

A316945


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)={k:(n,p,k) is admissible for some p}, and let a(n) = A(n).


3



0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 4, 4, 6, 7, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74
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OFFSET

1,13


COMMENTS

John Conway proposed an interesting math puzzle in the 1960s, which is now generally known as "Conway's wizard problem." Here is the problem.
Last night I sat behind two wizards on a bus and overheard the following:
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.
Red Wizard: How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?
Blue Wizard: No, you could not.
Red Wizard: Aha! At last, I know how old you are!
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?
This problem posed by Conway looks at different multisets that correspond to the same ordered triple, which motivated the study of this sequence.


LINKS



FORMULA

a(n) = 0 for 1 <= n <= 11, a(12) = 1, a(13) = 2, a(14) = 4, a(15) = 4, a(16) = 6, a(17) = 7, a(18) = 7, and a(n) = n  10 for n >= 19.


EXAMPLE

For n = 3, the only partitions of 3 are {3}, {1,2}, and {1,1,1}. Hence, there is no admissible triple (3,p,k).
For n = 4, the only partitions of 4 are {4}, {1,3}, {2,2},{1,1,2}, and {1,1,1,1}. Hence, there is no admissible triple (4,p,k).
For n = 12, the only admissible triple (12,p,k) is when p = 48 and k = 4. This is achieved by the following multisets: {1,3,4,4} and {2,2,2,6}. Thus a(12) = 1.


MATHEMATICA

Table[Count[
Table[intpart = IntegerPartitions[sum, {n}];
DuplicateFreeQ[
Table[Product[intpart[[i]][[j]], {j, n}], {i,
Length[intpart]}]], {n, sum}], False], {sum, 50}]


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



