HP Forums

Quote:Please welcome our newest member, 1332881954

1332881954 has 2 representations as a sum of 2 squares:
\(1332881954 = 6277^2+35965^2 = 23927^2+27575^2\)

Welcome to this forum
Thomas

And my HP Prime enthusiastically assures me that the prime factorization is:

ifactor(1332881954) = 2*349*1909573



A quick google search shows that our new member is registered on a number of alternative sites, most related to calculators.

Welcome to the MoHPC!

Welcome to the MoHPC!

\[\frac{3}{2}\cdot \pi ^{18}+\frac{6!-5!+\frac{\pi ^{-\pi }}{9}}{7}\cdot \pi ^{4}\]

Welcome indeed.

If you're an incipient CAS sufferer, welcome and condolences.

(07-11-2015 05:30 PM)Gerson W. Barbosa Wrote: [ -> ]\[\frac{3}{2}\cdot \pi ^{18}+\frac{6!-5!+\frac{\pi ^{-\pi }}{9}}{7}\cdot \pi ^{4}\]

I'm always amazed by your pandigital expressions. Do you use a program to find them?

Cheers
Thomas

(07-12-2015 11:08 AM)Thomas Klemm Wrote: [ -> ]

(07-11-2015 05:30 PM)Gerson W. Barbosa Wrote: [ -> ]\[\frac{3}{2}\cdot \pi ^{18}+\frac{6!-5!+\frac{\pi ^{-\pi }}{9}}{7}\cdot \pi ^{4}\]

I'm always amazed by your pandigital expressions. Do you use a program to find them?

Hi, Thomas!

No program, only a WP 34S set to double precision and about one hour's time. The pandigital form was almost an accident. At a certain point I noticed I had an expression with only 5 and 9 missing, but with two zeros, so I thought it would not be too difficult to go pandigital:


\[\frac{3\pi ^{18}}{2}+\frac{600\pi ^{4}}{7}+e^{-\pi }\]

These will give results in excess of three on HP 10-digit calculators, like the HP-15C. Although inaccurate, the following expression will give the expected result on the Voyagers:

\[\frac{3\pi ^{3(3 + 3)}}{3-\frac{3}{3}}+\frac{3^{3}\pi ^{3+\frac{3!}{3}}}{.33\times 3}\]

No program has been needed to find this one either :-)

Cheers,

Gerson.

(07-12-2015 11:08 AM)Thomas Klemm Wrote: [ -> ]

(07-11-2015 05:30 PM)Gerson W. Barbosa Wrote: [ -> ]\[\frac{3}{2}\cdot \pi ^{18}+\frac{6!-5!+\frac{\pi ^{-\pi }}{9}}{7}\cdot \pi ^{4}\]

I'm always amazed by your pandigital expressions. Do you use a program to find them?

Cheers
Thomas

Thomas - Does 'pandigital' imply each digit is used only once?
You guys amaze me... thanks for that!

(07-12-2015 03:22 PM)rprosperi Wrote: [ -> ]Does 'pandigital' imply each digit is used only once?

Quote:From Ancient Greek πᾶν (pân), neuter form of πᾶς (pâs, “all, every”)

pan

Quote:From Latin digitālis, from digitus (“finger, toe”) + -alis (“-al”).

digital

Quote:You guys amaze me...

Some of us just know how to spend their time:

(07-12-2015 01:47 PM)Gerson W. Barbosa Wrote: [ -> ]No program, only a WP 34S set to double precision and about one hour's time.

(07-11-2015 10:11 PM)PANAMATIK Wrote: [ -> ]Now, after this is done, I probably don't know what to do with the coming sunday

We're eagerly awaiting the result.

Cheers
Thomas

(07-12-2015 03:22 PM)rprosperi Wrote: [ -> ]Does 'pandigital' imply each digit is used only once?

