(OT) Pandigital expression (HP-48,49,50g)

[Gerson W. Barbosa]

Today is Sunday, but it's also the national date for some members of this forum. The following is kind of a celebration. Notice all digits from 0 through 9 are present and have been used only once. Perhaps FIX 6 should be used for proper day format.

Congratulations !

_{Edited to correct a language mistake}

Edited: 14 July 2013, 4:52 p.m.

[hugh steers]

hmm. i'm out by nearly a millenium.

> (0!+sqrt(2)+14*14/75!/(10^6+89))^3

14.07106781186547524400845

must be this mystery (-,75) thing?

any hints.

[Thomas Klemm]

Quote:

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must be this mystery (-,75) thing?

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For you probably rather (-.75)! = 3.62560990822.

Cheers

Thomas

[Gerson W. Barbosa]

Gamma(1/4) would be nicer, but then 1 and 4 had already been used.

DECIMAL POINT IS COMMA :-)

Cheers,

Gerson.

[Massimo Gnerucci (Italy)]

Quote:

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DECIMAL POINT IS COMMA

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Ditto!

forever and ever amen.

[hugh steers]

thank you!

the flux capacity is now fixed!

> (0!+sqrt(2)+14*14/-.75!/(10^6+89))^3

14.07201300094483189388136

for a while i was thinking (-1,75)! ie (-1+75i)! which doesnt end up in our spacetime even :-)

[Gerson W. Barbosa]

Wolfram Alpha will understand (0!+Sqrt[2]+Sq[14]/((-.75)!*(Alog[6]+89)))^3.
Originally pandigital expressions would involve only arithmetic operators, but I think the use of Sq, Sqrt etc., when available, makes things a bit more interesting. Sure this is a futile puzzle, but I didn't spend more than thirty minutes on this one :-)

[Gerson W. Barbosa]

Too bad most programming languages don't have this feature. Back in the day I changed a few bytes in the ROM of my MSX computer. The result was fair enough :-)

[Didier Lachieze]

Thanks, this is a nice one! I'm curious about the methodology you used to get it...

[Victor Koechli]

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I'm curious about the methodology you used to get it...

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Me too, especially after learning that it only took you 30 minutes to find it. Nicely done!

[Gerson W. Barbosa]

keystrokes                       display          comments

14.072013 STO A                14.072013    

LN                             2.64418793126      ~ pi^2/6 + 1

pi ENTER * 6 / 1 + - 1/X +/-   1340.23897721      nothing interesting after trying a

                                                  few functions and multiples

RCL A sqrt sqrt sqrt sqrt      1.1797018602       again, nothing interesting here

RCL A 3 1/x y^x                2.41426761738      ~ sqrt(2) + 1  -- this looks promising

2 sqrt - 1 - 1/x STO B         18499.6728333 

3 *                            55499.0184999      here we have

                                                  (sqrt(2) + 1 + 3/55499)^3 = 14.0720130004

                                                  but the 5-digit constant is almost as long

                                                  as the number we want to represent, also 

                                                  it is not interesting. So let's try other 

                                                  multiples

RCL B ENTER ENTER ENTER +      36999.3456666

+ + + + + + + + + + + + +      277495.092497

...                                               (very fast keystrokes, I may have missed

                                                   some interesting results)

+ + + + + + + + + +  STO C     3625935.87485      the first four digits match those of 

                                                  gamma(1/4)

4 1/x 1 - x! /                 1000089.9067       now we have 

                                                  (196/(gamma(1/4)*(10^6 + 90)) + sqrt(2) + 1)^3

                                                  = 14.0720129999

                                                  Again, not interesting enough, but after

                                                  noticing 196 = 14^2 and gamma(1/4) = (-0.75)! 

                                                  we can try a pandigital expression. There are 

                                                  repeated digits (0, 1 and 2) and 8 is missing. 

                                                  Replacing 90 with 89 solves the latter and 

                                                  eliminates one repeated 0, 1 can be written as 

                                                  0! and 14^2 as Sq(14). Also Alog(x) can be used

                                                  for 10^x, so we finally have

                                                  (Sq(14)/((-.75)!*(Alog(6) + 89)) + Sqrt(2) + 0!)^3

14 ENTER * .75 +/- x! / 6 10x

89 + / 2 sqrt + 0 x! + 3 yx 

DISP FIX 6                     14.072013          = 14.0720130009

Calculator: HP-32SII
Shifts have been omitted in the keystrokes listing above

[Thomas Klemm]

Quote:

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3625935.87485 the first four digits match those of gamma(1/4)

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Of course we all know that by heart. Let's be honest: who would not think immediately of that?
Gerson, you're just amazing!

Cheers

Thomas

[Gerson W. Barbosa]

Quote:

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Quote:

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3625935.87485 the first four digits match those of gamma(1/4)

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Of course we all know that by heart.

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Well, at least the first few digits of a few constants we all do.

Not exactly a scientific methodology, but I guess W|A is not capable of this kind of thing [yet] :-)

Cheers,

Gerson.

[Victor Koechli]

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Of course we all know that by heart.

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You literally took the words out of my mouth. I'm still amazed by Gerson's procedure!

[Gerson W. Barbosa]

A similar "method" yielded 'e*XROOT(12,e^(-3*4)+5.6789)' four years ago (this can be appended to '0+' to include all 10 digits, in ascending order!). And that took only 5 minutes :-)
It was only a matter of luck in both occasions, though. No idea for this year's Pi Approximation Day...

Cheers,

Gerson.

[Gilles Carpentier]

:D

Thank's !

I tried to find another pandigital expression for this without success

[Gerson W. Barbosa]

I don't think I would have found this one if I had started by trying to find it in the beginning. As I said, I was lucky I came up with an almost ready-made pandigital expression :-)

Cheers,

Gerson.

[Gilles Carpentier]

I'm late but :

'SQ(4!-1)*2*7*(3*6+9-8)-5^0'

In french date format of course ;)

[Gerson W. Barbosa]

Très bien !
And you already have the ones for the next two years :-)

[Gilles Carpentier]

To find this, I use the FACTORS command of the 50G (or the ifactors of the Prime):

140713=140714-1

140713=2*7*19*23²-1

Edited: 19 July 2013, 6:01 a.m.