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.

A345597 Numbers that are the sum of ten fourth powers in four or more ways.

Original entry on oeis.org

1620, 2660, 2725, 2740, 2835, 2855, 2870, 2900, 2915, 2920, 2935, 2950, 2965, 2980, 3000, 3015, 3030, 3045, 3095, 3110, 3160, 3175, 3190, 3205, 3220, 3240, 3255, 3270, 3285, 3335, 3350, 3415, 3430, 3445, 3460, 3479, 3510, 3525, 3544, 3559, 3574, 3589, 3639
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			2660 is a term because 2660 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 6^4 + 6^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 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, 10):
        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])

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])

A345635 Numbers that are the sum of ten fifth powers in three or more ways.

Original entry on oeis.org

8194, 21940, 52419, 52450, 52481, 52661, 52692, 52903, 53442, 53473, 53684, 54465, 54520, 54551, 54582, 54762, 54793, 55004, 55543, 55574, 55691, 55722, 55785, 55933, 56566, 56714, 57644, 57675, 57886, 58667, 58815, 60194, 60225, 60436, 60768, 61217, 62295
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			21940 is a term because 21940 = 1^5 + 2^5 + 2^5 + 5^5 + 5^5 + 5^5 + 5^5 + 5^5 + 5^5 + 5^5 = 1^5 + 3^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 = 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5.
		

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 >= 3])
        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])

A346349 Numbers that are the sum of ten fifth powers in exactly four ways.

Original entry on oeis.org

55543, 55574, 55785, 56566, 58667, 63318, 72349, 73002, 85186, 86506, 87287, 87529, 88310, 103134, 111498, 113599, 114591, 118250, 119031, 120351, 120382, 120593, 121374, 123475, 128126, 134475, 134878, 135201, 137157, 142008, 142219, 143000, 143211, 143506
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345636 at term 92 because 200009 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 6^5 + 6^5 + 6^5 + 9^5 + 10^5 = 1^5 + 3^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 8^5 + 8^5 + 10^5 = 1^5 + 3^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 8^5 + 10^5 = 2^5 + 2^5 + 4^5 + 4^5 + 6^5 + 8^5 + 8^5 + 8^5 + 8^5 + 9^5 = 1^5 + 2^5 + 3^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5.

Examples

			55543 is a term because 55543 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5.
		

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 == 4])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-5 of 5 results.