An elementary approach to Somos-4 sequences
Alexey Ustinov
Pacific National University;
Institute of Applied Mathematics (Khabarovsk),
Russian Academy of Sciences

October 1–5, 2018
Alexey Ustinov (Khabarovsk)

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Somos–(4)Somos–(4) sequence {sn } is defined by initial datas1 = s2 = s3 = s4 = 1
and recurrence relation
sn+2 sn−2 = sn+1 sn−1 + sn2 .
It begings with. . . , 2, 1, 1, 1, 1, 2, 3, 7, 23, 59, 314, 1529, . . .
(Obviously sn = s5−n .)Alexey Ustinov (Khabarovsk)

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Somos–(6)Somos first introduced the sequence Somos-(6) such thats1 = s2 = s3 = s4 = s5 = s6 = 1
andsn+3 sn−3 = sn+2 sn−2 + sn+1 sn−1 + sn2 .
He raised the question whether all the terms are integer:

Somos M. Problem 1470. Crux Mathematicorum, 15: 7 (1989), p. 208.
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Integrality

The integrality of Somos–(4) and Somos–(5) was proved by Janice
Malouf, Enrico Bombieri and Dean Hickerson (1990).
The integrality of Somos–(6) was proved by Dean Hickerson (April
1990).
The integrality of Somos–(7) was proved by Ben Lotto (May 1990).
D. Gale: The proof, rather than illuminating the phenomenon, makes it
if anything more mysterious. . . One is reminded of the proof of the
four-color theorem.

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Integrality

The integrality of Somos–(4) and Somos–(5) was proved by Janice
Malouf, Enrico Bombieri and Dean Hickerson (1990).
The integrality of Somos–(6) was proved by Dean Hickerson (April
1990).
The integrality of Somos–(7) was proved by Ben Lotto (May 1990).
D. Gale: The proof, rather than illuminating the phenomenon, makes it
if anything more mysterious. . . One is reminded of the proof of the
four-color theorem.

Alexey Ustinov (Khabarovsk)

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Somos–(8)Somos–(8) is a sequence with initial datas1 = s2 = s3 = s4 = s5 = s6 = s7 = s8 = 1
satisfying recurrence relation
sn+4 sn−4 = sn+3 sn−3 + sn+2 sn−2 + sn+1 sn−1 + sn2 .
Somos–(8) is NOT an integer sequence:. . . , 1, 1, 1, 1, 1, 1, 1, 1, 4, 7, 13, 25, 61, 187, 775, 5827, 14815,

420514,...
7[It is a wild object with no properties.]

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Somos sequences
Definition
For integer k ≥ 4 Somos–k sequence is a sequence generated byquadratic recurrence relation of the form
[k/2]

sn+k sn =X

αj sn+k−j sn+j ,j=1

where αj are constants and s0 , . . . , sk−1 are initial data.
In particular Somos-4 is defined by initial data s0 , s1 , s2 , s3 andfourth-order recurrence
sn+2 sn−2 = αsn+1 sn−1 + βsn2 ,Somos-6 is defined by initial data s0 , . . . , s5 and sixth-order recurrence
sn+3 sn−3 = αsn+2 sn−2 + βsn+1 sn−1 + γsn2 .
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Somos sequences
Definition
For integer k ≥ 4 Somos–k sequence is a sequence generated byquadratic recurrence relation of the form
[k/2]

sn+k sn =X

αj sn+k−j sn+j ,j=1

where αj are constants and s0 , . . . , sk−1 are initial data.
In particular Somos-4 is defined by initial data s0 , s1 , s2 , s3 andfourth-order recurrence
sn+2 sn−2 = αsn+1 sn−1 + βsn2 ,Somos-6 is defined by initial data s0 , . . . , s5 and sixth-order recurrence
sn+3 sn−3 = αsn+2 sn−2 + βsn+1 sn−1 + γsn2 .
Alexey Ustinov (Khabarovsk)

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Somos sequences
Definition
For integer k ≥ 4 Somos–k sequence is a sequence generated byquadratic recurrence relation of the form
[k/2]

sn+k sn =X

αj sn+k−j sn+j ,j=1

where αj are constants and s0 , . . . , sk−1 are initial data.
In particular Somos-4 is defined by initial data s0 , s1 , s2 , s3 andfourth-order recurrence
sn+2 sn−2 = αsn+1 sn−1 + βsn2 ,Somos-6 is defined by initial data s0 , . . . , s5 and sixth-order recurrence
sn+3 sn−3 = αsn+2 sn−2 + βsn+1 sn−1 + γsn2 .
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A sigma-function solution for Somos–4
The integrality of Somos–(4)· · · (7) may be proved by elementary
methods. But elementary proofs don’t spread any light on the nature of
Somos sequences: there is some elliptic curve hidden behind
Somos–4 and hyperelliptic curve of genus 2 behind Somos–6.
A solution of general Somos–4 recurrence relation was given by C.
Swart (2003) and A. Hone (2005)
sn = AB nσ(z0 + nz),
σ(z)n2where z, z0 ∈ C∗ , andσ(z) = σΓ (z) = zYw∈Γ\{0}

