a(n) = a(n-2) + a(n-4) ; 1,2,3,4.

[Davide Rotondo]

Hi to all, I used to think that a recursive sequence following the rule a(n) = a(n-2) + a(n-4) would always yield two distinct constants when dividing a(n) by a(n-1), but I analyzed the sequence 1,2,3,4,4,6,7,10,11,16,18,26,..., which is a sequence following the rule a(n) = a(n-2) + a(n-4) with starting numbers 1,2,3,4, and I found that dividing a(n)/a(n-1) more and more precisely yields a constant equal to 1/sqrt(80) = sqrt(5)/20. Note that the sequence is a Lucas sequence interspersed with a double Fibonacci sequence. What do you think?
Are there other similar sequences with this property?

Davide

[Gareth McCaughan]

This will always give you two interleaved Fibonacci-ish sequences. What
does each of those sequences look like?

It will have terms of the form A.a^n + B.b^n where a,b are the two roots
of x^2-x-1=0 and A,B are constants. If we take a=1.618... and
b=-0.618... then "in the long term" the terms are just A.a^n + much
smaller unless A=0 in which case the terms are just B.b^n.

So when we have two of these, with constants say A1,B1 and A2,B2 with
A1,A2 nonzero, "in the long term" successive terms look like A1.a^n,
A2.a^n, A1.a^(n+1), A2.a^(n+1), etc., with ratios alternating between
(increasingly accurately) A2/A1, A1.a/A2.

So you'll get a consistent ratio exactly when A1.a/A2 = A2/A1 or
(A2/A1)^2 = a, and the ratio will be sqrt(a).

This _isn't_ the ratio you say you get from your sequence beginning
1,2,3,4, and indeed if I compute say the first 30 terms then the last
few numbers and their ratios to their predecessors look like

2207 1.118034447821682
3194 1.4472134118713185
3571 1.1180338134001253
5168 1.4472136656398769

which doesn't match your report. Am I missing something? How did you
analyse the sequence to find that the ratios converge to 1/sqrt(80)?

Making the dubious assumption that the error here is yours rather than
mine (if that's wrong then you can probably safely ignore the rest of
this email), can we get a sequence with integer entries and a single
limiting ratio? I don't think so. Given the first two terms of a
sequence satisfying the Fibonacci recurrence, say p,q, we have A+B=p,
aA+bB=q, and hence (if I've got the algebra right; even if not the right
thing will have the same sort of shape) (A,B) = (bp-q, q-ap) / (b-a); in
particular A,B will both be rational + rational.sqrt(5). So if both our
interleaved sequences have integer entries, their A,B will both be
rational + rational.sqrt(5) and hence so will the ratio of their As.
(Or, if those happen to be zero, of their Bs.) But sqrt(a) is not of
that form.

(Proof: if (u + v sqrt(5))^2 = a = (1+sqrt(5))/2 then u^2+5v^2 + 2uv
sqrt(5) = (1+sqrt(5)/2); if u,v are rational then this means u^2+5v^2 =
1/2 and 2uv = 1/2 but this is impossible by the AM-GM inequality.)

--
g

[M F Hasler]

On Thu, Jan 8, 2026 at 4:22 AM Davide Rotondo <david...@gmail.com> wrote:

Hi to all, I used to think that a recursive sequence following the rule a(n) = a(n-2) + a(n-4) would always yield two distinct constants when dividing a(n) by a(n-1), but I analyzed the sequence 1,2,3,4,4,6,7,10,11,16,18,26,..., which is a sequence following the rule a(n) = a(n-2) + a(n-4) with starting numbers 1,2,3,4, and I found that dividing a(n)/a(n-1) more and more precisely yields a constant equal to 1/sqrt(80) = sqrt(5)/20. Note that the sequence is a Lucas sequence interspersed with a double Fibonacci sequence. What do you think?

I think the ratio must be > 1 since the sequence is increasing.

So it *might* be the inverse of what you say,  sqrt(80) ~ 9 (since 9² = 81).

The rule a(n) = a(n-2) + a(n-4) quite obviously yields two independent Fibonacci-type sequences

a(2k) = f(k) and a(2k+1) = g(k), and we know simple explicit formulas for these,

as for any 2nd order linear recurrence with constant coefficients, s(n+1) = b s(n) + c s(n-1).

In particular, if  b² + 4c > 0,  then  s(n) = s* r^n + s' r'^n  with r, r' = (b +- sqrt(b² + 4c))/2

the roots of the characteristic equation x² = b x + c  (so  r = Phi = (sqrt(5)+1)/2 for b=c=1).

From there it's easy to see that a(n+1) / a(n) ~ g* / f*  if  n is even, else  r f* / g*.

The two are equal iff  (g* / f*)² = r.

Are there other similar sequences with this property?

What do you mean by "similar"? 

Obviously, we can multiply the sequence (<=> initial values) with any constant and still get the same ratio.

More generally, as the above shows, for b=c=1, it is necessary and sufficient to have g* / f* = sqrt(Phi), 

so you can choose for example any a(1) and a(3) which determines g and thus g* and thus f* and thus the entire sequence f and in particular a(2) and a(4).

 - Maximilian

[Davide Rotondo]

Dear seq fans, any news about this question?

Please let me know 

.

All good

Davide

[Davide Rotondo]

Please forgive me my result was incorrect. 

