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%I #14 Oct 08 2018 18:11:14
%S 1,0,0,2,0,0,0,0,2,0,0,1,0,0,0,2,0,0,0,0,0,0,0,2,2,0,0,0,0,0,0,0,0,0,
%T 0,2,0,0,0,2,0,0,0,0,1,0,0,0,2,0,0,0,0,0,0,0,0,0,0,2,0,0,0,2,0,0,0,0,
%U 0,0,0,2,0,0,0,0,0,0,0,0,2,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,2
%N Number of solutions to n=x*y*z, x+y-z=1 with ordered triples (x,y,z), x,y,z>=1.
%C Record values start a(1)=1, a(4)=2, a(112)=3, a(144)=6, a(23400)=8, a(28224)=10.
%C If n is a perfect square, a(n)>0.
%C Larger record indices are listed in A181485, and associated values in A181486. - _M. F. Hasler_, Oct 23 2010
%C First occurrence of k: 2, 1, 4, 112, 480, 43120, 144, 218880, 23400, ??, 28224, ??, 373464, ??, 247104, ??, 604800, ??, 83010312, ??, 26812800, ..., . - _Robert G. Wilson v_, Oct 30 2010
%C a(388778796252000) = 38.
%H Antti Karttunen, <a href="/A171919/b171919.txt">Table of n, a(n) for n = 1..28224</a>
%e For n=4, the a(4)=2 solutions are (x,y,z)=(1,2,2) and (2,1,2).
%e For n=12, the a(12)=1 solution is (x,y,z)=(2,2,3).
%p 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 return a; end proc: seq(A(n),n=1..100) ; # _R. J. Mathar_, Oct 23 2010
%t f[n_] := Block[{c = 0, cong = {0, 1, 4, 9, 12, 16, 21, 24, 25, 36, 37, 40, 45, 49, 52, 57}, dvs = Divisors@ n, dvt, j = 1, k, lmt1, lmt2}, If[ MemberQ[ cong, Mod[n, 60]], lmtj = Length@ dvs + 1; While[j < lmtj, dvt = Divisors[ n/dvs[[j]]]; k = 1; lmtk = Length@ dvt + 1; While[k < lmtk, If[ dvs[[j]] + dvt[[k]] == n/(dvs[[j]]*dvt[[k]]) + 1, c++ ]; k++ ]; j++ ]]; c]; Array[f, 105] (* _Robert G. Wilson v_, Oct 24 2010 *)
%o (PARI) A171919(n)={ my(c=0,t); fordiv(n, z, fordiv( t=n/z, y, y>z & break; y*(z+1-y)==t & c++) ); c} /* can be improved by restricting to x<=y and counting twice if x<y */ \\ _M. F. Hasler_, Oct 23 2010
%Y Cf. A181485, A181486.
%K nonn
%O 1,4
%A _Georgi Guninski_, Oct 23 2010
%E Some more values from _M. F. Hasler_, Oct 23 2010
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