

A268267


Orbit periods of the iteration s(1)=0, s(m+1)=(2*i*Pi*n)/log(3)+log(1/(12^s(m)))/log(3).


0



2, 1, 4, 2, 1, 3, 3, 1, 2, 1, 6, 2, 1, 3, 3, 1, 2, 4, 1, 2, 1, 4, 5, 1, 3, 3, 1, 2, 1, 5, 2, 1, 3, 3, 1, 2, 5, 1, 2, 1, 7, 5, 1, 5, 4, 1, 2, 1, 4, 2, 1, 3, 3, 1, 2, 5, 1, 2, 1, 3, 3, 1, 5, 4, 1, 2, 1, 4, 2, 1, 3, 3, 1, 2, 6, 1, 2, 1, 3, 3, 1, 2, 4, 1, 2, 1, 4, 2, 1, 3, 3
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OFFSET

0,1


COMMENTS

Using the Reduce[1 + 1/2^a + 1/3^b == 0, {a, b}] command in Mathematica to solve the equation 1+1/2^a+1/3^b=0 gives the following conditional expression: b=(2*i*Pi*n)/log(3)+log(1/(12^a))/log(3), where n is an integer.
By setting n=0 and replacing "a" and "b" with "s", setting s(1)=0 as starting value and then iterating:
s(1)=0
s(2)=(2*i*Pi*n)/log(3)+log(1/(12^(s(1))))/log(3)
s(3)=(2*i*Pi*n)/log(3)+log(1/(12^(s(2))))/log(3)
s(4)=(2*i*Pi*n)/log(3)+log(1/(12^(s(3))))/log(3)
s(5)=(2*i*Pi*n)/log(3)+log(1/(12^(s(4))))/log(3)
...
s(m+1)=(2*i*Pi*n)/log(3)+log(1/(12^s(m)))/log(3)
The iteration converges to sequences with periods of orbit equal to the entries in this sequence a(n). The entries in the converged period are then zeros to equations as in the examples in the conjecture below:
a(0)=2, because after 1000 iterations, we have the periodic sequence of repeating convergents with period of the orbit equal to 2:
s(m+1) = 0.60158922138 + 2.05252038097 I
s(m+2) = 0.60158922138  2.05252038097 I
s(m+3) = 0.60158922138 + 2.05252038097 I
s(m+4) = 0.60158922138  2.05252038097 I
...
These two complex numbers are then solutions to:
Sum_{k=1..3} 1/n^(s(m+1)) + Sum_{k=1..3} 1/n^(s(m+2)) = 0
Conjecture:
In general for some suitable index value n to a(n), if:
a(n)=1 then:
Sum_{k=1..3} 1/n^(s(m+1))=0
a(n)=2 then:
Sum_{k=1..3} 1/n^(s(m+1))
+Sum_{k=1..3} 1/n^(s(m+2))=0
a(n)=3 then:
Sum_{k=1..3} 1/n^(s(m+1))
+Sum_{k=1..3} 1/n^(s(m+2))
+Sum_{k=1..3} 1/n^(s(m+3))=0
...
a(n)=k then:
Sum_{k=1..3} 1/n^(s(m+1))
+Sum_{k=1..3} 1/n^(s(m+2))
+Sum_{k=1..3} 1/n^(s(m+3))
+...
+Sum_{k=1..3} 1/n^(s(m+k))=0
and so on, by observation.
The index n of positions of ones in the sequence a(n) gives a linear looking plot. The sequences of positions of ones starts: 1, 4, 7, 9, 12, 15, 18, 20, 23, 26, 28, 31, 34, 37,... and is equal to the first 34 entries of A184589, but then begins to differ at n=35. The differences between the index positions of ones starts: 3, 3, 2, 3, 3, 3, 2, 3, 3, 2, 3, 3, 3,.. which appears to have initial periodic patterns that fail later on, and each time the pattern fails the periodic patterns get longer and longer. The values of n when a(n)=1, the function in the equation the iteration solves is the partial Riemann zeta function, here 1 + 1/2^s+ 1/3^s = 0.
The Mathematica program can treat all partial Riemann zeta functions but caution is needed when changing the parameters of the program in order for the formula to converge.


LINKS

Table of n, a(n) for n=0..90.


MATHEMATICA

(*Mathematica 8*)
Clear[k, m, s, kk, n, aa, b1, t, aaa, kkk, aaa]
Print["k can be varied to any integer greater than or equal to 2:"]
k = 3;
m = 1000;
s = 0;
Monitor[aaa =
Table[Table[
Do[s = (2 I \[Pi]*(kk))/Log[k] +
N[Round[Log[1/(Sum[1/n^s, {n, 1, k  1}])]/Log[k]*10^20]/
10^20, 20], {i, 1, m}];
aa = Table[
s = (2 I \[Pi]*(kk))/Log[k] +
N[Round[Log[1/(Sum[1/n^s, {n, 1, k  1}])]/Log[k]*10^20]/
10^20, 20], {i, 1, 100}]; , {kk, kkk, kkk}];
(*end*)
Flatten[
Position[Chop[Accumulate[Sum[1/n^(aa), {n, 1, k}]]],
0]][[1]], {kkk, 0, 90}], kkk]
b1 = Flatten[Position[aaa, 1]]  1;
Differences[b1];


CROSSREFS

Sequence in context: A187816 A088423 A006839 * A205395 A243070 A243060
Adjacent sequences: A268264 A268265 A268266 * A268268 A268269 A268270


KEYWORD

nonn


AUTHOR

Mats Granvik, Jan 30 2016


STATUS

approved



