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A160375
Given n, let S denote the set of numbers c_1*c_2*...*c_n where 1<=c_1<=c_2<=...<=c_n<=n; a(n) = number of members of S that have a unique representation of this form.
1
1, 3, 10, 16, 61, 81, 337, 477, 601, 901, 4291, 5798, 27314, 33671, 45732, 59397, 299745, 421363, 2090647, 2739022, 4597263, 5401826, 27510715, 23666955
OFFSET
1,2
COMMENTS
Number of combinations as in A001700.
From David A. Corneth, Sep 26 2016: (Start)
a(n + 1) / a(n) is fairly large if n + 1 is prime; for the given data, it's at least three. In the other cases it's less than 2.
Let p be a distinct product as described in the name. We look at the factors rather than the result. For n = 4, we see the product p = 1*2*3*3.
Let F(p) be a vector of size n which counts the frequency F_e of each e where 1 <= e <= n. For n = 4 and the product we find (1,1,2,0).
For n = 6, we can put the following restrictions on a vector F(p) = (f_1, f_2, f_3, f_4, f_5, f_6): Trivially, f_e >= 0, f_1+f_2+...+f_6 = 6.
Furthermore,
f_2 * f_3 = 0, as 2*3 = 1*6 and 1<=n=6 and 6<=n=6, so if f_2, f_3 > 0, the value of the product isn't unique, contradiction;
f_2 < 2, 2*2 = 1*4;
f_3 * f_4 = 0 as 3*4 = 2*6. (End)
LINKS
EXAMPLE
a(3) = 10 because there are 10 numbers that can be written as such a product in exactly one way:
1*1*1 = 1
1*1*2 = 2
1*1*3 = 3
1*2*2 = 4
1*2*3 = 6
2*2*2 = 8
1*3*3 = 9
2*2*3 = 12
2*3*3 = 18
3*3*3 = 27
There are 25 possible products of the numbers 1,2,3,4 (see A110713), but 9 of those products can be attained in multiple ways (e.g., 1*2*2*4 = 1*1*4*4), so a(4) = 25-9 = 16.
MATHEMATICA
Table[Count[Split@ Sort@ Map[Times @@ # &, Union@ Map[Sort, Tuples[Range@ n, n]]], w_ /; Length@ w == 1], {n, 8}] (* Michael De Vlieger, Sep 26 2016 *)
CROSSREFS
Sequence in context: A141497 A366395 A059911 * A300017 A376023 A176760
KEYWORD
more,nonn
AUTHOR
Mats Granvik, May 11 2009
EXTENSIONS
a(7)-a(13) from Nathaniel Johnston, Nov 29 2010
a(14)-a(24) from Gerhard Kirchner, Aug 30 2016
Definition edited by N. J. A. Sloane, Sep 27 2016
STATUS
approved