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.

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A335842 Nonnegative differences of positive cubes and positive tetrahedral numbers.

Original entry on oeis.org

0, 1, 4, 5, 7, 8, 17, 23, 26, 29, 31, 36, 41, 44, 49, 51, 54, 57, 60, 63, 68, 69, 77, 83, 90, 93, 96, 99, 105, 115, 121, 122, 123, 124, 132, 144, 148, 149, 151, 160, 169, 173, 178, 180, 181, 184, 188, 191, 196, 206, 211, 212, 215, 223, 226, 230, 258, 259, 274
Offset: 1

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Author

Ya-Ping Lu, Jun 26 2020

Keywords

Comments

The sequence is the difference between the cubic number (A000578) and the tetrahedral number (A000292) such that terms are of the form A000578(i) - A000292(j), where A000578(i) >= A000292(j) >= 0.
It appears that, for a(n) > 456, the number of terms up to a(n) in this sequence is smaller than the number of prime numbers less than or equal to a(n), or n < pi(a(n)), where pi is the prime counting function. See the figure attached in the Links section.

Examples

			a(1)=0 because c(1)-t(1) = 1-1 = 0;
a(2)=1 because c(11)-t(19) = 1331-1330 = 1;
a(5)=7 because c(2)-t(1) = 8-1 = 7, and c(3)-t(4) = 27-20 = 7;
a(18)=57 because c(7)-t(11) = 343-286 = 57, and c(8)-t(13) = 512-455 = 57;
a(26)=93 because c(2313)-t(4202) = 12374478297-12374478204 = 93.
		

Crossrefs

Programs

  • Python
    import math
    n_max = 10000000
    d_max = 10000
    list1 = []
    n = 1
    while n <= n_max:
      a_tetr = n*(n + 1)*(n + 2)//6
      m_min = math.floor(math.pow(a_tetr, 1/3))
      m = m_min
      a_cube_max = n*(n + 1)*(n + 2)//6 + d_max
      m_max = math.ceil(math.pow(a_cube_max, 1/3))
      while m <= m_max:
          a_cube = m**3
          d = a_cube - a_tetr
          if d >= 0 and d <= d_max and d not in list1:
              list1.append(d)
          m += 1
      n += 1
    list1.sort()
    print(list1)

Formula

The difference between the i-th cubic number, c(i), and j-th tetrahedral number, t(j), is d = i^3 - j*(j+1)*(j+2)/6, where i, j >=1 and c(i) >= t(j).
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