A048249 Number of distinct values produced from sums and products of n unity arguments.

1, 2, 3, 4, 6, 9, 11, 17, 23, 30, 44, 60, 80, 114, 156, 212, 296, 404, 556, 770, 1065, 1463, 2032, 2795, 3889, 5364, 7422, 10300, 14229, 19722, 27391, 37892, 52599, 73075, 101301, 140588, 195405, 271024, 376608, 523518, 726812, 1010576, 1405013, 1952498

Offset: 1

Author

Tony Bartoletti

Keywords

• nonn
• nice

Comments

Values listed calculated by exhaustive search algorithm.

For n+1 operands (n operations) there are (2n)!/((n!)((n+1)!)) possible postfix forms over a single operator. For each such form, there are 2^n ways to assign 2 operators (here, sum and product). Calculate results and eliminate duplicates.

Number of distinct positive integers that can be obtained by iteratively adding or multiplying together parts of an integer partition until only one part remains, starting with 1^n. - Gus Wiseman, Sep 29 2018

Examples

			a(3)=3 since (in postfix): 111** = 11*1* = 1, 111*+ = 11*1+ = 111+* = 11+1* = 2 and 111++ = 11+1+ = 3. Note that at n=7, the 11 possible values produced are the set {1,2,3,4,5,6,7,8,9,10,12}. This is the first n for which there are "skipped" values in the set.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..75
• Index to sequences related to the complexity of n
• Index entries for similar sequences

Crossrefs

Cf. A000792, A005520, A066739, A070960, A201163, A319850, A318949, A319855, A319856, A319909, A319910, A319911.

Programs

Maple

b:= proc(n) option remember; `if`(n=1, {1}, {seq(seq(seq(
     [f+g, f*g][], g=b(n-i)), f=b(i)), i=1..iquo(n, 2))})
    end:
a:= n-> nops(b(n)):
seq(a(n), n=1..35);  # Alois P. Heinz, May 05 2019

Mathematica

ReplaceListRepeated[forms_,rerules_]:=Union[Flatten[FixedPointList[Function[pre,Union[Flatten[ReplaceList[#,rerules]&/@pre,1]]],forms],1]];
Table[Length[Select[ReplaceListRepeated[{Array[1&,n]},{{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x+y]],{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x*y]]}],Length[#]==1&]],{n,10}] (* Gus Wiseman, Sep 29 2018 *)

Python

from functools import cache
@cache
def f(m):
    if m == 1: return {1}
    out = set()
    for j in range(1, m//2+1):
        for x in f(j):
            for y in f(m-j):
                out.update([x + y, x * y])
    return out
def a(n): return len(f(n))
print([a(n) for n in range(1, 40)]) # Michael S. Branicky, Aug 03 2022

Formula

Equals partial sum of "number of numbers of complexity n" (A005421). - Jonathan Vos Post, Apr 07 2006

Extensions

More terms from David W. Wilson, Oct 10 2001

a(43)-a(44) from Alois P. Heinz, May 05 2019