Ramanujan

devarajkandadai · 2008-12-22, 03:52

Hardy's cab number & Ramanujan

Today is Ramanujan's birth day and I thought I would give a small variation of the above,

As is well known 1979 is the smallest number which can be expressed as the sum of

two cubes of natural numbers in two different ways.

If one of the four Diophantine variables were to belong to Z we get

91 = 3^3 + 4^3 = 6^3 - 5^3.

Q: Is 91 the smallest number that can be expressed in this manner?

A.K. Devaraj

NBtarheel_33 · 2008-12-22, 04:19

Quote:

Originally Posted by devarajkandadai

Hardy's cab number & Ramanujan

As is well known 1979 is the smallest number which can be expressed as the sum of two cubes of natural numbers in two different ways.

Of course, you mean 1729. I don't believe 1979 is the sum of two cubes, even in only one way:

1979 - 1^3 = 1978 not a cube
1979 - 2^3 = 1971 not a cube
1979 - 3^3 = 1952 not a cube
1979 - 4^3 = 1915 not a cube
1979 - 5^3 = 1854 not a cube
1979 - 6^3 = 1763 not a cube
1979 - 7^3 = 1636 not a cube
1979 - 8^3 = 1467 not a cube
1979 - 9^3 = 1250 not a cube
1979 - 10^3 = 979 not a cube
1979 - 11^3 = 648 not a cube
1979 - 12^3 = 251 not a cube
1979 - 13^3 < 0

Yep, just what I thought

retina · 2008-12-22, 04:19

Quote:

Originally Posted by devarajkandadai

Hardy's cab number & Ramanujan

Today is Ramanujan's birth day and I thought I would give a small variation of the above,

As is well known 1979 is the smallest number which can be expressed as the sum of

two cubes of natural numbers in two different ways.

If one of the four Diophantine variables were to belong to Z we get

91 = 3^3 + 4^3 = 6^3 - 5^3.

Q: Is 91 the smallest number that can be expressed in this manner?

A.K. Devaraj

I doubt it, since you mention the set Z then I expect an infinitude of negative values will be smaller.

devarajkandadai · 2008-12-22, 09:30

Quote:

Originally Posted by NBtarheel_33

Of course, you mean 1729. I don't believe 1979 is the sum of two cubes, even in only one way:

1979 - 1^3 = 1978 not a cube
1979 - 2^3 = 1971 not a cube
1979 - 3^3 = 1952 not a cube
1979 - 4^3 = 1915 not a cube
1979 - 5^3 = 1854 not a cube
1979 - 6^3 = 1763 not a cube
1979 - 7^3 = 1636 not a cube
1979 - 8^3 = 1467 not a cube
1979 - 9^3 = 1250 not a cube
1979 - 10^3 = 979 not a cube
1979 - 11^3 = 648 not a cube
1979 - 12^3 = 251 not a cube
1979 - 13^3 < 0

Yep, just what I thought

Tks; the error was just a typo - the cab number was 1729.
Devaraj

Brian-E · 2008-12-22, 14:15

A nice page is http://euler.free.fr/taxicab.htm though unfortunately it does not seem to have been updated for about 2 years.

Your puzzle is defined by the author as cabtaxi(2).

Edit: I just noticed that the new results are now being presented on a new page http://cboyer.club.fr/Taxicab.htm

davieddy · 2008-12-22, 15:15

Quote:

Originally Posted by devarajkandadai

Tks; the error was just a typo - the cab number was 1729.
Devaraj

I don't wish to put a dampener on Hardy's famous anecdote,
but I suspect he might have moulded it so that the punchline
had maximum effect: the interesting property of 1729 raises
gasps of incredulity in the layman.
OTOH knowing that 9^3=729 and 12^3=1728 makes 1729
easy to remember.

David

spkarra · 2009-07-15, 01:33

Quote:

Originally Posted by devarajkandadai

Hardy's cab number & Ramanujan

Today is Ramanujan's birth day and I thought I would give a small variation of the above,

Well, I saw another astonishing quote from Ramanujan in 1913. According to him Factorial(n) + 1 = m^2, then n=4,5,7 only.
I wrote a JAVA program and did as follows:
1. For a given number, calculated the factorial
2. Added 1 to it
3. Calculated the square root of the result
4. Checked if its a perfect square root (ie. sqrt(4) = 2.000000)

Guess what, I tried upto n=22167 and DIDNOT find a perfect m.

It still amazes me, how Ramanujan could state this in 1913?

Eager to know more about Ramanujan's unbelievable conjectures.

-Sastry

Damian · 2009-07-17, 20:20

spkarra, reading your post I wondered wich roots would have the equation $\Gamma(x) = x^2$

Of course we have $x_0 = 1$
But plotting both functions, it seems there is another intersection near $x=3,5$

Is there any way of getting a closed form of that root?

