cp's OEIS Frontend

A319186 Total number of three disjoint subsets of divisors of k, each of ones adding to sigma(k)/3, where k are the terms listed in A204830.

%I A319186 #43 Dec 20 2018 23:53:19
%S A319186 1,1,5,77,53,33,14,21,21,8,1,1940,2,8217,230,4894
%N A319186 Total number of three disjoint subsets of divisors of k, each of ones adding to sigma(k)/3, where k are the terms listed in A204830.
%e A319186 a(1) = 1 because A204830(1) = 120, divisors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 and 1+2+3+4+5+6+8+10+12+15+24+30 = 20+40+60 = 120 = sigma(120)/3. Of course also 1+2+3+5+8+12+15+20+24+30 = 4+6+10+40+60 = 120 = sigma(120)/3 but these three subsets share {120} with the previous ones and therefore they are not disjoint.
%e A319186 a(13) = 2 because A204830(13) = 780, divisors of 780 are 1, 2, 3, 4, 5, 6, 10, 12, 13, 15, 20, 26, 30, 39, 52, 60, 65, 78, 130, 156, 195, 260, 390, 780: 1+3+52+78+260+390 = 2+5+6+10+12+13+15+20+26+30+39+60+65+130+156+195 = 4+780 = sigma(780)/3 and 5+6+10+12+13+15+26+39+52+60+65+130+156+195 = 2+4+20+30+78+260+390 = 1+3+780 = sigma(780)/3.
%e A319186 a(3) = 5 because A204830(3) = 240, divisors of 240 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240 and sigma(240)/3 = 248: it is easy to see that the total number of three disjoint subset is 5 because the only subsets containing 240 that sum to 248 are {1,2,5,240}, {1,3,4,240}, {2,6,240}, {3,5,240} and {8,240}.
%p A319186 with(numtheory): with(combstruct): P:=proc(q,h) local a,b,c,d,f,k,n,r;
%p A319186 for n from 1 to q do a:=sigma(n); b:=op(divisors(n));
%p A319186 if a mod h=0 and a>=h*n then k:=0; c:=1/h*a-n;
%p A319186 r:=select(m->m<=c,[b]); f:=iterstructs(Combination(r));
%p A319186 while not finished(f) do if c=add(d,d=nextstruct(f)) then k:=k+1; fi; od; lprint(n,k); fi; od; end: P(10^4,3);
%Y A319186 Cf. A023197, A083206, A204830.
%K A319186 nonn,more
%O A319186 1,3
%A A319186 _Paolo P. Lava_, Dec 17 2018