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.

A345589 Numbers that are the sum of nine fourth powers in five or more ways.

Original entry on oeis.org

3189, 4149, 4229, 4244, 4309, 4374, 4404, 4419, 4469, 4484, 4549, 4659, 4724, 4853, 4899, 5028, 5093, 5139, 5189, 5204, 5269, 5284, 5349, 5379, 5414, 5444, 5459, 5509, 5524, 5574, 5589, 5619, 5634, 5654, 5684, 5699, 5749, 5764, 5814, 5829, 5939, 6068, 6133
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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])

A345613 Numbers that are the sum of eight fifth powers in five or more ways.

Original entry on oeis.org

926372, 952653, 993573, 1133343, 1414591, 1431366, 1431397, 1447327, 1597928, 1637020, 1663391, 1697685, 1876624, 1933329, 1992377, 1993376, 1993666, 2033328, 2091879, 2175912, 2182160, 2231110, 2280544, 2280575, 2280786, 2281567, 2283668, 2329602, 2345563
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			952653 is a term because 952653 = 2^5 + 2^5 + 6^5 + 7^5 + 9^5 + 12^5 + 12^5 + 13^5 = 2^5 + 2^5 + 7^5 + 7^5 + 9^5 + 11^5 + 11^5 + 14^5 = 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5 = 3^5 + 4^5 + 4^5 + 6^5 + 10^5 + 10^5 + 13^5 + 13^5 = 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 9^5 + 10^5 + 15^5.
		

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, 8):
        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])

A345621 Numbers that are the sum of nine fifth powers in four or more ways.

Original entry on oeis.org

55542, 120350, 143507, 167241, 182549, 192233, 202890, 326685, 327986, 328247, 329028, 329809, 333257, 351722, 358474, 358968, 359210, 359538, 359813, 365404, 367071, 367313, 374034, 374846, 375627, 376619, 377158, 379259, 381157, 383910, 384765, 390396
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			120350 is a term because 120350 = 1^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 + 8^5 + 8^5 = 1^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 + 8^5 = 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 7^5 + 7^5 + 7^5 + 9^5 = 2^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 6^5 + 8^5 + 9^5.
		

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 >= 4])
        for x in range(len(rets)):
            print(rets[x])

A345623 Numbers that are the sum of nine fifth powers in six or more 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., Jun 20 2021

Keywords

Examples

			936607 is a term because 936607 = 1^5 + 1^5 + 2^5 + 7^5 + 10^5 + 11^5 + 11^5 + 12^5 + 12^5 = 1^5 + 3^5 + 4^5 + 7^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5 = 1^5 + 3^5 + 5^5 + 6^5 + 8^5 + 8^5 + 11^5 + 11^5 + 14^5 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 9^5 + 11^5 + 11^5 + 14^5 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 8^5 + 10^5 + 12^5 + 14^5 = 4^5 + 4^5 + 4^5 + 7^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5.
		

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])

A346340 Numbers that are the sum of nine fifth powers in exactly five ways.

Original entry on oeis.org

392063, 392274, 406559, 458875, 519237, 538291, 607947, 663871, 672024, 672055, 672266, 672297, 673586, 673797, 674578, 675390, 680041, 681330, 704582, 704822, 714299, 730260, 732603, 763027, 763324, 765873, 766417, 777820, 780099, 814082, 820887, 825678
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345622 at term 50 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.

Examples

			392063 is a term because 392063 = 2^5 + 2^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 2^5 + 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 2^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 1^5 + 1^5 + 1^5 + 3^5 + 8^5 + 9^5 + 10^5 + 10^5 + 10^5.
		

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 == 5])
        for x in range(len(rets)):
            print(rets[x])

A345637 Numbers that are the sum of ten fifth powers in five or more ways.

Original entry on oeis.org

200009, 220350, 235658, 329271, 329810, 330052, 359211, 359453, 359498, 360298, 367314, 368529, 374519, 374847, 375089, 375870, 376620, 376651, 377159, 377643, 380283, 382622, 384395, 384934, 387035, 388933, 391736, 392064, 392095, 392275, 392306, 392339
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			220350 is a term because 220350 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 3^5 + 9^5 + 11^5 = 1^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 + 8^5 + 8^5 + 10^5 = 1^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 + 8^5 + 10^5 = 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 7^5 + 7^5 + 7^5 + 9^5 + 10^5 = 2^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 6^5 + 8^5 + 9^5 + 10^5.
		

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, 10):
        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])
Showing 1-6 of 6 results.