Orgasmic Troll · 2009-07-17, 22:41

Quote:

Originally Posted by Damian

spkarra, reading your post I wondered wich roots would have the equation $\Gamma(x) = x^2$

Of course we have $x_0 = 1$
But plotting both functions, it seems there is another intersection near $x=3,5$

Is there any way of getting a closed form of that root?

3.5? Uhm. What? $\Gamma(x) - x^2 = 0$ has a root at 5.036722570531942...

Ahh, it looks like you were solving $\Gamma(x) - x = 0$ , which has a root at 3.562382285390843...

Damian · 2009-07-18, 03:23

Yes, I was solving those ecuations

$\Gamma(x) - x^1 = 0$
$\Gamma(x) - x^2 = 0$
$\Gamma(x) - x^3 = 0$
$\Gamma(x) - x^4 = 0$
$\Gamma(x) - x^5 = 0$

for $x > 1$

with I tried on wxMaxima:

find_root(x!-x^1, x, 1.01, 100);
find_root(x!-x^2, x, 1.01, 100);
find_root(x!-x^3, x, 1.01, 100);
find_root(x!-x^4, x, 1.01, 100);
find_root(x!-x^5, x, 1.01, 100);

and yielded the following numbers

2.0
3.562382285390898
5.036722570588711
6.464468490129385
7.861923212307213

I was wondering if there were a closed form for these numbers.

xilman · 2009-07-19, 20:26

Quote:

Originally Posted by spkarra

Well, I saw another astonishing quote from Ramanujan in 1913. According to him Factorial(n) + 1 = m^2, then n=4,5,7 only.
I wrote a JAVA program and did as follows:
1. For a given number, calculated the factorial
2. Added 1 to it
3. Calculated the square root of the result
4. Checked if its a perfect square root (ie. sqrt(4) = 2.000000)

Guess what, I tried upto n=22167 and DIDNOT find a perfect m.

It still amazes me, how Ramanujan could state this in 1913?

Eager to know more about Ramanujan's unbelievable conjectures.

-Sastry

Many years ago I verified this conjecture up to n = 100M. A simple algorithm is much more efficient than yours.

Using only pencil and paper it is straightforward to verify for n up to 1000 or so. Very tedious and requires attention to accuracy, but straightforward.

I won't immediately reveal how to test the conjecture for small n, but I will give a hint: quadratic residues.

Paul

2008-12-22, 03:52	#1
devarajkandadai May 2004 316₁₀ Posts	Ramanujan Hardy's cab number & Ramanujan Today is Ramanujan's birth day and I thought I would give a small variation of the above, As is well known 1979 is the smallest number which can be expressed as the sum of two cubes of natural numbers in two different ways. If one of the four Diophantine variables were to belong to Z we get 91 = 3^3 + 4^3 = 6^3 - 5^3. Q: Is 91 the smallest number that can be expressed in this manner? A.K. Devaraj

2008-12-22, 14:15	#5
Brian-E "Brian" Jul 2007 The Netherlands 2×11×149 Posts	A nice page is http://euler.free.fr/taxicab.htm though unfortunately it does not seem to have been updated for about 2 years. Your puzzle is defined by the author as cabtaxi(2). Edit: I just noticed that the new results are now being presented on a new page http://cboyer.club.fr/Taxicab.htm Last fiddled with by Brian-E on 2008-12-22 at 14:28 Reason: discovered new link

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2009-07-17, 20:20	#8
Damian May 2005 Argentina 2×3×31 Posts	spkarra, reading your post I wondered wich roots would have the equation $\Gamma(x) = x^2$ Of course we have $x_0 = 1$ But plotting both functions, it seems there is another intersection near $x=3,5$ Is there any way of getting a closed form of that root?

2009-07-18, 03:23	#10
Damian May 2005 Argentina 2×3×31 Posts	Yes, I was solving those ecuations $\Gamma(x) - x^1 = 0$ $\Gamma(x) - x^2 = 0$ $\Gamma(x) - x^3 = 0$ $\Gamma(x) - x^4 = 0$ $\Gamma(x) - x^5 = 0$ for $x > 1$ with I tried on wxMaxima: find_root(x!-x^1, x, 1.01, 100); find_root(x!-x^2, x, 1.01, 100); find_root(x!-x^3, x, 1.01, 100); find_root(x!-x^4, x, 1.01, 100); find_root(x!-x^5, x, 1.01, 100); and yielded the following numbers 2.0 3.562382285390898 5.036722570588711 6.464468490129385 7.861923212307213 I was wondering if there were a closed form for these numbers.