login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

Nonnegative differences of positive cubes and positive tetrahedral numbers.
1

%I #20 Aug 17 2020 23:17:47

%S 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,

%T 93,96,99,105,115,121,122,123,124,132,144,148,149,151,160,169,173,178,

%U 180,181,184,188,191,196,206,211,212,215,223,226,230,258,259,274

%N Nonnegative differences of positive cubes and positive tetrahedral numbers.

%C 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.

%C 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.

%H Ya-Ping Lu, <a href="/A335842/a335842.pdf">The number of terms up to a(n) and the number of prime numbers less than or equal to a(n)</a>

%F 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).

%e a(1)=0 because c(1)-t(1) = 1-1 = 0;

%e a(2)=1 because c(11)-t(19) = 1331-1330 = 1;

%e a(5)=7 because c(2)-t(1) = 8-1 = 7, and c(3)-t(4) = 27-20 = 7;

%e a(18)=57 because c(7)-t(11) = 343-286 = 57, and c(8)-t(13) = 512-455 = 57;

%e a(26)=93 because c(2313)-t(4202) = 12374478297-12374478204 = 93.

%o (Python)

%o import math

%o n_max = 10000000

%o d_max = 10000

%o list1 = []

%o n = 1

%o while n <= n_max:

%o a_tetr = n*(n + 1)*(n + 2)//6

%o m_min = math.floor(math.pow(a_tetr, 1/3))

%o m = m_min

%o a_cube_max = n*(n + 1)*(n + 2)//6 + d_max

%o m_max = math.ceil(math.pow(a_cube_max, 1/3))

%o while m <= m_max:

%o a_cube = m**3

%o d = a_cube - a_tetr

%o if d >= 0 and d <= d_max and d not in list1:

%o list1.append(d)

%o m += 1

%o n += 1

%o list1.sort()

%o print(list1)

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

%K nonn

%O 1,3

%A _Ya-Ping Lu_, Jun 26 2020