This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(3)=10 because 1=1, 2=2, 3=3, 4=1+3, 5=2+3, 6=2*3, 7=2*3+1, 8=(3+1)*2, 9=(1+2)*3, but there is no way to make 10 using 1, 2, and 3 at most once.
def a(n):
R = dict() # R[|s|-1][s] = reachable values using subset s
for i in range(n+1): R[i] = dict()
for i in range(1, n+1): R[0][(i,)] = {i}
reach = set(range(1, n+1))
for j in range(1, n):
for i in range((j+1)//2):
for s in R[i]:
for t in R[j-1-i]:
if set(s) & set(t) == set():
u = tuple(sorted(set(s) | set(t)))
if u not in R[len(u)-1]:
R[len(u)-1][u] = set()
for a in R[i][s]:
for b in R[j-1-i][t]:
R[len(u)-1][u].update([a+b, a*b])
reach.update([a+b, a*b])
k = n+1
while k in reach: k += 1
return k
print([a(n) for n in range(10)]) # Michael S. Branicky, May 30 2022
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.
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
a(4)=31 because we can obtain the positive integers 1,2,...,28 and 30,32,36 by using the integers {1, 2, 3, 4} at most once and the four operations. For example 30 = 3*2*(4+1).
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][(i,)] = {i}
reach = set(range(1, n+1))
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)
return len(set(r for r in reach if r > 0 and r.denominator == 1))
print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Jul 01 2022
a(3)=11 because using {1,2,3} we can write 1, 2, 3, 3+1=4, 3+2=5, 3*2=6, 3*2+1=7, 2^3=8, 3^2=9, (3^2)+1=10 but we cannot obtain 11 in the same way.
The Torrence article gives a description of how one can use Mathematica to investigate the sequence.
a(4)=2 because 29 and 31 are the smallest numbers that you cannot obtain using {1,2,3,4} and the four operations and 31-29 = 2.
a(3)=5 because using {x, x+1, x+2, x+3} we can get x+1-x=1, x+2-x=2, x+3-x=3, (x+2-x)*(x+3-(x+1))=4 but we cannot obtain 5 in the same way.
a(3)=7 because using two of the numbers {1,2,3} with the four operations we can obtain 1=1, 2=2, 3=3, 3+1=4, 3+2=5, 3*2=6 but we cannot obtain 7 in the same way.
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][(i,)] = {i}
reach = set(i for i in range(1, n+1)) if n > 1 else set()
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(1, 6)]) # Michael S. Branicky, Jul 29 2022
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