A071314 a(n) is the smallest number that cannot be obtained from the numbers {2^0,2^1,...,2^n} using each number at most once and the operators +, -, *, /. Parentheses are allowed, intermediate fractions are not allowed.

2, 4, 11, 27, 77, 595, 2471, 9643, 51787

Offset

0,1

Comments

The A309886 is a similar sequence, except: there we allow intermediate fractions, and we require all numbers to be used when building an expression. - Matej Veselovac, Aug 28 2019

For n>=2, the largest number that can be obtained in this manner is given by the following formula: (2^1 + 2^0)*(Product_{k=2..n} 2^k). This product notation is equivalent to the expression: (3/2)*2^(n*(n+1)/2). Thus, for n>=2, this sequence has an upper bound: (3/2)*2^(n*(n+1)/2) + 1. - Alejandro J. Becerra Jr., Apr 22 2020

Links

Table of n, a(n) for n=0..8.

G. Bannay, Countdown Problem

Index entries for similar sequences

Formula

a(n) <= A309886(n+1). - Michael S. Branicky, Jul 15 2022

Example

a(2) = 11 because using {1,2,4} and the four operations we can obtain all the numbers up to 10, for example 10=(4+1)*2, but we cannot obtain 11 in the same way.
a(6) <= 595 since the only way to make 595 is: (1 + 16 + 4/8)*(2 + 32), which requires the use of an intermediate fraction 4/8 in the calculation process, which is not allowed. - Matej Veselovac, Aug 28 2019
a(8) != 19351 = 1+((2+256)*(((4+16)*(128-8))/32). - Michael S. Branicky, Jul 15 2022

Prog

(Python)
def a(n):
    R = dict() # index of each reachable subset is [card(s)-1][s]
    for i in range(n+1): R[i] = dict()
    for i in range(n+1): R[0][(2**i, )] = {2**i}
    reach = set(2**i for i in range(n+1))
    for j in range(1, n+1):
        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 and b%a == 0: allowed.append(b//a)
                                if b != 0 and a%b == 0: allowed.append(a//b)
                                R[len(s12)-1][s12].update(allowed)
                                reach.update(allowed)
    k = 1
    while k in reach: k += 1
    return k
print([a(n) for n in range(6)]) # Michael S. Branicky, Jul 15 2022

Crossrefs

Cf. A060315, A309886.

Sequence in context: A243736 A316772 A123440 * A309886 A123441 A086441

Adjacent sequences: A071311 A071312 A071313 * A071315 A071316 A071317

Keyword

nonn,hard,more

Author

Koksal Karakus (karakusk(AT)hotmail.com), Jun 11 2002

Extensions

a(8) corrected by Michael S. Branicky, Jul 15 2022

Status

approved