# Most consecutive positive integers using two 1s

Using two 1s, try to come up with the most consecutive positive integers.

Allowed operations:

• Subtraction

• Multiplication

• Division

• Concatenation

• Square Root

• Factorial

• Floor and Ceiling Functions

• Decimal Point

• Isn't there a conjecture that with only factorial, square root and floor, and starting with any single number >1, you can form all positive integers? If so, this seems like one that could go on for ever... Commented Jan 23, 2017 at 3:52
• @GarethMcCaughan It'd have to be $>2$, because $2!=2$, floor decreases and square root makes closer to 1. Commented Jan 23, 2017 at 4:34
• @GarethMcCaughan ... and floor yields an integer. Commented Jan 23, 2017 at 7:04
• Oops, yes, I meant >2. Commented Jan 23, 2017 at 10:04
• Do you happen to recall the name of this conjecture? Commented Jan 24, 2017 at 0:48

Let's try (feel free to add on or correct, this is community wiki):

$$1=1\times1$$

$$2=1+1$$

$$3=\left\lfloor\sqrt{11}\right\rfloor$$

$$4=\left\lceil\sqrt{11}\right\rceil$$

$$5=\left\lfloor\sqrt{\sqrt{\left(\left\lfloor\sqrt{11}\right\rfloor!\right)!}}\right\rfloor$$

$$6=\left\lfloor\sqrt{11}\right\rfloor!$$

$$7=\left\lceil\sqrt{\sqrt{11}!}!\right\rceil$$

$$8=\left\lfloor\sqrt{\sqrt{\sqrt{11!}}}\right\rfloor$$

$$9=\left\lfloor\sqrt{11}!\right\rfloor$$

$$10=\left\lceil\sqrt{11}!\right\rceil$$

$$11=11$$
$$12=\left\lceil\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\left\lceil\sqrt{\sqrt{\sqrt{\sqrt{(\left\lceil\sqrt{11}\space\right\rceil!)!}}}}\right\rceil!}}}}}\right\rceil$$
$$13=\left\lceil\sqrt{\sqrt{\sqrt{\left\lceil\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\left\lceil\sqrt{\sqrt{\sqrt{\sqrt{(\left\lceil\sqrt{11}\space\right\rceil!)!}}}}\right\rceil!}}}}}\right\rceil}}}\right\rceil$$

$$14=\sqrt{\sqrt{\left\lfloor\sqrt{\sqrt{\left\lceil\sqrt{\sqrt{\sqrt{(1\div(.1))!}}}\right\rceil!}}\right\rfloor!}}$$

• I found a way for 12 but it's just a hassle to write it down. We can go all the way to infinity without much problem. Commented Jan 23, 2017 at 6:05
• Getting past 10 was the hard part.
– humn
Commented Jan 23, 2017 at 6:08
• @humn I simplified my 10 as well as the 7 and 9. Looks much better now. But I guess 11 is a good number to stop this... I am not going any higher lol. Commented Jan 23, 2017 at 6:17
• Anyone want to write a program for this...? Commented Jan 23, 2017 at 6:19
• This is turning into a ridiculous open-ended mathematics exercise. It seems likely that given an unlimited number of factorials, square roots, floor and ceilings, you could express the values of all positive integers. Commented Jan 23, 2017 at 13:46