OFFSET
1,1
COMMENTS
I was inspired by the 24-game. How many results can you get from two numbers by addition, subtraction, multiplication and division?
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms 1..2000 from Owen Whitby)
FORMULA
a(n) = A263995(n) + 2*A002088(n) - n = A263995(n) + A018805(n) - n + 1. To see this, terms of the form |i-j| are covered by terms of the form i+j or i*j with the exception of 0. Thus i+j, |i-j| and ij result in A263995(n)+1 distinct terms. Terms i/j where gcd(i,j) > 1 can be reduced to a term i'/j' where gcd(i',j')=1. The only terms left are i/j with gcd(i,j) = 1 for which there are A018805(n) such terms. We need to subtract n terms for the cases where j=1 since i/j=i/1 are integers. This results in the formula. - Chai Wah Wu, Aug 21 2024
EXAMPLE
For n = 3 sums 2, 3, 4, 5, 6 differences 0, 1, 2 multiplications 1, 2, 3, 4, 6, 9 divisions 1/2, 1/3, 2/3, 2, 3, 3/2 different results are 2, 3, 4, 5, 6, 0, 1, 9, 1/2, 1/3, 2/3, 3/2 or ordered 0, 1/3, 1/2, 2/3, 1, 3/2, 2, 3, 4, 5, 6, 9 so f(3) = 12.
MAPLE
A180622 := proc(n) s := {} ; for i from 1 to n do for j from 1 to n do s := s union {i+j} ; s := s union {abs(i-j)} ; s := s union {i*j} ; s := s union {i/j} ; s := s union {j/i} ; end do: end do: nops(s) ; end proc: seq(A180622(n), n=1..83) ; # R. J. Mathar, Sep 19 2010
MATHEMATICA
a180622[maxn_] := Module[{seq = {}, vals = {}, vnew, an, n1}, Do[vnew = {}; n1 = n - 1; Do[vnew = vnew~Join~{i + n, n - i, i*n, i/n, n/i}, {i, n1}]; vnew = vnew~Join~{n + n, 0, n*n, 1}; vals = Union[vals, vnew]; an = Length[vals]; AppendTo[seq, an], {n, maxn}]; seq] (* Owen Whitby, Nov 03 2010 *)
PROG
(Python)
from fractions import Fraction
def A180622(n): return len(set(range((n<<1)+1))|set().union(*({i*j, Fraction(i, j), Fraction(j, i)} for i in range(1, n+1) for j in range(1, i+1)))) # Chai Wah Wu, Aug 21 2024
(Python)
# faster program
from functools import lru_cache
from sympy import primepi
def A180622(n):
@lru_cache(maxsize=None)
def f(n): # based on second formula in A018805
if n == 0:
return 0
c, j = 0, 2
k1 = n//j
while k1 > 1:
j2 = n//k1 + 1
c += (j2-j)*(f(k1)-1)
j, k1 = j2, n//j2
return (n*(n-1)-c+j)
return len({i*j for i in range(1, n+1) for j in range(1, i+1)})+f(n)+primepi(n<<1)-primepi(n)-n # Chai Wah Wu, Aug 21 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Hein van Winkel (hein65(AT)duizendknoop.com), Sep 12 2010
EXTENSIONS
More terms from R. J. Mathar, Sep 19 2010
STATUS
approved