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.
%I A335842 #20 Aug 17 2020 23:17:47 %S A335842 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 A335842 93,96,99,105,115,121,122,123,124,132,144,148,149,151,160,169,173,178, %U A335842 180,181,184,188,191,196,206,211,212,215,223,226,230,258,259,274 %N A335842 Nonnegative differences of positive cubes and positive tetrahedral numbers. %C A335842 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 A335842 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 A335842 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 A335842 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 A335842 a(1)=0 because c(1)-t(1) = 1-1 = 0; %e A335842 a(2)=1 because c(11)-t(19) = 1331-1330 = 1; %e A335842 a(5)=7 because c(2)-t(1) = 8-1 = 7, and c(3)-t(4) = 27-20 = 7; %e A335842 a(18)=57 because c(7)-t(11) = 343-286 = 57, and c(8)-t(13) = 512-455 = 57; %e A335842 a(26)=93 because c(2313)-t(4202) = 12374478297-12374478204 = 93. %o A335842 (Python) %o A335842 import math %o A335842 n_max = 10000000 %o A335842 d_max = 10000 %o A335842 list1 = [] %o A335842 n = 1 %o A335842 while n <= n_max: %o A335842 a_tetr = n*(n + 1)*(n + 2)//6 %o A335842 m_min = math.floor(math.pow(a_tetr, 1/3)) %o A335842 m = m_min %o A335842 a_cube_max = n*(n + 1)*(n + 2)//6 + d_max %o A335842 m_max = math.ceil(math.pow(a_cube_max, 1/3)) %o A335842 while m <= m_max: %o A335842 a_cube = m**3 %o A335842 d = a_cube - a_tetr %o A335842 if d >= 0 and d <= d_max and d not in list1: %o A335842 list1.append(d) %o A335842 m += 1 %o A335842 n += 1 %o A335842 list1.sort() %o A335842 print(list1) %Y A335842 Cf. A000292, A000578, A175034, A175035, A335761. %K A335842 nonn %O A335842 1,3 %A A335842 _Ya-Ping Lu_, Jun 26 2020