[math-fun] Fibonacci power triangle
(Everything that follows is unproven conjecture. It's probably old news to Fibonacci aficionados.) It occurred to me that squares of Fibonacci numbers should have a simple recurrence relation. After a bit of futzing, I found 2 2 2 2 F' = 2 F + 2 'F - ''F (I use ' to mean "next term" and prefix ' to mean "previous term".) So 169 = 2*64 + 2*25 - 9. Another way to say this is that the sequence of Fibonacci squares can be dotted with the vector (-1 2 2 -1) to produce 0. The vector for the plain Fibonacci sequence is (1 1 -1). Experimenting showed the vector for Fibonacci cubes to be (-1 -3 6 3 -1). I worked out fourth and fifth powers, and developed the Fibonacci-Pascal triangle, built from the vectors. 1 -1 1 1 -1 -1 2 2 -1 -1 -3 6 3 -1 1 -5 -15 15 5 -1 1 8 -40 -60 40 8 -1 -1 13 104 -260 -260 104 13 -1 The rule for the signs seems clear enough. The magnitudes turn out to be Fibonacci-Binomials. The outer diagonals are 1 1 1 1 ...; the next diagonals are just the plain Fibs; and the next diagonals are products of adjacent Fibs: 104 = 8*13, etc. The next diagonals turn out to be products of three adjacent Fibs divided by 2: 260 = 5*8*13/2. The right way to look at this seems to be 5*8*13/(1*1*2), a kind of binomial coefficent with the regular integers replaced by Fibonacci numbers. The next diagonals will be products of four adjacent Fibs divided by 1*1*2*3, and so on. This can probably be connected with q-binomial coefficients. Some other curiosities: (-1 2 2 -1) also works for F*F' (-1 -3 6 3 -1) works for F*F'*F'' (1 -5 -15 15 5 -1) works for F*F'*F''*F''' Fib(2N+1) = (1 1) dotted with F^2; Fib(2N) = (-1 0 1) dot F^2. Fib(3N) = (-1 1 1) dotted with F^3. Since (-1 2 2 -1) works with both F^2 and F*F', it will also work for squares of the Lucas sequence. Continuing this idea a bit further, the vectors work with powers of generalized Fibonacci sequences, where the starting pair of numbers is A,B, and the terms continue A+B, A+2B, 2A+3B, ... Rich rcs@cs.arizona.edu
I'm copying this to OEIS editors, in the hope that the essence is, or will be, included in OEIS. R. On Sun, 5 Jan 2003, Richard Schroeppel wrote:
(Everything that follows is unproven conjecture. It's probably old news to Fibonacci aficionados.)
It occurred to me that squares of Fibonacci numbers should have a simple recurrence relation. After a bit of futzing, I found
2 2 2 2 F' = 2 F + 2 'F - ''F
(I use ' to mean "next term" and prefix ' to mean "previous term".)
So 169 = 2*64 + 2*25 - 9.
Another way to say this is that the sequence of Fibonacci squares can be dotted with the vector (-1 2 2 -1) to produce 0. The vector for the plain Fibonacci sequence is (1 1 -1). Experimenting showed the vector for Fibonacci cubes to be (-1 -3 6 3 -1). I worked out fourth and fifth powers, and developed the Fibonacci-Pascal triangle, built from the vectors.
1 -1
1 1 -1
-1 2 2 -1
-1 -3 6 3 -1
1 -5 -15 15 5 -1
1 8 -40 -60 40 8 -1
-1 13 104 -260 -260 104 13 -1
The rule for the signs seems clear enough. The magnitudes turn out to be Fibonacci-Binomials. The outer diagonals are 1 1 1 1 ...; the next diagonals are just the plain Fibs; and the next diagonals are products of adjacent Fibs: 104 = 8*13, etc. The next diagonals turn out to be products of three adjacent Fibs divided by 2: 260 = 5*8*13/2. The right way to look at this seems to be 5*8*13/(1*1*2), a kind of binomial coefficent with the regular integers replaced by Fibonacci numbers. The next diagonals will be products of four adjacent Fibs divided by 1*1*2*3, and so on. This can probably be connected with q-binomial coefficients.
