A175035 -id:A175035 - cp's OEIS frontend

A230044 Nonnegative numbers k such that k plus a perfect square is a triangular number.

0, 1, 2, 3, 5, 6, 9, 10, 11, 12, 14, 15, 17, 19, 20, 21, 24, 27, 28, 29, 30, 32, 35, 36, 39, 41, 42, 44, 45, 46, 50, 51, 53, 54, 55, 56, 57, 62, 65, 66, 69, 71, 72, 74, 75, 77, 78, 80, 82, 84, 87, 89, 90, 91, 95, 96, 100, 101, 104, 105, 107, 109, 110, 111, 116, 117, 119, 120, 122, 126, 127, 128

Offset: 1

Ralf Stephan, Oct 06 2013

nonn

Negative k are in A175035.
Numbers such that the Diophantine equation y^2 + y - 2x^2 = 2n, y > 0 has a solution. Empirically, solutions (x,y) don't exceed (5n,5n) for n < 10^5. Record quotients y/n are at n = 2, 3, 12, 45, 1225, 6806, ...
Conjecture: these are the sorted distinct terms of A064784.
n is in this sequence iff 8n+1 is in A035251, that is, every prime p == 3 or 5 (mod 8) dividing 8n+1 appears to an even power. - Max Alekseyev, Oct 14 2013

			28 is triangular, and 25 is a square <= 28, and 28-25=3, so 3 is in sequence.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A035251, A064784, A175035.

B=bnfinit(z^2-8); is(n)=#bnfisintnorm(B,8*n+1) \\ Max Alekseyev, Oct 13 2013

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

Original entry on oeis.org

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

A175034 Offsets i such that i + n*(n+1)/2 is never a perfect square for any n>0.

Original entry on oeis.org

2, 5, 7, 11, 12, 14, 17, 18, 20, 23, 27, 29, 31, 32, 37, 38, 40, 41, 42, 44, 47, 50, 51, 52, 56, 57, 59, 62, 65, 67, 68, 69, 70, 73, 74, 77, 82, 83, 84, 86, 87, 88, 92, 95, 96, 98, 101, 102, 104, 107, 109, 110, 112, 113, 117, 119, 122, 125, 126, 127, 128, 131, 132, 135, 137, 139

Offset: 1

Ctibor O. Zizka, Nov 10 2009

Complement of A175035.

Cf. A001108, A006451, A154138, A154139, A154140, A154141

Extended by R. J. Mathar, Nov 26 2009

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).

A365905 "2-peloton numbers": Numbers that appear at least twice in A365904.

Original entry on oeis.org

15, 36, 43, 49, 64, 66, 78, 85, 99, 100, 118, 120, 134, 141, 151, 159, 168, 169, 190, 204, 210, 211, 219, 225, 241, 246, 253, 256, 270, 274, 279, 283, 288, 295, 309, 321, 323, 325, 345, 351, 355, 358, 364, 372, 376, 379, 386, 393, 394, 400, 405, 406, 423, 429, 435, 438, 440, 456, 463, 474, 484, 498

Offset: 1

Joan Llobera Querol, Sep 22 2023

nonn

Called "peloton" numbers after the original sequence idea in first link: the difference of a rhombus (a square number) and a triangular number, placed as points on a triangular grid, form the shape of a peloton in bicycle racing.

Contains all elements of A001110 other than 0 and 1.

			15 can be obtained as T(4,1) or T(5,4) following notation in A365904.
36 can be obtained as T(6,0) or T(8,7).

Eric Snyder, Table of n, a(n) for n = 1..10000
Zach Wissner-Gross, Can You Shape the Peloton?, Fiddler on the Proof, Sep 22, 2023.

Cf. A175035 (numbers appear at least once), A365904.

isok(n) = sum(m=sqrtint(n), (sqrtint(8*n+1)-1)\2, ispolygonal(m^2-n,3)) > 1 \\ Andrew Howroyd, Sep 24 2023
(Python/SageMath)
nmax, m, Out = 300, 2, []
Z = [ n^2 - (k^2 + k)/2 for n in [2..nmax] for k in [0..n-1] ]
for i in Z:
    if Z.count(i) >= m: Out.append(i)
Out=sorted(list(set(Out)))
for j in [1..10000]: print(j+1, Out[j])
\\ Eric Snyder, Sep 29 2023

A326917 Nonnegative numbers of the form 8*T(x) - T(y) with 0 <= x, 0 <= y, where T() denotes a triangular number.

Original entry on oeis.org

0, 2, 3, 5, 7, 8, 9, 12, 14, 15, 18, 20, 21, 23, 24, 25, 27, 29, 32, 33, 34, 35, 38, 42, 44, 45, 47, 48, 52, 53, 54, 57, 59, 60, 62, 63, 65, 70, 71, 74, 75, 77, 78, 79, 80, 84, 88, 89, 90, 92, 93, 96, 98, 99, 102, 104, 105, 107, 110, 113, 114, 115, 117, 119

Offset: 1

Ralf Steiner, Oct 21 2019

nonn

When incremented by 1 this is also the difference between an odd square (1 + 8*T) and a triangular number T.

			8*A000217(1) - A000217(2) = 8*1 - 3 = 5 = a(4).

Cf. A000217 (T), A175035, A016754 (odd squares).

T[n_] := n (n + 1)/2;Select[Union[Flatten[Table[8 T[x] - T[y], {x, 0, 15}, {y, 0, 100}]]],115 >= # >= 0 &]

a(n) = A175035(n) - 1.

cp's OEIS Frontend

A230044 Nonnegative numbers k such that k plus a perfect square is a triangular number.

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

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A175034 Offsets i such that i + n*(n+1)/2 is never a perfect square for any n>0.

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A335842 Nonnegative differences of positive cubes and positive tetrahedral numbers.

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A365905 "2-peloton numbers": Numbers that appear at least twice in A365904.

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Programs

PARI

A326917 Nonnegative numbers of the form 8*T(x) - T(y) with 0 <= x, 0 <= y, where T() denotes a triangular number.

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