A217043 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.
1, 2, 4, 11, 34, 152, 1143, 8285, 98863, 657309
Offset: 1
Examples
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.
Links
- Gilles Bannay, Countdown Problem
Programs
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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
Extensions
a(10) corrected by Clément Morelle, Jun 12 2025