[seqfan] Re: n = x*y*z , x + y - z = 1

Richard Mathar mathar at strw.leidenuniv.nl
Sat Oct 23 14:25:26 CEST 2010

One would probably start at a lower level, with a sequence equivalent to
a characteristic function (I am adding n=1 and n=12 etc):

%I A000001
%S A000001 1,0,0,2,0,0,0,0,2,0,0,1,0,0,0,2,0,0,0,0,0,0,0,2,2,0,0,0,0,0,0,
%T A000001 0,0,0,0,2,0,0,0,2,0,0,0,0,1,0,0,0,2,0,0,0,0,0,0,0,0,0,0,2,0,0,
%U A000001 0,2,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,2,0,0,2,0,0,0,0,0,0,0,0,0
%N A000001 Number of solutions to n=x*y*z, x+y-z=1 with ordered triples (x,y,z), x,y,z>=1.
%C A000001 Record values start a(1)=1, a(4)=2, a(112)=3, a(144)=6, a(23400)=8, a(28224)=10.
%C A000001 If n is a perfect square, a(n)>0.
%e A000001 For n=4, the a(4)=2 solutions are (x,y,z)=(1,2,2) and (2,1,2).
%e A000001 For n=12, the a(12)=1 solution is (x,y,z)=(2,2,3).
%p A000001 A := proc(n) local a,dvs,x,y,z,dvsyz; a :=0 ; dvs := numtheory[divisors](n) ; for x in dvs do yz := n/x ; dvsyz := numtheory[divisors](yz) ; for y in dvsyz do z := yz/y ; if x+y-z=1 then a := a+1 ; fi; end do; end do:
%p A000001 return a; end proc: seq(A(n),n=1..100) ;  # R. J. Mathar, Oct 23 2010
%Y A000001
%K A000001 nonn
%O A000001 1,4
%A A000001 Georgi Guninski (guninski(AT)guninski.com), Oct 23 2010

%I A000002
%S A000002 1,4,9,12,16,24,25,36,40,45,49,60,64,72,81,84,100,105,112,121,144,
%T A000002 160,169,180,189,196,216,220,225,240,256,264,280,289,297,300,312,
%U A000002 324,352,360,361,364,385,396,400,420,429,432,441,480,484
%N A000002 Numbers n with at least one solution to n=x*y*z, x+y-z=1 with ordered triples (x,y,z), x,y,z>=1.
%C A000002 Supersequence of A000290, i.e., all perfect squares are in the sequence.
%e A000002 n=1 allows a solution (x,y,z)=(1,1,1), and is in the sequence.
%e A000002 n=4 allows solutions (x,y,z)=(1,2,2) and (2,1,2) and is in the sequence.
%K A000002 nonn
%O A000002 1,2
%A A000002 Georgi Guninski (guninski(AT)guninski.com), Oct 23 2010

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