1−z  z + 1 ( z )2
ew 2 w
w

is Weierstrass sigma-function associated to plane lattice Γ.
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A sigma-function solution for Somos–4
The integrality of Somos–(4)· · · (7) may be proved by elementary
methods. But elementary proofs don’t spread any light on the nature of
Somos sequences: there is some elliptic curve hidden behind
Somos–4 and hyperelliptic curve of genus 2 behind Somos–6.
A solution of general Somos–4 recurrence relation was given by C.
Swart (2003) and A. Hone (2005)
sn = AB nσ(z0 + nz),
σ(z)n2where z, z0 ∈ C∗ , andσ(z) = σΓ (z) = zYw∈Γ\{0}

1−z  z + 1 ( z )2
ew 2 w
w

is Weierstrass sigma-function associated to plane lattice Γ.
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Integrality condition
Theorem (Fomin and Zelevinsky, 2002)
For a Somos–k sequences (k = 4, 5, 6 and 7) all of the terms in thesequences are Laurent polynomials in these initial data whose
coefficients are in Z[α1 , . . . , α[k/2] ], so thatsn ∈ Z[α1 , . . . , α[k/2] , s1±1 , . . . , sk±1 ] for all n ∈ Z.
Fomin S. and Zelevinsky A. “The Laurent Phenomenon”, Adv.
Appl. Math. 28 (2002) 119–144.
Integrality of original Somos–(k) sequences follows from the theoremwith
α1 = · · · = α[k/2] = s1 = · · · = sk = 1.
But this Theorem is not a final step.
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Finite rank sequences
The main property of Somos–4 and Somos–6: they have finite rank.

Definition
The sequence {sn }∞n=−∞ has a (finite) rank r if maximal rank of two
infinite matices∞∞(sm+n sm−n )m,n=−∞,(sm+n+1 sm−n )m,n=−∞

is r .

Definition (1)The sequence {sn }∞n=−∞ has a (finite) rank r if r is a least possible ksuch that for all integer m and n
sm+n sm−n =kX

fj (m)gj (n),sm+n+1 sm−n =j=1
Alexey Ustinov (Khabarovsk)kX

efj (m)gej (n).

j=1
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Finite rank sequences: examples
Example (1)Let sn = AB n . Thensm+n sm−n = A2 B 2m = f (m)g(n),wheref (m) = A2 B 2m

and g(n) = 1.

(And almost the same for sm+n+1 sm−n .) So the sequence sn = AB n
has rank 1.

Example (2)The sequence sn = n has rank 2 because
sm+n sm−n = m2 − n2 ,Alexey Ustinov (Khabarovsk)

and sm+n+1 sm−n = m(m + 1) − n(n + 1).
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Finite rank sequences: examples
Example (3)sn = nk has rank 2k.

Example (4)sn = sin n has rank 2: sin(m + n) sin(m − n) = 12 (cos 2m − cos 2n) .

Example (5)For Fibonacci sequence sn = Fn we have1L2m − (−1)m+n L2n ,5

1
Fm+n+1 Fm−n =L2m+1 − (−1)m+n L2n+1 ,5
Fm+n Fm−n =where Ln = Fn−1 + Fn+1 are Lucas numbers. So Fibonacci sequence
has rank 2 as well.
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Finite rank sequences: non-trivial examples

Example (6)The sequence sn = σ(z0 + zn) has rank 2 because of addition formulaσ(u − v )σ(u + v ) = −℘(u)σ(u)2 σ(v )2 + ℘(v )σ(u)2 σ(v )2 .

Example (7)The same forsn = θj (z0 + zn)
because we know addition theorems of the form
θ1 (y + z)θ1 (y − z)θ42 = θ32 (y )θ22 (z) − θ22 (y )θ32 (z) . . .

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Finite rank sequences: Somos examples
Example (Ma, 2010; conjectured by Gosper & Schroeppel, 2007)
Somos-4 has rank 2.
It follows from general formulasn = AB nσ(z0 + nz).
σ(z)n2

Example (Hone, 2016; conj. by Gosper & Schroeppel, 2007)
Somos-6 has rank 4.
It follows from general formula for sn in terms of the Kleiniansigma-function of genus two.

