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

%I #15 Dec 01 2024 11:44:25

%S 1,1,3,12,83,1323,63090,14736464

%N a(n) = number of solutions (x_1, x_2, ..., x_n) to Product_{i=1..n} (1 + 1/x_i) = any integer.

%C Number of ways any integer is a product of n superparticular ratios, without regard to order. A superparticular ratio is a ratio of the form (m+1)/m.

%e For n = 2, a(2) = 3, three solutions, {x_1, x_2} = {2, 3} = 2; {1, 2} = 3; {1, 1} = 4.

%e In other words, a(2) = 3 since 2 can be written as (3/2)(4/3), 3 can be written as (2/1)(3/2), and 4 can be written as (2/1)^2, but no other integers are the product of two superparticular ratios.

%e a(3) = 12 since 2 can be written in 5 ways, 3 can be written in 3 ways, and 4, 5, 6, and 8 can be written in 1 way each, as the product of three superparticular ratios.

%o (PARI) f(n, m, p, q)={ \\ the number of solutions in which product of the ratios is equal to p/q

%o if(p<=q, return(0));

%o if(n==1, return(if(q%(p-q)==0, 1, 0)));

%o x=floor(1/((p/q)^(1/n)-1)); \\ x is the maximum of the least denominator of the ratios

%o my(ans=0);

%o for(i=m, x, ans=ans+f(n-1, i, p*i, q*(i+1)));

%o \\ m indicates that the solutions require the denominators of all ratios not to be less than m

%o \\ discard the ratio (i+1)/i

%o return(ans);

%o };

%o a(n)={

%o my(ans=0);

%o for(i=2, 2^n, ans=ans+f(n, 1, i, 1));

%o return(ans);

%o }; \\ _Yifan Xie_, Nov 21 2024

%Y Cf. A085098, A118086, A277608.

%K hard,nonn,more

%O 0,3

%A _Keith F. Lynch_, Sep 05 2024