A071313 a(n) is the smallest number that cannot be obtained from the numbers {1,3,...,2*n-1} using each number at most once and the operators +, -, *, /, where intermediate subexpressions must be integers.

2, 5, 11, 41, 92, 733, 4337, 28972, 195098, 1797746

Offset

1,1

Comments

If noninteger subexpressions are permitted, a(5) = 122 and not 92 since 92 = (3+7)*(9 + 1/5). - Michael S. Branicky, Jul 01 2022

Links

Table of n, a(n) for n=1..10.

Gilles Bannay, Countdown Problem

Index entries for similar sequences

Example

a(2)=5 because using {1,3} and the four operations we can obtain 1=1, 3-1=2, 3=3, 3+1=4 but we cannot obtain 5 in the same way.

Prog

(Python)
def a(n):
    R = dict() # index of each reachable subset is [card(s)-1][s]
    for i in range(n): R[i] = dict()
    for i in range(1, n+1): R[0][(2*i-1, )] = {2*i-1}
    reach = set(range(1, 2*n, 2))
    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 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(1, 9)]) # Michael S. Branicky, Jul 01 2022

Crossrefs

Cf. A060315, A071848.

Sequence in context: A106886 A237814 A007700 * A172297 A128231 A088148

Adjacent sequences: A071310 A071311 A071312 * A071314 A071315 A071316

Keyword

hard,more,nonn

Author

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

Extensions

a(10) from Michael S. Branicky, Jul 01 2022

Status

approved