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A382547 a(n) is the smallest positive integer s that can be partitioned into n distinct positive integers whose product is s * 100^(n-1), or 0 if no such s exists.

%I A382547 #25 Jul 06 2025 00:46:46
%S A382547 1,405,525,644,762,882,1038,1155,1302,1428,1638,1863,2079,2187,2457,
%T A382547 2673,3078,3213,3402,3861,4374,5103,5103,6174
%N A382547 a(n) is the smallest positive integer s that can be partitioned into n distinct positive integers whose product is s * 100^(n-1), or 0 if no such s exists.
%C A382547 a(n) >= A380887(n) in case of a(n) > 0.
%C A382547 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.
%C A382547 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.
%C A382547 Programs used for A380887 can be adapted for this sequence.
%H A382547 Mathematics StackExchange user Servaes, <a href="https://math.stackexchange.com/questions/5070389/">What is the maximum number of distinct prices whose product equals their sum?</a>, answer dated May 29 2025.
%e A382547 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.
%o A382547 (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]);
%o A382547 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
%Y A382547 Cf. A380887, A381187, A383026.
%K A382547 nonn,fini,hard,more
%O A382547 1,2
%A A382547 _Markus Sigg_, Mar 31 2025
%E A382547 a(17)-a(21) from _Markus Sigg_, Apr 21 2025
%E A382547 a(22)-a(24) from _Jinyuan Wang_, May 14 2025