OFFSET
1,2
COMMENTS
a(n) >= A380887(n) in case of a(n) > 0.
There are only finitely many positive a(n): If x_1 < ... < x_n are positive integers with the required properties, then x_k >= k, and (n-1)! * x_n <= x_1 * ... * x_n = 100^(n-1) * (x_1 + ... + x_n) <= 100^(n-1) * n * x_n gives (n-1)! <= 100^(n-1) * n, hence n <= 274. In fact, n <= 273 must hold, see Mathematics StackExchange link. A more elaborate argumentation in the same discussions shows n <= 269.
By restricting the search space, solution tuples have been found for 25 <= n <= 42. These tuples are not guaranteed to have the smallest possible sum and thus only give upper bounds for a(n). For example, the tuple (1, 2, 3, 4, 5, 6, 8, 10, 12, 14, 16, 20, 25, 30, 32, 40, 50, 60, 64, 75, 80, 100, 120, 125, 128, 150, 160, 200, 225, 250, 400, 625, 800, 1000, 1250, 2500, 3125, 5000, 6250, 12500, 78125, 1953125) shows a(42) <= 2066715.
Programs used for A380887 can be adapted for this sequence.
LINKS
Mathematics StackExchange user Servaes, What is the maximum number of distinct prices whose product equals their sum?, answer dated May 29 2025.
EXAMPLE
a(2) = 405 because 180 + 225 = 405 and 180 * 225 = 405 * 100^1 and no positive integer smaller than 405 exists with the requested properties.
PROG
(PARI) dfs(rs, rp, i, r, tp) = if(r==1, return(rs==rp&&setsearch(d, rs)>i)); if((rs/r)^r<=rp, return(0)); for(j=i+1, oo, if(tp>rp, return(0)); if(rp%d[j]==0, if(dfs(rs-d[j], rp/d[j], j, r-1, tp/d[j]), return(1))); tp*=d[j+r]/d[j]);
a(n) = if(n>1, my(p); for(s=100*n, oo, d=divisors(p=s*100^(n-1)); if(dfs(s, p, 0, n, prod(i=1, n, d[i])), return(s))), 1); \\ Jinyuan Wang, May 14 2025
CROSSREFS
KEYWORD
nonn,fini,hard,more
AUTHOR
Markus Sigg, Mar 31 2025
EXTENSIONS
a(17)-a(21) from Markus Sigg, Apr 21 2025
a(22)-a(24) from Jinyuan Wang, May 14 2025
STATUS
approved