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.

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])

A345598 Numbers that are the sum of ten fourth powers in five or more ways.

Original entry on oeis.org

2935, 3110, 3175, 3190, 3205, 3270, 3445, 3814, 3940, 4150, 4165, 4180, 4195, 4215, 4230, 4245, 4260, 4290, 4310, 4325, 4375, 4390, 4405, 4420, 4435, 4455, 4470, 4485, 4500, 4550, 4565, 4615, 4630, 4660, 4675, 4695, 4725, 4740, 4774, 4805, 4854, 4869, 4870
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			3110 is a term because 3110 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 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, 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])

A345856 Numbers that are the sum of ten fourth powers in exactly four ways.

Original entry on oeis.org

1620, 2660, 2725, 2740, 2835, 2855, 2870, 2900, 2915, 2920, 2950, 2965, 2980, 3000, 3015, 3030, 3045, 3095, 3160, 3220, 3240, 3255, 3285, 3335, 3350, 3415, 3430, 3460, 3479, 3510, 3525, 3544, 3559, 3574, 3589, 3639, 3654, 3685, 3700, 3719, 3765, 3784, 3799
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345597 at term 11 because 2935 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 7^4.

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])

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])

A345807 Numbers that are the sum of ten cubes in exactly five ways.

Original entry on oeis.org

288, 349, 382, 384, 401, 403, 408, 410, 414, 415, 417, 421, 429, 443, 454, 455, 462, 475, 482, 487, 496, 501, 520, 543, 544, 545, 546, 548, 555, 556, 558, 565, 566, 570, 573, 574, 575, 576, 578, 580, 582, 586, 587, 591, 592, 593, 594, 600, 609, 610, 611, 617
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345553 at term 14 because 436 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 7^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3.
Likely finite.

Examples

			349 is a term because 349 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 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, 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])

A346350 Numbers that are the sum of ten fifth powers in exactly five 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, 392275, 392339, 392517, 392581
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345637 at term 29 because 392095 = 2^5 + 2^5 + 2^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 1^5 + 1^5 + 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 7^5 + 10^5 + 12^5 = 2^5 + 2^5 + 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 2^5 + 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 2^5 + 2^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 1^5 + 1^5 + 1^5 + 2^5 + 3^5 + 8^5 + 9^5 + 10^5 + 10^5 + 10^5.

Examples

			200009 is a term 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.
		

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.