Re: [math-fun] Triangular+Triangular = Factorial
Very interesting, Richard -- and Ed ! This suggests a few questions to me, possible hard ones: 1) Can we prove there are infinitely many solutions to T[x] + T[y] = z! ? How about a probabilistic heuristic? 2) Generally, given reasonably simple functions F,G: Z+ -> Z+ when can we prove there are infinitely many solutions to F(x) + F(y) = G(z) ? Likewise, what about a probabilistic "proof" ? --Dan ---------------------------------------------- Richard wrote: << . . . Solutions of x(x+1)/2 + y(y+1)/2 = z! are solutions of (2x+1)^2 + (2y+1)^2 = 8(z!) + 2 and the first few values of z for which there are solutions can easily be ascertained: (x,y,z) = (0,1,0), (0,1,1), (1,1,2), (0,3,3), (2,2,3), (2,6,4), (0,15,5), (5,14,5), (45,89,7), (89,269,8), (210,825,9), (760,2610,10), (1770,2030,10), . . .
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
What it boils down to is: Are there infinitely many 4*n! + 1 which only have factors of shape 4k + 1 (or possibly some factors 4k + 3, but to an even power). I'm inclined to think ``yes'', but this is far beyond our reach. It may even be true that there are infinitely many primes 4*n! + 1 ???? R. On Fri, 10 Sep 2010, Dan Asimov wrote:
Very interesting, Richard -- and Ed !
This suggests a few questions to me, possible hard ones:
1) Can we prove there are infinitely many solutions to
T[x] + T[y] = z! ? How about a probabilistic heuristic?
2) Generally, given reasonably simple functions
F,G: Z+ -> Z+
when can we prove there are infinitely many solutions to
F(x) + F(y) = G(z) ?
Likewise, what about a probabilistic "proof" ?
--Dan ----------------------------------------------
Richard wrote:
<< . . . Solutions of
x(x+1)/2 + y(y+1)/2 = z!
are solutions of (2x+1)^2 + (2y+1)^2 = 8(z!) + 2 and the first few values of z for which there are solutions can easily be ascertained:
(x,y,z) = (0,1,0), (0,1,1), (1,1,2), (0,3,3), (2,2,3), (2,6,4), (0,15,5), (5,14,5), (45,89,7), (89,269,8), (210,825,9), (760,2610,10), (1770,2030,10), . . .
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Using typical probability arguments, there are infinitely many primes K = 4 N! + 1, but they are pretty thin: Ignoring what we know about no-small-factors, the likelihood that 4 N! + 1 is prime is 1/log K, very roughly 1/N log N. The sum of this is about loglog N. By this calculation, we get a new prime every time N is raised to the power e. The numbers K are very thin indeed. We can correct for the fact that our number shape has no divisors <=N (in fact, N+1 and N+2 are also impossible for N>3) by bumping up the likelihood by prod(P/(P-1)) for P<=N (and P prime of course). IIRC this is O(logN), making our summand O(1/N), and the sum O(logN). This is comparable to Mersenne primes in sparsity for K, and MP exponents for sparsity in N. I.e., we get a new prime every time N doubles (and K squares). The likelihood that a number K of this shape is the sum of two squares but not necessarily prime, (again IIRC) is O(1/sqrt log K). The approximate sum of 1/sqrt(N logN) seems to be O(sqrt(N/logN)), indicating quite a few N are winners. [No, I can't prove any of this.] Rich -------------- Quoting Richard Guy <rkg@cpsc.ucalgary.ca>:
What it boils down to is:
Are there infinitely many 4*n! + 1 which only have factors of shape 4k + 1 (or possibly some factors 4k + 3, but to an even power).
I'm inclined to think ``yes'', but this is far beyond our reach. It may even be true that there are infinitely many primes 4*n! + 1 ???? R.
On Fri, 10 Sep 2010, Dan Asimov wrote:
Very interesting, Richard -- and Ed !
This suggests a few questions to me, possible hard ones:
1) Can we prove there are infinitely many solutions to
T[x] + T[y] = z! ? How about a probabilistic heuristic?
2) Generally, given reasonably simple functions
F,G: Z+ -> Z+
when can we prove there are infinitely many solutions to
F(x) + F(y) = G(z) ?
Likewise, what about a probabilistic "proof" ?
--Dan ----------------------------------------------
Richard wrote:
<< . . . Solutions of
x(x+1)/2 + y(y+1)/2 = z!
are solutions of (2x+1)^2 + (2y+1)^2 = 8(z!) + 2 and the first few values of z for which there are solutions can easily be ascertained:
(x,y,z) = (0,1,0), (0,1,1), (1,1,2), (0,3,3), (2,2,3), (2,6,4), (0,15,5), (5,14,5), (45,89,7), (89,269,8), (210,825,9), (760,2610,10), (1770,2030,10), . . .
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (3)
-
Dan Asimov -
rcs＠xmission.com -
Richard Guy