As Thomas has explained above, this hybrid compound noun means "all digits". Therefore, pandigital doesn't necessary imply only one instance of each digit, but it's better in this kind of play. Even better when the digits are in order :-)

\[e \sqrt[12]{e^{-3\times 4}+5.67890}\]

But this was a very lucky one, I wonder if I will ever come up with something like that again! Also, this one works nicely on the Voyagers, even on the HP-12C, which lacks π.

Gerson.

(07-12-2015 04:26 PM)Gerson W. Barbosa Wrote: [ -> ]\[e \sqrt[12]{e^{-3\times 4}+5.67890}\]

Genius! 3.141592654571540303409367866518360732739326280463671619184479...

(07-12-2015 05:04 PM)Thomas Klemm Wrote: [ -> ]

(07-12-2015 04:26 PM)Gerson W. Barbosa Wrote: [ -> ]\[e \sqrt[12]{e^{-3\times 4}+5.67890}\]

Genius! 3.141592654571540303409367866518360732739326280463671619184479...

Thanks, but I think you'll change your mind when you know about my higher mathematical methodology

(O.T.) Happy Pi Approximation Day! 'e*XROOT(12,e^(-3*4)+5.6789)' (N.T.) (Message #3 in this Old Forum thread)

(07-12-2015 04:26 PM)Gerson W. Barbosa Wrote: [ -> ]

(07-12-2015 03:22 PM)rprosperi Wrote: [ -> ]Does 'pandigital' imply each digit is used only once?

As Thomas has explained above, this hybrid compound noun means "all digits". Therefore, pandigital doesn't necessary imply only one instance of each digit, but it's better in this kind of play. Even better when the digits are in order :-)

\[e \sqrt[12]{e^{-3\times 4}+5.67890}\]

But this was a very lucky one, I wonder if I will ever come up with something like that again! Also, this one works nicely on the Voyagers, even on the HP-12C, which lacks π.

Gerson.

I have not looked ahead to see if anyone else has replied yet, but you forgot the part that makes it so amazing: Showing the PI=(equation); though maybe that's left to be a surprise. This is totally amazing (I was going to say F^%* amazing, but not sure who I might offend). Really, very, very cool. How long did this one take? Now I'm relatively sure Thomas has something to say as well...

(07-12-2015 05:59 PM)Gerson W. Barbosa Wrote: [ -> ]Thanks, but I think you'll change your mind when you know about my higher mathematical methodology

Well, it's still genius, even if 6 years old.

This goes into my collection of amazing equations that are cool just to look at.

Thank you both for the smile on a nice Sunday.

(07-12-2015 06:24 PM)rprosperi Wrote: [ -> ]This goes into my collection of amazing equations that are cool just to look at.

Beauty-wise, nothing surpasses the following real identity, by a real genius:

\[e^{i\pi }+1=0\]

This is what it is, just a mere approximation from someone with idle time to spare:

\[\pi \simeq e \sqrt[12]{e^{-3\times 4}+5.67890}\]

(07-12-2015 06:20 PM)rprosperi Wrote: [ -> ]How long did this one take?

Only five minutes, but as I said then, just a matter of luck.

Best regards,

Gerson.

(07-12-2015 06:20 PM)rprosperi Wrote: [ -> ]Now I'm relatively sure Thomas has something to say as well...

It's a repost but nonetheless relevant.

Cheers
Thomas

(07-12-2015 07:14 PM)Gerson W. Barbosa Wrote: [ -> ]Beauty-wise, nothing surpasses the following real identity, by a real genius:

\[e^{i\pi }+1=0\]

I totally agree, a thing of visual and mathematical beauty. Always will be #1 on my list.

Thanks Thomas. While just about any post from XKCD is worth a look, this one is new to me. Wow, 2 in 1 day!

(07-12-2015 08:36 PM)rprosperi Wrote: [ -> ]While just about any post from XKCD is worth a look, this one is new to me.

It's an old one from 2007-1-31.