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%I #26 Jan 09 2023 07:41:25
%S 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
%T 0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
%U 0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0
%N Number of partitions of n into 3 distinct positive cubes.
%C In other words, number of solutions to the equation n = x^3 + y^3 + z^3 with x > y > z > 0. - _Antti Karttunen_, Aug 29 2017
%H Antti Karttunen, <a href="/A025469/b025469.txt">Table of n, a(n) for n = 0..17073</a>
%H <a href="/index/Su#ssq">Index entries for sequences related to sums of cubes</a>
%F a(n) = A025465(n) - A025468(n). - _Antti Karttunen_, Aug 29 2017
%e From _Antti Karttunen_, Aug 29 2017: (Start)
%e For n = 36 there is one solution: 36 = 27 + 8 + 1, thus a(36) = 1.
%e For n = 1009 there are two solutions: 1009 = 10^3 + 2^3 + 1^3 = 9^3 + 6^3 + 4^3, thus a(1009) = 2. This is also the first point where sequence attains value greater than one.
%e (End)
%p A025469 := proc(n)
%p local a, x, y, zcu ;
%p a := 0 ;
%p for x from 1 do
%p if 3*x^3 > n then
%p return a;
%p end if;
%p for y from x+1 do
%p if x^3+2*y^3 > n then
%p break;
%p end if;
%p zcu := n-x^3-y^3 ;
%p if zcu > y^3 and isA000578(zcu) then
%p a := a+1 ;
%p end if;
%p end do:
%p end do:
%p end proc:
%p seq(A025469(n),n=1..80) ; # _R. J. Mathar_, Jun 15 2018
%t Table[Count[IntegerPartitions[n, {3}], _?(And[UnsameQ @@ #, AllTrue[#, IntegerQ[#^(1/3)] &]] &)], {n, 105}] (* _Michael De Vlieger_, Aug 29 2017 *)
%o (PARI) A025469(n) = { my(s=0); for(x=1,n,if(ispower(x,3),for(y=x+1,n-x,if(ispower(y,3),for(z=y+1,n-(x+y),if((ispower(z,3)&&(x+y+z)==n),s++)))))); (s); }; \\ _Antti Karttunen_, Aug 29 2017
%Y Cf. A025465 (not necessarily distinct), A025468, A025419 (greedy inverse).
%Y Cf. A024975 (positions of nonzero terms), A024974 (positions of terms > 1), A025399-A025402.
%K nonn
%O 0,1010
%A _David W. Wilson_
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