2
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Using two 1s, try to come up with the most consecutive positive integers.

Allowed operations:

  • Addition

  • Subtraction

  • Multiplication

  • Division

  • Concatenation

  • Square Root

  • Radical

  • Factorial

  • Floor and Ceiling Functions

  • Decimal Point

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  • 5
    $\begingroup$ 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... $\endgroup$ – Gareth McCaughan Jan 23 '17 at 3:52
  • $\begingroup$ @GarethMcCaughan It'd have to be $>2$, because $2!=2$, floor decreases and square root makes closer to 1. $\endgroup$ – boboquack Jan 23 '17 at 4:34
  • $\begingroup$ @GarethMcCaughan ... and floor yields an integer. $\endgroup$ – Rosie F Jan 23 '17 at 7:04
  • $\begingroup$ Oops, yes, I meant >2. $\endgroup$ – Gareth McCaughan Jan 23 '17 at 10:04
  • $\begingroup$ Do you happen to recall the name of this conjecture? $\endgroup$ – greenturtle3141 Jan 24 '17 at 0:48
5
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Let's try (feel free to add on or correct, this is community wiki):

$1=1\times1$

$2=1+1$

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

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

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

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

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

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

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

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

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

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  • 1
    $\begingroup$ 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. $\endgroup$ – stack reader Jan 23 '17 at 6:05
  • $\begingroup$ Getting past 10 was the hard part. $\endgroup$ – humn Jan 23 '17 at 6:08
  • 2
    $\begingroup$ @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. $\endgroup$ – stack reader Jan 23 '17 at 6:17
  • 1
    $\begingroup$ Anyone want to write a program for this...? $\endgroup$ – greenturtle3141 Jan 23 '17 at 6:19
  • $\begingroup$ 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. $\endgroup$ – wildBillMunson Jan 23 '17 at 13:46

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