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 A348536 #34 Apr 07 2025 14:28:41
%S A348536 0,0,1,1,0,2,0,1,1,0,0,3,0,0,1,1,0,2,0,1,1,0,0,3,0,0,1,1,0,2,0,1,1,0,
%T A348536 0,3,0,0,1,1,0,2,0,1,1,0,0,3,0,0,1,1,0,2,0,1,1,0,0,3,0,0,1,1,0,2,0,1,
%U A348536 1,0,0,3,0,0,1,1,0,2,0,1,1,0,0,3,0,0,1,1,0,2,0,1,1,0,0
%N A348536 Number of partitions of n into 3 parts that divide n.
%C A348536 From _David A. Corneth_, Oct 08 2022: (Start)
%C A348536 Proof of formula: suppose we have a partition d_1 + d_2 + d_3 = n where 0 < d_1 <= d_2 <= d_3 and d_1 | n, d_2 | n and d_3 | n.
%C A348536 Then let m_i * d_i = n for i in (1, 2, 3).
%C A348536 Dividing d_1 + d_2 + d_3 = n by n gives d_1/n + d_2/n + d_3/n = n/n, i.e., 1/m_1 + 1/m_2 + 1/m_3 = 1. There are 3 solutions to that equation namely (m_1, m_2, m_3) in {(2, 4, 4), (2, 3, 6), (3, 3, 3)}. The lcm of these numbers is 12 so the sequence is periodic with a period of 12 and then calculating the first 12 terms defines the sequence.
%C A348536 Alternative name: number of divisors of n from {3, 4, 6}. (End)
%H A348536 David A. Corneth, <a href="/A348536/b348536.txt">Table of n, a(n) for n = 1..10000</a>
%H A348536 <a href="/index/Par#part">Index entries for sequences related to partitions</a>
%F A348536 a(n) = Sum_{j=1..floor(n/3)} Sum_{i=j..floor((n-j)/2)} c(n/j) * c(n/i) * c(n/(n-i-j)), where c(n) = 1 - ceiling(n) + floor(n).
%F A348536 a(n + 12) = a(n) where n >= 1. - _David A. Corneth_, Oct 08 2022
%e A348536 a(12) = 3 via 2 + 4 + 6 = 3 + 3 + 6 = 4 + 4 + 4. - _David A. Corneth_, Oct 08 2022
%t A348536 Block[{c}, c[n_] := 1 - Ceiling[n] + Floor[n]; Array[Sum[Sum[c[#/j]*c[#/i]*c[#/(# - i - j)], {i, j, Floor[(# - j)/2]} ], {j, Floor[#/3]} ] &, 105]] (* _Michael De Vlieger_, Oct 21 2021 *)
%t A348536 Table[Count[IntegerPartitions[n,{3}],_?(Mod[n,#]=={0,0,0}&)],{n,100}] (* _Harvey P. Dale_, Apr 07 2025 *)
%o A348536 (PARI) a(n) = [0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 0, 3][(n-1)%12 + 1] \\ _David A. Corneth_, Oct 08 2022
%o A348536 (PARI) a(n) = { my(d = divisors(n), res = 0); d = d[^#d]; forvec(x = vector(2, i, [1, #d]), s = d[x[1]] + d[x[2]]; if(n - s >= d[x[2]], if(n % (n - s) == 0, print([d[x[1]], d[x[2]], n-s]); res++ ) ) , 1 ); res } \\ _David A. Corneth_, Oct 08 2022
%Y A348536 Cf. A002966, A069905, A355200.
%K A348536 nonn,easy
%O A348536 1,6
%A A348536 _Wesley Ivan Hurt_, Oct 21 2021