%I #24 Dec 29 2022 06:46:17
%S 1,2,4,11,34,152,1143,8285,98863,1211572
%N a(1) = 1; a(n+1) is the smallest integer >=0 that cannot be obtained from the integers {a(1), ..., a(n)} using each number at most once and the operators +, -, *, / and accepting fractional intermediate results.
%H Gilles Bannay, <a href="https://web.archive.org/web/20061201125224/http://gilles.bannay.free.fr/jeux_us.html">Countdown Problem</a>
%e 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.
%e a(7)=1143 because 1142 = (152+((34-4)*(11*(2+1)))), and 1143 is impossible.
%e 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.
%o (Python)
%o from fractions import Fraction
%o def a(n, v):
%o R = dict() # index of each reachable subset is [card(s)-1][s]
%o for i in range(n): R[i] = dict()
%o for i in range(n): R[0][(v[i],)] = {v[i]}
%o reach = set(v)
%o for j in range(1, n):
%o for i in range((j+1)//2):
%o for s1 in R[i]:
%o for s2 in R[j-1-i]:
%o if set(s1) & set(s2) == set():
%o s12 = tuple(sorted(set(s1) | set(s2)))
%o if s12 not in R[len(s12)-1]:
%o R[len(s12)-1][s12] = set()
%o for a in R[i][s1]:
%o for b in R[j-1-i][s2]:
%o allowed = [a+b, a*b, a-b, b-a]
%o if a != 0: allowed.append(Fraction(b, a))
%o if b != 0: allowed.append(Fraction(a, b))
%o R[len(s12)-1][s12].update(allowed)
%o reach.update(allowed)
%o k = 1
%o while k in reach: k += 1
%o return k
%o alst = [1]
%o [alst.append(a(n, alst)) for n in range(1, 6)]
%o print(alst) # _Michael S. Branicky_, Jul 01 2022
%Y Cf. A060315, A071115 (disallows intermediate fractions).
%K nonn,more
%O 1,2
%A _Clément Morelle_, Sep 25 2012