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 1-7 of 7 results.

A345590 Numbers that are the sum of nine fourth powers in six or more ways.

Original entry on oeis.org

4469, 4484, 5444, 5459, 5524, 5589, 5699, 5764, 6629, 6659, 6674, 6694, 6724, 6739, 6789, 6804, 6854, 6869, 6884, 6899, 6914, 6934, 6949, 6964, 6979, 7014, 7029, 7044, 7094, 7109, 7154, 7219, 7269, 7284, 7334, 7348, 7349, 7413, 7459, 7478, 7494, 7523, 7524
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			4484 is a term because 4484 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4.
		

Crossrefs

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, 9):
        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])

A345838 Numbers that are the sum of eight fourth powers in exactly six ways.

Original entry on oeis.org

6723, 6788, 6853, 6898, 6963, 7028, 7938, 8068, 8178, 8308, 8483, 8963, 9173, 9348, 9413, 9493, 9668, 9763, 9828, 10003, 10132, 10258, 10277, 10307, 10628, 10708, 10738, 10788, 10933, 10978, 11108, 11123, 11188, 11347, 11363, 11428, 11492, 11652, 11668, 11843
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345581 at term 8 because 8003 = 1^4 + 1^4 + 1^4 + 2^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 9^4 = 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 7^4.

Examples

			6788 is a term because 6788 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.
		

Crossrefs

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 == 6])
        for x in range(len(rets)):
            print(rets[x])

A345847 Numbers that are the sum of nine fourth powers in exactly five ways.

Original entry on oeis.org

3189, 4149, 4229, 4244, 4309, 4374, 4404, 4419, 4549, 4659, 4724, 4853, 4899, 5028, 5093, 5139, 5189, 5204, 5269, 5284, 5349, 5379, 5414, 5509, 5574, 5619, 5634, 5654, 5684, 5749, 5814, 5829, 5939, 6068, 6133, 6179, 6308, 6419, 6564, 6594, 6614, 6644, 6709
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345589 at term 9 because 4469 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 8^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4.

Examples

			4149 is a term because 4149 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 = 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4.
		

Crossrefs

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, 9):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v == 5])
        for x in range(len(rets)):
            print(rets[x])

A345849 Numbers that are the sum of nine fourth powers in exactly seven ways.

Original entry on oeis.org

6739, 6854, 6979, 7029, 7044, 7094, 7109, 7269, 7284, 7844, 7909, 7939, 8004, 8149, 8194, 8244, 8309, 8389, 8434, 8628, 8739, 8868, 8979, 9059, 9189, 9254, 9414, 9509, 9524, 9668, 9684, 9734, 9814, 9829, 9843, 9844, 9908, 9909, 9924, 9989, 10019, 10038, 10084
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345591 at term 2 because 6804 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

Examples

			6804 is a term because 6804 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.
		

Crossrefs

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, 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])

A345798 Numbers that are the sum of nine cubes in exactly six ways.

Original entry on oeis.org

472, 498, 505, 507, 524, 596, 598, 605, 636, 643, 655, 661, 662, 669, 672, 676, 681, 688, 690, 692, 696, 706, 718, 722, 725, 728, 729, 731, 732, 737, 739, 742, 748, 749, 750, 751, 756, 765, 772, 782, 783, 785, 787, 788, 791, 793, 794, 800, 801, 802, 808, 810
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345545 at term 9 because 624 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 = 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3.
Likely finite.

Examples

			498 is a term because 498 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3.
		

Crossrefs

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**3 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 == 6])
        for x in range(len(rets)):
            print(rets[x])

A345858 Numbers that are the sum of ten fourth powers in exactly six ways.

Original entry on oeis.org

3175, 4150, 4230, 4390, 4405, 4455, 4470, 4500, 4550, 4565, 4630, 4725, 4740, 4915, 4980, 5094, 5109, 5155, 5190, 5205, 5220, 5270, 5285, 5350, 5365, 5395, 5430, 5475, 5635, 5655, 5735, 5910, 5955, 6020, 6069, 6084, 6149, 6195, 6214, 6324, 6389, 6435, 6500
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345599 at term 8 because 4485 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 8^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4.

Examples

			4150 is a term because 4150 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 = 1^4 + 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 7^4.
		

Crossrefs

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, 10):
        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])

A346341 Numbers that are the sum of nine fifth powers in exactly six ways.

Original entry on oeis.org

926404, 936607, 952896, 985421, 993574, 993605, 993816, 1075779, 1123321, 1133344, 1134367, 1151406, 1160105, 1166111, 1177144, 1206514, 1209669, 1209847, 1215545, 1225630, 1251130, 1264929, 1265320, 1278611, 1414834, 1422367, 1422609, 1430384, 1431367
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345623 at term 30 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

			926404 is a term because 926404 = 2^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 8^5 + 10^5 + 15^5 = 2^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 10^5 + 15^5 = 2^5 + 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 2^5 + 7^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 2^5 + 6^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5 = 1^5 + 1^5 + 4^5 + 4^5 + 7^5 + 11^5 + 12^5 + 12^5 + 12^5.
		

Crossrefs

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 == 6])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-7 of 7 results.