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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. 3
1, 2, 4, 11, 34, 152, 1007, 7335, 85761, 812767 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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
LINKS
Gilles Bannay, Countdown Problem
EXAMPLE
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.
PROG
(Python)
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
CROSSREFS
Cf. A060315, A217043 (allows intermediate fractions).
Sequence in context: A344489 A238425 A358075 * A357891 A217043 A249696
KEYWORD
hard,more,nonn
AUTHOR
Koksal Karakus (karakusk(AT)hotmail.com), May 27 2002
STATUS
approved

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Last modified July 27 09:20 EDT 2024. Contains 374642 sequences. (Running on oeis4.)