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A316946 A triple of positive integers (n,p,k) is admissible if there exist at least two different multisets of k positive integers, {x_1,x_2,...,x_k} and {y_1,y_2,...,y_k}, such that x_1+x_2+...+x_k = y_1+y_2+...+y_k = n and x_1x_2...x_k = y_1y_2...y_k = p. For each n, let A(n) = {p:(n,p,k) is admissible for some k}, and let a(n) = |A(n)|.

%I A316946 #20 Feb 24 2023 03:09:27
%S A316946 0,0,0,0,0,0,0,0,0,0,0,1,2,5,6,10,14,19,26,33,43,54,68,87,106,129,157,
%T A316946 187,226,269,319,378,445,521,610,712,825,952,1099,1261,1443,1655,1889,
%U A316946 2148,2440,2769,3135,3542,4000,4494,5049,5661,6346,7099,7938,8857,9862,10972,12190,13532,15000,16611,18366
%N A316946 A triple of positive integers (n,p,k) is admissible if there exist at least two different multisets of k positive integers, {x_1,x_2,...,x_k} and {y_1,y_2,...,y_k}, such that x_1+x_2+...+x_k = y_1+y_2+...+y_k = n and x_1x_2...x_k = y_1y_2...y_k = p. For each n, let A(n) = {p:(n,p,k) is admissible for some k}, and let a(n) = |A(n)|.
%C A316946 John Conway proposed an interesting math puzzle in the 1960s, which is now generally known as the "Conway's wizard problem." Here is the problem.
%C A316946 Last night I sat behind two wizards on a bus and overheard the following:
%C A316946 Blue Wizard: I have a positive integer number of children, whose ages are positive integers. The sum of their ages is the number of this bus, while the product is my own age.
%C A316946 Red Wizard: How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?
%C A316946 Blue Wizard: No, you could not.
%C A316946 Red Wizard: Aha! At last, I know how old you are!
%C A316946 Apparently the Red Wizard had been trying to determine the Blue Wizard's age for some time. Now, what was the number of the bus?
%C A316946 This problem posed by Conway looks at different multisets that correspond to the same ordered triple, which motivated the study of this sequence.
%H A316946 Jay Bennett, <a href="https://www.popularmechanics.com/science/math/a27415/riddle-of-the-week-34-two-wizards-ride-a-bus/">Riddle of the week #34: Two wizards ride a bus</a>, Popular Mechanics. Hearst Communications, Inc., 4 Aug. 2017. 12 Jun. 2018 Accessed.
%H A316946 John B. Kelly, <a href="http://dx.doi.org/10.1090/S0002-9939-1964-0168542-2">Partitions with equal products</a>, Proc. Amer. Math. Soc. 15 (1964), 987-990.
%e A316946 a(15) = 6 since A(15) = {36,40,48,72,96,144}:
%e A316946 p = 36 [9, 2, 2, 1, 1], [6, 6, 1, 1, 1]
%e A316946 p = 40 [10, 2, 2, 1], [8, 5, 1, 1]
%e A316946 p = 48 [6, 2, 2, 2, 1, 1, 1], [4, 4, 3, 1, 1, 1, 1]
%e A316946 p = 72 [9, 2, 2, 2], [8, 3, 3, 1], [6, 6, 2, 1],
%e A316946 p = 96 [8, 3, 2, 2], [6, 4, 4, 1], [6, 2, 2, 2, 2, 1], [4, 4, 3, 2, 1, 1]
%e A316946 p = 144 [6, 3, 2, 2, 2], [4, 4, 3, 3, 1].
%t A316946 Do[repeats = {};  Do[intpart = IntegerPartitions[sum, {n}];   prod = Tally[Table[Times @@ intpart[[i]], {i, Length[intpart]}]];   repeatprod = Select[prod, #[[2]] > 1 &];   If[repeatprod != {},    repeats = Join[repeats, Transpose[repeatprod][[1]]]], {n, 3,    sum - 8}]; output = DeleteDuplicates[repeats];  Print[sum, " ", Length[output]], {sum, 12, 100}]
%Y A316946 Cf. A060277, A316945.
%K A316946 nonn
%O A316946 1,13
%A A316946 _Byungchul Cha_, _Adam Claman_, _Joshua Harrington_, _Ziyu Liu_, _Barbara Maldonado_, _Alexander M. Miller_, _Ann Palma_, _Wing Hong Tony Wong_, _Hongkwon V. Yi_, Jul 20 2018