How many consecutive integers can you make using only four digits?

Question

In the spirit of the classic four fours, I wonder what's the optimal set of four numbers?

Your goal is to make the most consecutive integers using four digits of your choice. Pick four: $0,1,2,3,4,5,6,7,8,9$ ( You can pick multiple instances of the same digit )

When constructing an integer:

• all of your four candidates must be used exactly once (order/placing of digits is irrelevant)
• You may use basic arithmetic operations $+,-,\times,\div$ and parentheses $()$
• You may use $a^b$ and $\sqrt[a]{b}$ but at the expense of 2 numbers as you can see
• You may not form new numbers, i.e. $ab$ is not allowed

If we were to use four $4$s, the best we could do would be up to $9$:

 0  =  4 ÷ 4 × 4 − 4
 1  =  4 ÷ 4 + 4 − 4
 2  =  4 −(4 + 4)÷ 4  
 3  = (4 × 4 − 4)÷ 4
 4  =  4 + 4 ×(4 − 4)
 5  = (4 × 4 + 4)÷ 4
 6  = (4 + 4)÷ 4 + 4
 7  =  4 + 4 − 4 ÷ 4
 8  =  4 ÷ 4 × 4 + 4
 9  =  4 ÷ 4 + 4 + 4
*10  =  4 ÷√4 + 4 ×√4 
*10  =  (44 − 4) ÷ 4

Number 10 can't be done and is an example of failing, since it would require either:

• expenseless roots; $\sqrt{4}$ isn't allowed. ( $\sqrt[2]{4}$ is, which requires you to use $4$ and $2$ )

• number formation which isn't allowed either.

Zero does not necessarily need to be included, you can start at either $0$ or $1$.

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For the purposes of freedom of puzzling, if you think you can top your solution for a chosen set of digits, by starting at any other positive integer, you can add that to your answer below your initial solution. (I suspect this is unlikely)

If you want, you can extend your consecutive list to negative integers but this is strictly optional and not necessary in any way, other than for the purposes of fulfillment and mathematical euphoria.

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Example

There is an example on Puzzling.SE using digits $2,2,4,5$:

https://puzzling.stackexchange.com/a/23078/32666

But this can be expanded since the given example uses only basic arithmetic operations, not including potentiation and roots. I also suspect It could be done better using another set.

I tried this by hand and I'm stuck at number $29$ using this example set, and at number $34$ using $9,8,3,2$.

Answer 1

Here is my result for $1,4,6,7$ from $1$ to $53$.

$1^{4-6-7}=1$

$1^4-6+7=2$

$(1-4)/(6-7)=3$

$1+4+6-7=4$

$1-4/(6-7)=5$

$1+4-6+7=6$

$(1^{4})^{6}*7=7$

$1^{4-6}+7=8$

$-1*(4-6-7)=9$

$1-4+6+7=10$

$1^6*(4+7)=11$

$1^6+4+7=12$

$1^4*(6+7)=13$

$1^4+6+7=14$

$(1-6)*(4-7)=15$

$-1+4+6+7=16$

$1*(4+6)+7=17$

$1+4+6+7=18$

$1-6*(4-7)=19$

$-4*(1^7-6)=20$

$-1+4*7-6=21$

$1*(4*7-6)=22$

$1+4*7-6=23$

$1^7*4*6=24$

$1^7+4*6=25$

$4*(1+7)-6=26$

$-4*(1-6)+7=27$

$1^6*4*7=28$

$1^6+4*7=29$

$-1+4*6+7=30$

$1*(4*6+7)=31$

$1+4*6+7=32$

$-1+4*7+6=33$

$1*(4*7+6)=34$

$1+4*7+6=35$

$(7-1)^{6-4}=36$

$-1-4+6*7=37$

$-1*(4-6*7)=38$

$1-4+6*7=39$

$4-6*(1-7)=40$

$6+7*(4+1)=41$

$1^4*6*7=42$

$1^4+6*7=43$

$-4+6*(1+7)=44$

$-1+4+6*7=45$

$1*(4+6*7)=46$

$1+4+6*7=47$

$-1+7^{6-4}=48$

$1*7^{6-4}=49$

$1+7^{6-4}=50$

$-1+4*(6+7)=51$

$1*4*(6+7)=52$

$1+4*(6+7)=53$

Answer 2

Here's a list of all numbers from 1 to 43 using 2, 3, 6 and 7. ^ is the power operator.

