log(m)'s close to integers, various log bases
Neil Fernandez
primeness at borve.demon.co.uk
Tue Jan 21 06:01:13 CET 2003
In message <BA5289FA.1B3F%abcloitre at wanadoo.fr>, cloitre
<abcloitre at wanadoo.fr> writes
>le 21/01/03 5:10, Neil Fernandez à primeness at borve.demon.co.uk a écrit :
>
>> In message <Pine.LNX.4.44.0301202025080.17633-100000 at eva117.cs.ualberta.
>> ca>, Jim Nastos <nastos at cs.ualberta.ca> writes
>>> On Tue, 21 Jan 2003, Neil Fernandez wrote:
[snip]
>It appears that this sequence 2,3,4,7,11... is actually Lucas sequence for
>n>1 (a(n)=a(n-)+a(n-2))
>
>3,4,7,11,18,29,47,76,123,199,322,521,843,1364,2207,3571,5778,9349,15127,2447
>6,39603,64079,103682,167761,271443,439204,710647,1149851,1860498,3010349,487
>0847,7881196,12752043,20633239,33385282,54018521,87403803
>
>If so sign are always alterning.
Hi Jim, hi Benoit,
yes, it is the Lucas sequence beginning 1,3, i.e the Lucas numbers
(A000204), except that the 1 has been changed to a 2.
Proof
The formula for the nth Lucas number is L(n)=(phi^n)+(1-phi)^n.
For n>1, every Lucas number is included in the 'abs(frac(log_phi(n))) is
a minimum' sequence (call it A) because the absolute value of the second
term in the formula for L(n) is strictly decreasing, so L(n) gets closer
and closer to phi^n as n increases, so log_phi(L(n)) gets closer and
closer to n; and because no other term can be included in A because it
wouldn't be as close to a power of phi as the Lucas numbers on either
side of it.
Neil
--
Neil Fernandez
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