OFFSET
1,2
LINKS
Gilles Bannay, Countdown Problem
EXAMPLE
a(4)=11 because we can write 4+1=5, 4+2=6, 4+2+1=7, 4*2=8, 4*2+1=9, (4+1)*2=10 by using 1, 2 and 4, but we cannot construct 11 this way.
a(7)=1143 because 1142 = (152+((34-4)*(11*(2+1)))), and 1143 is impossible.
a(7) is not 1007 because it can be constructed as 1007 = 152*(11-(34+1)/(4*2)); the fractional intermediate result 35/8, for example, is accepted in the composition.
PROG
(Python)
from fractions import Fraction
def a(n, v):
R = dict() # index of each reachable subset is [card(s)-1][s]
for i in range(n): R[i] = dict()
for i in range(n): R[0][(v[i], )] = {v[i]}
reach = set(v)
for j in range(1, n):
for i in range((j+1)//2):
for s1 in R[i]:
for s2 in R[j-1-i]:
if set(s1) & set(s2) == set():
s12 = tuple(sorted(set(s1) | set(s2)))
if s12 not in R[len(s12)-1]:
R[len(s12)-1][s12] = set()
for a in R[i][s1]:
for b in R[j-1-i][s2]:
allowed = [a+b, a*b, a-b, b-a]
if a != 0: allowed.append(Fraction(b, a))
if b != 0: allowed.append(Fraction(a, b))
R[len(s12)-1][s12].update(allowed)
reach.update(allowed)
k = 1
while k in reach: k += 1
return k
alst = [1]
[alst.append(a(n, alst)) for n in range(1, 6)]
print(alst) # Michael S. Branicky, Jul 01 2022
CROSSREFS
KEYWORD
nonn,more,changed
AUTHOR
Clément Morelle, Sep 25 2012
EXTENSIONS
a(10) corrected by Clément Morelle, Jun 12 2025
a(11) from Clément Morelle, Jun 16 2026
STATUS
approved