Some other curiosities: (-1 2 2 -1) also works for F*F' (-1 -3 6 3 -1) works for F*F'*F'' (1 -5 -15 15 5 -1) works for F*F'*F''*F'''
Fib(2N+1) = (1 1) dotted with F^2; Fib(2N) = (-1 0 1) dot F^2.
Fib(3N) = (-1 1 1) dotted with F^3.
Since (-1 2 2 -1) works with both F^2 and F*F', it will also work for squares of the Lucas sequence. Continuing this idea a bit further, the vectors work with powers of generalized Fibonacci sequences, where the starting pair of numbers is A,B, and the terms continue A+B, A+2B, 2A+3B, ...
Rich rcs@cs.arizona.edu
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There are various number theoretic problems in the Gaussian Integers that I haven't been able to solve. http://www.mathpuzzle.com/Gaussians.html If any of these matters are resolved, or if you have solutions to any of them, please let me know. --Ed Pegg Jr, www.mathpuzzle.com
as to: <CITE http://www.mathpuzzle.com/Gaussians.html> (3+13i)^3 + (7+i)^3 = (3+10i)^3 + (1+10i)^3 and (6+3i)^4 + (2+6i)^4 = (4+2i)^4 + (2+i)^4 Is there something similar for Gaussian fifth powers? See http://euler.free.fr/details.htm for details on this problem in the integers. I offer $20 for a nontrivial solution with proper Gaussian integers </CITE http://www.mathpuzzle.com/Gaussians.html> I have a parametric solution for the fifth power (at home). I'll dig it up and send the solution to the list. all the best, jj * ed pegg <ed@mathpuzzle.com> [Jan 06. 2003 20:26]:
There are various number theoretic problems in the Gaussian Integers that I haven't been able to solve.
http://www.mathpuzzle.com/Gaussians.html
If any of these matters are resolved, or if you have solutions to any of them, please let me know.
--Ed Pegg Jr, www.mathpuzzle.com
-- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
p^5 + q^5 == r^5 + s^5 == v [p, q, r, s] v [1 + 3/2*I, 1 - 3/2*I, 3/2, 1/2] 61/8 [1 + 7/5*I, 1 - 7/5*I, 1 + 1/5*I, 1 - 1/5*I] 152/125 [1 + 9/4*I, 1 - 9/4*I, 11/4, -3/4] 20101/128 [1 + 27/10*I, 1 - 27/10*I, 33/10, -13/10] 387641/1000 [1 + 51/10*I, 1 - 51/10*I, 59/10, -39/10] 6247001/1000 [1 + 17/13*I, 1 - 17/13*I, 1 + 7/13*I, 1 - 7/13*I] -84488/28561 Note that the solutions are semi-trivial: the sum of the lhs. terms is equal to those on the rhs. My parametric gives square roots in general. Need more time to check whether I can repair that. (My documentation and code about the stuff is extremely messy). I am pretty sure there are solutions for higher powers, too. If there is interest, I'll do a search when time permits. * Joerg Arndt <jj@suse.de> [Jan 09. 2003 13:09]:
as to: <CITE http://www.mathpuzzle.com/Gaussians.html> (3+13i)^3 + (7+i)^3 = (3+10i)^3 + (1+10i)^3 and (6+3i)^4 + (2+6i)^4 = (4+2i)^4 + (2+i)^4 Is there something similar for Gaussian fifth powers? See http://euler.free.fr/details.htm for details on this problem in the integers. I offer $20 for a nontrivial solution with proper Gaussian integers </CITE http://www.mathpuzzle.com/Gaussians.html>
I have a parametric solution for the fifth power (at home). I'll dig it up and send the solution to the list.
all the best, jj
* ed pegg <ed@mathpuzzle.com> [Jan 06. 2003 20:26]:
There are various number theoretic problems in the Gaussian Integers that I haven't been able to solve.
http://www.mathpuzzle.com/Gaussians.html
If any of these matters are resolved, or if you have solutions to any of them, please let me know.
