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 1-3 of 3 results.

A025375 Numbers that are the sum of 4 nonzero squares in 10 or more ways.

Original entry on oeis.org

198, 202, 210, 234, 246, 247, 250, 252, 255, 258, 262, 268, 270, 273, 274, 279, 282, 285, 290, 292, 294, 295, 297, 298, 300, 301, 303, 306, 307, 310, 313, 315, 318, 319, 322, 324, 325, 327, 330, 333, 335, 338, 339, 340, 342, 343, 345, 346, 348, 350, 351, 354, 355, 357
Offset: 1

Views

Author

Keywords

Crossrefs

Formula

{n: A025428(n) >= 10}. - R. J. Mathar, Jun 15 2018

A345156 Numbers that are the sum of four third powers in exactly ten ways.

Original entry on oeis.org

21896, 36225, 48825, 51506, 52416, 53200, 58338, 58968, 60480, 66024, 67851, 70434, 70525, 71155, 72819, 76923, 78624, 78912, 85995, 87507, 88641, 90181, 90783, 91728, 93555, 97552, 98280, 98560, 99008, 99225, 99792, 100170, 103040, 104104, 104265, 104958
Offset: 1

Views

Author

David Consiglio, Jr., Jun 09 2021

Keywords

Comments

Differs from A345155 at term 3 because 46872 = 1^3 + 16^3 + 22^3 + 30^3 = 2^3 + 11^3 + 17^3 + 33^3 = 3^3 + 3^3 + 4^3 + 35^3 = 3^3 + 4^3 + 26^3 + 29^3 = 3^3 + 5^3 + 23^3 + 31^3 = 4^3 + 10^3 + 24^3 + 30^3 = 5^3 + 17^3 + 23^3 + 29^3 = 6^3 + 10^3 + 20^3 + 32^3 = 11^3 + 11^3 + 21^3 + 31^3 = 11^3 + 14^3 + 17^3 + 32^3 = 19^3 + 21^3 + 21^3 + 25^3.

Examples

			21896 is a term because 21896 = 1^3 + 11^3 + 19^3 + 22^3  = 2^3 + 2^3 + 12^3 + 26^3  = 2^3 + 3^3 + 19^3 + 23^3  = 2^3 + 5^3 + 15^3 + 25^3  = 3^3 + 10^3 + 16^3 + 24^3  = 3^3 + 17^3 + 19^3 + 19^3  = 4^3 + 6^3 + 20^3 + 22^3  = 5^3 + 8^3 + 14^3 + 25^3  = 7^3 + 11^3 + 17^3 + 23^3  = 8^3 + 9^3 + 19^3 + 22^3.
		

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, 4):
        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])

A025385 Numbers that are the sum of 4 distinct nonzero squares in exactly 10 ways.

Original entry on oeis.org

351, 375, 398, 406, 422, 429, 434, 441, 442, 446, 466, 501, 515, 527, 537, 539, 549, 551, 562, 573, 586, 588, 589, 611, 647, 653, 659, 667, 673, 684, 697, 709, 817, 820, 829, 836, 844, 853, 940, 1060, 1108, 1172, 1588, 1592, 1624, 1688, 1736, 1768, 1784, 1864, 2248
Offset: 1

Views

Author

Keywords

Crossrefs

Formula

{n: A025443(n) = 10}. - R. J. Mathar, Jun 15 2018
Showing 1-3 of 3 results.