A085098 a(n) = number of solutions (x_1, x_2, ..., x_n) to Product_{i=1..n} (1 + 1/x_i) = 2.

1, 1, 5, 43, 876, 49513, 13005235

Offset: 1

Author

Philippe Deléham, Aug 10 2003

Keywords

• hard
• more
• nonn

Comments

Number of ways 2 is a product of n superparticular ratios, without regard to order. A superparticular ratio is a ratio of the form m/(m-1). The question relates to music theory, in that various permutations of these products result in scales. - Gene Ward Smith, Apr 11 2006

Examples

			For n = 1, a(1) = 1, one solution: {x_1} = {1}.
For n = 2, a(2) = 1, one solution: {x_1, x_2} = {2, 3}.
For n = 3, a(3) = 5, five solutions: {x_1, x_2, x_3} = {3, 4, 5}, {3, 3, 8}, {2, 4, 15}, {2, 5, 9}, {2, 6, 7}.
In other words, a(3) = 5 since 2 can be written as (4/3)^2 (9/8), (4/3)(5/4)(6/5), (3/2)(7/6)(8/7), (3/2)(6/5)(10/9) or (3/2)(5/4)(16/15) but in no other way using superparticular rations.
		

Crossrefs

Cf. A118086.

Programs

Maple

spsubdiv := proc(r::rational,n::integer) # Maple program by David Canright local i,j,l,s; # option remember; if n=1 then if numer(r)=denom(r)+1 then [r] else ( NULL ) end if; else s := NULL; for i from floor(1/(r-1))+1 while (1+1/i)^n >= r do l := [spsubdiv( r/(1+1/i), n-1 )]; for j to nops(l) do if op(1,op(j,l)) <= (1+1/i) then s := s, [(1+1/i),op(op(j,l))]; end if od; od; s; end if; end: spl := proc(r, n) [spsubdiv(r, n)] end: spcount := proc(r, n) nops(spl(r, n)) end:

Extensions

a(5) corrected and a(6) computed by Gene Ward Smith, Apr 11 2006

a(7) was found by Hugo van der Sanden, Dec 14 2007

Edited by N. J. A. Sloane, Oct 18 2008 at the suggestion of Max Alekseyev