563089, 21727552, 22237510, 22256608, 22438990, 22545600, 22686818, 23106589, 23122550, 23189782, 23221517, 23287858, 23346048, 23477344, 23798742, 23847285, 23931325, 24138358, 24385108, 24394139
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345617 at term 2 because 15539667 = 1^5 + 7^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 27^5 = 1^5 + 4^5 + 7^5 + 9^5 + 13^5 + 13^5 + 13^5 + 27^5 = 1^5 + 1^5 + 7^5 + 7^5 + 10^5 + 16^5 + 19^5 + 26^5 = 1^5 + 1^5 + 2^5 + 10^5 + 12^5 + 17^5 + 18^5 + 26^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 23^5 + 24^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 22^5 + 24^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 19^5 + 20^5 + 24^5 = 3^5 + 10^5 + 12^5 + 12^5 + 18^5 + 18^5 + 20^5 + 24^5 = 6^5 + 9^5 + 11^5 + 11^5 + 15^5 + 21^5 + 22^5 + 22^5 = 3^5 + 5^5 + 10^5 + 19^5 + 19^5 + 20^5 + 20^5 + 21^5.

Examples

			8742208 is a term 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.

Links

Crossrefs

Cf. A345617, A345841, A346286, A346333, A346335, A346344.

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

A346259 Numbers that are the sum of seven fifth powers in exactly ten ways.

Original entry on oeis.org

134581976, 189642309, 219063107, 235438301, 252277376, 275782407, 300919884, 308188849, 309631268, 315635200, 327287951, 335530174, 342030094, 358852218, 379913293, 384699424, 387538625, 391133568, 395423876, 405307926, 421322507, 423673757, 425588250
Offset: 1

Views

Author

David Consiglio, Jr., Jul 12 2021

Keywords

Comments

Differs from A345643 at term 7 because 281935070 = 17^5 + 17^5 + 18^5 + 21^5 + 23^5 + 26^5 + 48^5 = 7^5 + 17^5 + 20^5 + 23^5 + 24^5 + 32^5 + 47^5 = 7^5 + 13^5 + 13^5 + 26^5 + 30^5 + 36^5 + 45^5 = 1^5 + 13^5 + 21^5 + 21^5 + 33^5 + 37^5 + 44^5 = 6^5 + 7^5 + 13^5 + 31^5 + 34^5 + 36^5 + 43^5 = 4^5 + 8^5 + 16^5 + 29^5 + 31^5 + 41^5 + 41^5 = 6^5 + 8^5 + 12^5 + 28^5 + 37^5 + 38^5 + 41^5 = 3^5 + 6^5 + 15^5 + 32^5 + 35^5 + 38^5 + 41^5 = 7^5 + 24^5 + 25^5 + 32^5 + 34^5 + 37^5 + 41^5 = 13^5 + 20^5 + 21^5 + 34^5 + 35^5 + 36^5 + 41^5 = 8^5 + 24^5 + 26^5 + 31^5 + 31^5 + 40^5 + 40^5.

Examples

			134581976 is a term 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.

Links

Crossrefs

Cf. A345643, A345832, A346286, A346335, A346365.

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 == 10])
    for x in range(len(rets)):
        print(rets[x])

A346345 Numbers that are the sum of nine fifth powers in exactly ten ways.

