Four Nines

[R. Gerbicz]

Originally Posted by petrw1

Could you rerun your program without it and see how many we lose.

OK, made a rerun; without percent symbol: the smallest missing value is at n=103, up to n=1000 there are 445 missing values, and up to n=10000 there are 8682 missing values.

[xilman]

If you allow me the standard notation lg(x) defined as the binary logarithm of x, log_n(x) defined as logarithm of x to base n, the square root symbol and the minus-symbol I can generate all integers with only two nines.

I'll build up the proof in easy stages.

First, note that

log₉(√9) = 1/2,
log₉(√√9) = 1/4,
log₉(√√√9) = 1/8

and, in general,

log₉(√...√9) = 1/2^N where the √ symbol has been repeated N times.

Second, lg(1/2^N) = -N

So far I've created all the negative integers. Preceding the expressions given above with a '-' generates all the strictly positive integers.

The only remaining case to be solved is zero, which is just 9-9.

The case 103, in this notation, is

-lg(log₉(√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√√9))

If you disallow lg(x), replace it with log _log[SUB]√9(9)[/SUB](x) and now you've a solution which uses four nines.

[petrw1]

Originally Posted by R. Gerbicz

OK, made a rerun; without percent symbol: the smallest missing value is at n=103, up to n=1000 there are 445 missing values, and up to n=10000 there are 8682 missing values.

Cool thanks.

You found the following 9 that we had solutions for using the % (could be once people fouud the solution with the % they didn't look further)

140	 9/((√(9)!)%) - 9/.9	                    ((((sqrt(9))!)!)-((sqrt(9))!))/(((sqrt(9))!)-.9)
150	 ((9*9)/9%)/√(9)!	                    9/(.9/(9+((sqrt(9))!)))
152	 9/((√(9)!)%) + √(9)!/√(9)	            ((((sqrt(9))!)!)+((((sqrt(9))!)!)*.9))/9
170	 (√(9)!)/((√(9))%) - 9/(√(9%))	            (((sqrt(9))!)+((9!)/(((sqrt(9))!)!)))/(sqrt(9))
199	 99 + 9/9%	                            sqrt(((9+(9!))/9)-(((sqrt(9))!)!))
206	 (9+9)/9%+√(9)!	                            (((sqrt(9))!)^(sqrt(9)))-(9/.9)
220	 ((√(9)!)!/√(9)! + 9/9%	                    ((9+(sqrt(9)))!)/((9!)*((sqrt(9))!))
273	 (√(9)/9% - √(9)) * 9	                    ((((sqrt(9))!)!)+99)/(sqrt(9))
306	 9 / (√(9))% + √(9) + √(9)	            9+((sqrt(9))*99)

We stopped at 315 ... with a few sparse exceptions.

And at first look it seemed you missed one...however my combined list had 1 mistake.
295 = √(9)!^ √(9) + 9*9
should be 297 (though we had another solution for that one.

Now how about four 4's. That puzzle should be somewhere in this forum too.

[xilman]

Originally Posted by petrw1

Now how about four 4's. That puzzle should be somewhere in this forum too.

The solution I posted works exactly the same as for the four 9's. It works for all integers >1.

I must make it clear that I didn't invent the technique. I have an idea that it originates with PAM Dirac but may have misremembered. However, it appears not to be widely known so I take up opportunities to publicize it.

[R. Gerbicz]

Originally Posted by xilman

However, it appears not to be widely known so I take up opportunities to publicize it.

See for example: https://en.wikipedia.org/wiki/Four_fours, so it is well known. And we can do an even better, both of these formulas are using unary negative, but we can avoid this:

n=log_c{log_x{b}}, where we use four b values:
c=log_{sqrt(b)}{b} and x=sqrt(sqrt(...(sqrt(b))...) (here sqrt iterated n times)
with this c=2 and x=b^(1/2^n), hence log_c{log_x{b}}=log_2(2^n)=n.

Originally Posted by petrw1

Now how about four 4's. That puzzle should be somewhere in this forum too.

I'll update that, though up to n=999 there is no new solution at: http://www.mersenneforum.org/showthr...t=4756&page=10