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[Stephen Lucas]

Davide, just sent an email agreeing with you. I can probably work out how to make them match later.

--

Stephen Lucas, Professor
Department of Mathematics and Statistics
MSC 1911, James Madison University, Harrisonburg, VA 22807 USA
Phone 540 568 5104, Fax 540 568 6857, Web http://educ.jmu.edu/~lucassk/
Email lucassk at jmu dot edu (Work) stephen.k.lucas at gmail dot com (Other)

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[Davide Rotondo]

Maybe I'm focused. Please take a look at this photo of my book. What i claim now isn't that A sequence that follow the rule a(n)=a(n-2)+a(n-4) ; a(n)/a(n-1) approximates one constant, but that in the infinite composite of the photo more you go below with the row more the two constant will match each other like if there is an acceleration of approximation.

What do you think?

Davide

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[Davide Rotondo]

Hi dear, if we take in consideration more and more row of the infinite composite it can be noticed that the two constant approximates to one constant equal to square root of the golden ratio. What do you think?

Davide

[Gareth McCaughan]

On 10/01/2026 10:30, Davide Rotondo wrote:

Maybe I'm focused. Please take a look at this photo of my book. What i claim now isn't that A sequence that follow the rule a(n)=a(n-2)+a(n-4) ; a(n)/a(n-1) approximates one constant, but that in the infinite composite of the photo more you go below with the row more the two constant will match each other like if there is an acceleration of approximation.

What do you think?

I think that you've shown us a screenshot of a lot of numbers and not explained what they are.

--
g

[Davide Rotondo]

Sorry, you are right Gareth.

I started from the primitive line  ... 0  5  0  -3  0  2  0  -1  0  1  0  0  0  1  0  1  0  2  0  3  0  5  0  8  ... that is the Fibonacci sequence interspaced with 0. Note that follow the rule a(n) = a(n-2)+ a(n-4).

Then the other lines  are obtained using the Pascal triangle rule that every number is the sum of two above. Note that still every line follows the rule a(n) = a(n-2)+ a(n-4).

What I claim is that if we go deeper and deeper with the lines the ratio a(n)/a(n-1) approximates more and more one constant that is equal to square root of the golden ratio.

Ok like this? Please let me know.

Davide

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[h...@crypt.org]

Davide Rotondo <david...@gmail.com> wrote:
:Content-Type: image/png; name="Screenshot 2026-01-10 alle 11.21.37.png"

Gareth McCaughan <gareth.m...@pobox.com> wrote:
Content-Type: image/png; name="Screenshot 2026-01-10 alle 11.21.37.png"

Davide Rotondo <david...@gmail.com> wrote:
:Content-Type: image/png; name="Screenshot 2026-01-10 alle 11.21.37.png"

Please be moderate in sending multi-megabyte attachments to the list;
consider in particular whether something smaller would suffice.

When responding to messages with attachments, please do not include
the same attachments in the response.

Hugo

[Daniel Mondot]

Now that you explained it, it looks kind of obvious to me.

If, instead of interleaving one zero between 2 fibonacci numbers, you interleaved 2 numbers (i.e. you get 0,0,1,0,0,1,0,0,2,0,0,3,0,0,5, etc..), and again you create lines that are the sum of the lines above, the ratio of 2 adjacent numbers will tend to the cube root of the golden ratio (1.17398...)

Interleave 4 zeros, and you get the 4th root of the golden ratio....

Daniel.

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[Gareth McCaughan]

On 10/01/2026 14:27, Davide Rotondo wrote:
> Sorry, you are right Gareth.
> I started from the primitive line  ... 0  5  0  -3  0  2  0  -1  0  1
>  0  0  0  1  0  1  0  2  0  3  0  5  0  8  ... that is the Fibonacci
> sequence interspaced with 0. Note that follow the rule a(n) = a(n-2)+
> a(n-4).
> Then the other lines  are obtained using the Pascal triangle rule that
> every number is the sum of two above. Note that still every line
> follows the rule a(n) = a(n-2)+ a(n-4).
>
> What I claim is that if we go deeper and deeper with the lines the
> ratio a(n)/a(n-1) approximates more and more one constant that is
> equal to square root of the golden ratio.

This is correct. Your initial padded-Fibonacci sequence is a linear
combination of sequences (a^n) where the a are roots of x^4-x^2-1, i.e.,
are square roots of roots of x^2-x-1, i.e., are +-phi and two
smaller-magnitude imaginary numbers. (If you pad with period p rather
than period 2, you'll get the p complex pth roots of phi, and p
smaller-magnitude complex numbers.)

The Pascal-triangle construction turns this into the same linear
combination of sequences sum {k} (k choose r) a^(n-k) = (1+a)^k a^(n-k).

When k and n-k are both large, since both |1+a| and |a| are larger for
the root a=sqrt(phi) than for any of the other roots, the (n,k) entry
therefore just looks like a constant times (1+a)^k a^(n-k) for that root
and we can ignore the others, which means that successive terms on each
row have ratio close to a.

If you pad with period p>2 then you get 2p roots, p of which are of the
form w phi^1/p and p of which are of the form w (phi-1)^1/p where in
each case |w|=1, and once again the largest values of |1+a| and |a| come
from the root equal to phi^1/p, so again when k,n-k are large the ratio
converges to phi^1/p.

--
g