Original entry on oeis.org

4157156, 4492410, 4510461, 4915538, 5005474, 5015506, 5179747, 5219655, 5756794, 6323426, 6326519, 6382443, 6423394, 6705284, 6793170, 6861218, 7101038, 7147645, 7147656, 7148679, 7266240, 7280391, 7283268, 7314187, 7413493, 7422352, 7531076, 7651645, 7693425
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345627 at term 5 because 4948274 = 2^5 + 4^5 + 4^5 + 5^5 + 6^5 + 8^5 + 9^5 + 15^5 + 21^5 = 1^5 + 3^5 + 5^5 + 5^5 + 8^5 + 8^5 + 8^5 + 15^5 + 21^5 = 1^5 + 2^5 + 2^5 + 5^5 + 5^5 + 11^5 + 11^5 + 17^5 + 20^5 = 8^5 + 9^5 + 9^5 + 10^5 + 10^5 + 10^5 + 12^5 + 16^5 + 20^5 = 4^5 + 8^5 + 8^5 + 10^5 + 12^5 + 12^5 + 15^5 + 16^5 + 19^5 = 4^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 15^5 + 19^5 = 4^5 + 4^5 + 9^5 + 13^5 + 13^5 + 13^5 + 14^5 + 15^5 + 19^5 = 6^5 + 6^5 + 9^5 + 9^5 + 12^5 + 12^5 + 14^5 + 18^5 + 18^5 = 1^5 + 8^5 + 8^5 + 12^5 + 12^5 + 14^5 + 14^5 + 17^5 + 18^5 = 1^5 + 8^5 + 9^5 + 9^5 + 13^5 + 14^5 + 16^5 + 17^5 + 17^5 = 3^5 + 7^5 + 7^5 + 10^5 + 12^5 + 16^5 + 16^5 + 16^5 + 17^5.

Examples

			4157156 is a term because 4157156 = 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 9^5 + 21^5 = 1^5 + 1^5 + 3^5 + 4^5 + 5^5 + 5^5 + 8^5 + 8^5 + 21^5 = 1^5 + 4^5 + 4^5 + 8^5 + 10^5 + 12^5 + 12^5 + 16^5 + 19^5 = 1^5 + 4^5 + 4^5 + 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 19^5 = 5^5 + 5^5 + 5^5 + 5^5 + 7^5 + 9^5 + 15^5 + 17^5 + 18^5 = 3^5 + 3^5 + 5^5 + 6^5 + 9^5 + 10^5 + 16^5 + 16^5 + 18^5 = 1^5 + 1^5 + 5^5 + 5^5 + 13^5 + 13^5 + 15^5 + 15^5 + 18^5 = 2^5 + 3^5 + 4^5 + 4^5 + 10^5 + 14^5 + 16^5 + 16^5 + 17^5 = 11^5 + 11^5 + 12^5 + 12^5 + 12^5 + 12^5 + 13^5 + 16^5 + 17^5 = 2^5 + 2^5 + 2^5 + 5^5 + 12^5 + 15^5 + 16^5 + 16^5 + 16^5.

Links

Crossrefs

Cf. A345627, A345852, A346335, A346344, A346355.

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 == 10])
    for x in range(len(rets)):
        print(rets[x])

A345618 Numbers that are the sum of eight fifth powers in ten or more ways.

Original entry on oeis.org

15539667, 22932525, 24393600, 24650406, 24952961, 24953742, 25054306, 25142513, 25201550, 25423794, 26001294, 26851552, 27396567, 27988486, 28609075, 29309819, 29558650, 31052406, 31794336, 32223105, 32527286, 32610600, 32807777, 32890541, 32998317, 33015125
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			15539667 is a term because 15539667 = 1^5 + 1^5 + 2^5 + 10^5 + 12^5 + 17^5 + 18^5 + 26^5 = 1^5 + 1^5 + 7^5 + 7^5 + 10^5 + 16^5 + 19^5 + 26^5 = 1^5 + 4^5 + 7^5 + 9^5 + 13^5 + 13^5 + 13^5 + 27^5 = 1^5 + 7^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 27^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 23^5 + 24^5 = 3^5 + 5^5 + 10^5 + 19^5 + 19^5 + 20^5 + 20^5 + 21^5 = 3^5 + 10^5 + 12^5 + 12^5 + 18^5 + 18^5 + 20^5 + 24^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 22^5 + 24^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 19^5 + 20^5 + 24^5 = 6^5 + 9^5 + 11^5 + 11^5 + 15^5 + 21^5 + 22^5 + 22^5.

Crossrefs

Cf. A345585, A345617, A345627, A345643, A346335.

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 >= 10])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-5 of 5 results.

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