[math-fun] phi(N)*sigma(N)<N^2, phi(N)+sigma(N)>=2*N
The formula for phi(N) is N * product (1 - 1/P) product taken over primes P that divide N. The formula for sigma(N) is N * product (1 + 1/P + 1/P^2 + ... 1/P^E) with P dividing N and P^E being the highest power of P that divides N. So obviously sigma(N) < N * product(1 + 1/P + 1/P^2 + ... ) = N^2 / phi(N). Q.E.D. Also obviously sigma(N) >= N * product ( 1 + 1/P ) so phi(N) + sigma(N) >= 2*N since the cross terms in the products all either cancel out or come out to something positive. [The avoidance of phi+sigma=2*N+5 is not so trivial, though.] -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Puzzle: What is the next pair in this sequence of integer pairs: (5, 9) (10, 16) (213, 303) (382, 542) (7277, 10293) (13018, 18412) (247245, 349659) ? Hint: They are all related in the same way to 256. —Dan
On 11/22/2015 4:07 PM, Dan Asimov wrote:
Puzzle: What is the next pair in this sequence of integer pairs:
(5, 9) (10, 16) (213, 303) (382, 542) (7277, 10293) (13018, 18412) (247245, 349659)
?
Hint: They are all related in the same way to 256.
(442270, 625466), perhaps? Although I don't see the relation to 256. Never mind, now I do. -- Fred W. Helenius fredh@ix.netcom.com
Wow, Fred, you nailed it! I would love to know how you found the next pair without reference to 256, but for now let's just see who else can explain the sequence one way or another. —Dan
On Nov 22, 2015, at 1:42 PM, Fred W. Helenius <fredh@ix.netcom.com> wrote:
On 11/22/2015 4:07 PM, Dan Asimov wrote:
Puzzle: What is the next pair in this sequence of integer pairs:
(5, 9) (10, 16) (213, 303) (382, 542) (7277, 10293) (13018, 18412) (247245, 349659)
?
Hint: They are all related in the same way to 256.
(442270, 625466), perhaps? Although I don't see the relation to 256.
Never mind, now I do.
After exchanging a few emails with Dan, I've reduced the problem to the integer solutions (x,y) of the generalised Pell equation: 8 y^2 - (5 + 4 x)^2 = 23 An explicit formula - well, two of them, as the equation has two components - can be found: For n>=0, x=1/8*((3-4*sqrt(2))*(3-2*sqrt(2))^(2*n+1)+(3+4*sqrt(2))*(3+2*sqrt(2))^(2*n+1)-10); y=1/8*((8-3*sqrt(2))*(3-2*sqrt(2))^(2*n+1)+(8+3*sqrt(2))*(3+2*sqrt(2))^(2*n+1)); For n>0, x=1/8*(-(3+4*sqrt(2))*(3-2*sqrt(2))^(2*n)+(-3+4*sqrt(2))*(3+2*sqrt(2))^(2*n)-10); y=1/8*((8+3*sqrt(2))*(3-2*sqrt(2))^(2*n)+(8-3*sqrt(2))*(3+2*sqrt(2))^(2*n)); The next solutions are (8399093,11878113) (15024202,21247432) (285321957,403506183) (510380638,721787222) One could thus argue that (-2,2) and (-3,3) should also be solutions (and many others with negative first terms) - as long as one accepts negative bases. Cheers, Seb On 22 November 2015 at 22:47, Dan Asimov <asimov@msri.org> wrote:
Wow, Fred, you nailed it!
I would love to know how you found the next pair without reference to 256, but for now let's just see who else can explain the sequence one way or another.
—Dan
On Nov 22, 2015, at 1:42 PM, Fred W. Helenius <fredh@ix.netcom.com> wrote:
On 11/22/2015 4:07 PM, Dan Asimov wrote:
Puzzle: What is the next pair in this sequence of integer pairs:
(5, 9) (10, 16) (213, 303) (382, 542) (7277, 10293) (13018, 18412) (247245, 349659)
?
Hint: They are all related in the same way to 256.
(442270, 625466), perhaps? Although I don't see the relation to 256.
Never mind, now I do.
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participants (4)
-
Dan Asimov -
Fred W. Helenius -
Seb Perez-D -
Warren D Smith