A217043 -id:A217043 - cp's OEIS frontend

A071115 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 +, -, *, /, where intermediate subexpressions must be integers.

1, 2, 4, 11, 34, 152, 1007, 7335, 85761, 812767

Offset: 1

Koksal Karakus (karakusk(AT)hotmail.com), May 27 2002

a(n+1) > 2*a(n) + 2 for n > 3 since a(n) may be added to every number possible at the previous step (at least 1..a(n)-1) and a(n), 2*a(n), 2*a(n)+1, and 2*(a(n)+1) are also present. - Michael S. Branicky, Jan 30 2023

			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 do the same thing for 11.

Gilles Bannay, Countdown Problem
Index entries for similar sequences

Cf. A060315, A217043 (allows intermediate fractions).

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 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
alst = [1]
[alst.append(a(n, alst)) for n in range(1, 8)]
print(alst) # Michael S. Branicky, Jul 01 2022

A357891 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 exactly once and the operators +, -, *, /.

Original entry on oeis.org

1, 2, 4, 11, 34, 152, 1079, 6610, 93221

Offset: 1

Rainer Rosenthal and Hugo Pfoertner, Nov 01 2022

Cf. A071115, A217043, A358075.

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)
    k = 1
    while k in R[n-1][tuple(v)]: k += 1
    return k
alst = [1]
[alst.append(a(n, alst)) for n in range(1, 6)]
print(alst) # Michael S. Branicky, Nov 01 2022

a(9) from Michael S. Branicky, Nov 10 2022

cp's OEIS Frontend

A071115 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 +, -, *, /, where intermediate subexpressions must be integers.

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