(6-((3+7)/2))
(6*(3/(2+7)))
(2/((7-3)/6))
((6*3)-(2*7))
((7-(3-6))/2)
(((7-6)*2)*3)
(7*((3*2)/6))
(2^(3*(7-6)))
((3^(7-6))^2)
((3^2)+(7-6))
(((7+3)/2)+6)
(((7+6)+2)-3)
(((6/2)+7)+3)
(7+(6-(2-3)))
(((6*2)-7)*3)
(7-(3-(6*2)))
((6-3)+(7*2))
((3+(2+6))+7)
(((3*2)+6)+7)
(2*((7-3)+6))
(6-((2-7)*3))
(7+((2*6)+3))
((6+3)+(2*7))
((6/(2/7))+3)
((7*3)+(6-2))
(6+((3+7)*2))
((2+7)*(6-3))
((2/(3/7))*6)
(6+((3*7)+2))
(2*((7*3)-6))
(7+((6+2)*3))
(3-(7-(6^2)))
(3*((7-2)+6))
(((6/2)^3)+7)
(((6-3)+2)*7)
(6*(2-(3-7)))
((6*(3+2))+7)
((7-(2/3))*6)
((3+(7/2))*6)
((6^2)-(3-7))
((7*6)-(3-2))
(((3-2)*6)*7)
(3-(2-(6*7)))

As you can guess, I wrote a program to find this. It's not so hard to make one, but I'm disclosing it anyway. It puts the output into spoiler code blocks automatically ;)

The program below doesn't use the root operator because most of the time it's useless. If you want to include it, add 'r' to the operators array, but beware of overflow errors.

I stole the RPN parser from here.

import random

operators = ['-', '+', '*', '/', '^']

def parse(rpn):

    stack = []

    for val in rpn:
        if val in operators:
            if len(stack) < 2:
                return -1
            op1 = stack.pop()
            op2 = stack.pop()
            if val=='-': result = op2 - op1
            if val=='+': result = op2 + op1
            if val=='*': result = op2 * op1
            if val=='/':
                if op1 != 0:
                    result = op2 / op1
                else:
                    return -1
            if val=='^':
                if op2 > 0 and op1 < 6:
                    result = op2 ** op1
                else:
                    return -1
            if val=='r':
                if op2 > 0 and op1 != 0:
                    result = op2 ** (1/op1)
                else:
                    return -1

            stack.append(result)
        else:
            stack.append(float(val))

    if len(stack)!= 1:
        return -1
    return stack.pop()


def randrpn(digits):
    rpn = digits[:]
    rpn.append(random.choice(operators))
    rpn.append(random.choice(operators))
    rpn.append(random.choice(operators))
    random.shuffle(rpn)
    return rpn


def randdigits():
    d = []
    d.append(random.choice(range(10)))
    d.append(random.choice(range(10)))
    d.append(random.choice(range(10)))
    d.append(random.choice(range(10)))
    return d

def rpn_to_infix(rpn):
    stack = []
    infix = ""
    for nxt in rpn:
        if nxt in operators:
            op1 = stack.pop()
            op2 = stack.pop()
            expr = '(' + op2 + nxt + op1 + ')'
            stack.append(expr)
        else:
            stack.append(str(nxt))
    return '>! `' + stack.pop() + '`<br>'

d = randdigits()
results = []
formulas = []
for i in range(300000):
    r = randrpn(d)
    res = parse(r)
    if (res >= 0 and res.is_integer() and not res in results):
        results.append(res)
        formulas.append(rpn_to_infix(r))

for i in range(1, 1000):
    if float(i) in results:
        print formulas[results.index(float(i))]
    else:
        break