A175034 -id:A175034 - cp's OEIS frontend

A175032 a(n) is the difference between the n-th triangular number and the next perfect square.

0, 0, 1, 3, 6, 1, 4, 8, 0, 4, 9, 15, 3, 9, 16, 1, 8, 16, 25, 6, 15, 25, 3, 13, 24, 36, 10, 22, 35, 6, 19, 33, 1, 15, 30, 46, 10, 26, 43, 4, 21, 39, 58, 15, 34, 54, 8, 28, 49, 0, 21, 43, 66, 13, 36, 60, 4, 28, 53, 79, 19, 45, 72, 9, 36, 64, 93, 26, 55, 85, 15, 45, 76, 3, 34, 66, 99, 22

Offset: 0

Ctibor O. Zizka, Nov 09 2009

All terms are from {0} U A175035. No terms are from A175034.

The sequence consists of ascending runs of length 3 or 4. The first run starts at n = 1 and thereafter the k-th run starts at n = A214858(k - 1). - John Tyler Rascoe, Nov 05 2022

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A001109, A214858.
Cf. A000217, A080819.
Cf. A175034, A175035.
Cf. sequences where a(m)=k: A001108 (0), A006451 (1), A154138 (3), A154139 (4), A154140 (6), A154141 (8), A154142 (9), A154143 (10), A154144 (13), A154145 (15), A154146 (16), A154147 (19), A154148 (21), A154149 (22), A154150(24), A154151 (25), A154151 (26), A154153(28), A154154 (30).

Ceiling[Sqrt[#]]^2-#&/@Accumulate[Range[0,80]] (* Harvey P. Dale, Aug 25 2013 *)

a(n) = my(t=n*(n+1)/2); if (issquare(t), 0, (sqrtint(t)+1)^2 - t); \\ Michel Marcus, Nov 06 2022

a(n) = (ceiling(sqrt(n*(n+1)/2)))^2 - n*(n+1)/2. - Ctibor O. Zizka, Nov 09 2009

a(n) = A080819(n) - A000217(n). - R. J. Mathar, Aug 24 2010

Erroneous formula variant deleted and offset set to zero by R. J. Mathar, Aug 24 2010

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

Ya-Ping Lu, Jun 26 2020

nonn

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.

			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.

Ya-Ping Lu, The number of terms up to a(n) and the number of prime numbers less than or equal to a(n)

Cf. A000292, A000578, A175034, A175035, A335761.

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)

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).

cp's OEIS Frontend

A175032 a(n) is the difference between the n-th triangular number and the next perfect square.

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