--Ed Pegg Jr, www.mathpuzzle.com
-- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
Fred Helenius found (15+14i)^5 + (5-18i)^5 = (18-7i)^5 + (2+3i)^5 --- Joerg Arndt <jj@suse.de> wrote:
p^5 + q^5 == r^5 + s^5 == v
[p, q, r, s] v
[1 + 3/2*I, 1 - 3/2*I, 3/2, 1/2] 61/8 [1 + 7/5*I, 1 - 7/5*I, 1 + 1/5*I, 1 - 1/5*I] 152/125 [1 + 9/4*I, 1 - 9/4*I, 11/4, -3/4] 20101/128 [1 + 27/10*I, 1 - 27/10*I, 33/10, -13/10] 387641/1000 [1 + 51/10*I, 1 - 51/10*I, 59/10, -39/10] 6247001/1000 [1 + 17/13*I, 1 - 17/13*I, 1 + 7/13*I, 1 - 7/13*I] -84488/28561
Note that the solutions are semi-trivial: the sum of the lhs. terms is equal to those on the rhs.
My parametric gives square roots in general. Need more time to check whether I can repair that. (My documentation and code about the stuff is extremely messy).
I am pretty sure there are solutions for higher powers, too. If there is interest, I'll do a search when time permits.
* Joerg Arndt <jj@suse.de> [Jan 09. 2003 13:09]:
as to: <CITE http://www.mathpuzzle.com/Gaussians.html> (3+13i)^3 + (7+i)^3 = (3+10i)^3 + (1+10i)^3 and (6+3i)^4 + (2+6i)^4 = (4+2i)^4 + (2+i)^4 Is there something similar for Gaussian fifth powers? See http://euler.free.fr/details.htm for details on this problem in the integers. I offer $20 for a nontrivial solution with proper Gaussian integers </CITE http://www.mathpuzzle.com/Gaussians.html>
I have a parametric solution for the fifth power (at home). I'll dig it up and send the solution to the list.
all the best, jj
* ed pegg <ed@mathpuzzle.com> [Jan 06. 2003 20:26]:
There are various number theoretic problems in the Gaussian Integers that I haven't been able to solve.
http://www.mathpuzzle.com/Gaussians.html
If any of these matters are resolved, or if you have solutions to any of them, please let me know.
--Ed Pegg Jr, www.mathpuzzle.com
-- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com
http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
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--- Ed Pegg Jr <ed@mathpuzzle.com> wrote:
Fred Helenius found (15+14i)^5 + (5-18i)^5 = (18-7i)^5 + (2+3i)^5 --- Joerg Arndt <jj@suse.de> wrote:
Whoops ... I meant that as a private message to Joerg. As it turns out, Fred Helenius found many things, and W. Edwin Clarke found many references. If you want to know any of the number theoretic properties of Gaussian integers, it's probably mentioned at http://www.mathpuzzle.com/Gaussians.html . And I did an update ... usual batch of mathy results. --Ed Pegg Jr, www.mathpuzzle.com
Thanks for the info. Note that, as for my solutions, with x^5 + y^5 = u^5 + v^5 then (x+y) = (u+v) My solution can be selected from a parametric that in general gives sqrt-terms. Shall I write up a sketch and post it to the list? all the best, jj * Ed Pegg Jr <ed@mathpuzzle.com> [Feb 06. 2003 08:41]:
--- Ed Pegg Jr <ed@mathpuzzle.com> wrote:
Fred Helenius found (15+14i)^5 + (5-18i)^5 = (18-7i)^5 + (2+3i)^5 --- Joerg Arndt <jj@suse.de> wrote:
Whoops ... I meant that as a private message to Joerg. As it turns out, Fred Helenius found many things, and W. Edwin Clarke found many references. If you want to know any of the number theoretic properties of Gaussian integers, it's probably mentioned at http://www.mathpuzzle.com/Gaussians.html .
And I did an update ... usual batch of mathy results. --Ed Pegg Jr, www.mathpuzzle.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
participants (5)
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ed pegg -
Ed Pegg Jr
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Joerg Arndt -
Richard Guy -
Richard Schroeppel