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Somos-4: the main property
Magic determinant

General Somos-4 has rank 2. It means that Somos–4 sequencesatisfy “magic determinant property” or “addition formula” (Ma, 2010)

m1
m2
m3

n1
sm1 +n1 sm1 −n1
sm2 +n1 sm2 −n1
sm3 +n1 sm3 −n1n2
sm1 +n2 sm1 −n2
sm2 +n2 sm2 −n2
sm3 +n2 sm3 −n2n3
sm1 +n3 sm1 −n3
sm2 +n3 sm2 −n3 = 0,
sm3 +n3 sm3 −n3where mi , ni (i = 1, 2, 3) are arbitrary integers or half-integers. It is themain property because another properties of Somos-4 sequence
follow from “magic determinant”.

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Magic determinant
Applications: additional recurrences

The identity
n
1
0

k
1
0
sn+k sn−k sn+1 sn−1 sn2
s1+k s1−ks2 s0
s12 = 0,sk s−ks1 s−1s02

means that Somos-4 is Somos-k for arbitrary k ≥ 4: for some αk , βk
sn+k sn−k = αk sn+1 sn−1 − βk sn2 .
General “magic determinant property” follows from this equation bysimple algebraic manipulations.

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Magic determinant
Applications: additional recurrences

The identity
n
1
0

k
1
0
sn+k sn−k sn+1 sn−1 sn2
s1+k s1−ks2 s0
s12 = 0,sk s−ks1 s−1s02

means that Somos-4 is Somos-k for arbitrary k ≥ 4: for some αk , βk
sn+k sn−k = αk sn+1 sn−1 − βk sn2 .
General “magic determinant property” follows from this equation bysimple algebraic manipulations.

Alexey Ustinov (Khabarovsk)

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The main goal

For Somos-4 sequence we must havesn+k sn−k = αk sn+1 sn−1 − βk sn2 ,sn+k+1 sn−k = α
ek sn+2 sn−1 − βek sn+1 sn ,Purely algebraic proof of this formula was given by van der Poorten
and Swart (2006). But it was based on some tricky symmetry.
Our goal is to obtain direct proof by induction.

Alexey Ustinov (Khabarovsk)

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The main goal

For Somos-4 sequence we must havesn+k sn−k = αk sn+1 sn−1 − βk sn2 ,sn+k+1 sn−k = α
ek sn+2 sn−1 − βek sn+1 sn ,Purely algebraic proof of this formula was given by van der Poorten
and Swart (2006). But it was based on some tricky symmetry.
Our goal is to obtain direct proof by induction.

Alexey Ustinov (Khabarovsk)

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The main goal

For Somos-4 sequence we must havesn+k sn−k = αk sn+1 sn−1 − βk sn2 ,sn+k+1 sn−k = α
ek sn+2 sn−1 − βek sn+1 sn ,Purely algebraic proof of this formula was given by van der Poorten
and Swart (2006). But it was based on some tricky symmetry.
Our goal is to obtain direct proof by induction.

Alexey Ustinov (Khabarovsk)

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Gauge transformations: 2 degrees of freedom
Somos-4 sequences
sn+2 sn−2 = αsn+1 sn−1 + βsn2
are invariant under the two-parameter abelian group of gaugetransformations defined by
sn → esn = A · B n · sn .
Thus it is natural to introduce the gauge-invariant variables
fn =sn−1 sn+1
sn2

satisfying
fn+1 fn2 fn−1 = αfn + β.

Alexey Ustinov (Khabarovsk)

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Gauge transformations: 2 degrees of freedom
Somos-4 sequences
sn+2 sn−2 = αsn+1 sn−1 + βsn2
are invariant under the two-parameter abelian group of gaugetransformations defined by
sn → esn = A · B n · sn .
Thus it is natural to introduce the gauge-invariant variables
fn =sn−1 sn+1
sn2

satisfying
fn+1 fn2 fn−1 = αfn + β.

Alexey Ustinov (Khabarovsk)

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First non-trivial case
The equation
sn+3 sn−2 sn+2 sn−1 sn+1 sn
s4 s−1
s3 s0
s2 s1 = 0s3 s−2s2 s−1s1 s0
is equivalent to
fn fn−1 + α

1
1+
fn fn−1β= f2 f1 + α+
fn fn−1



1
1
+
f2 f1


+

β
.
f2 f1

So the quantity
T = fn fn−1 + α

1
1+
fn fn−1+β
fn fn−1

must be conserved. (This equation defines an elliptic curve in the(fn , fn−1 ) plane.)
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First non-trivial case
The equation
sn+3 sn−2 sn+2 sn−1 sn+1 sn
s4 s−1
s3 s0
s2 s1 = 0s3 s−2s2 s−1s1 s0
is equivalent to
fn fn−1 + α

1
1+
fn fn−1β= f2 f1 + α+
fn fn−1



1
1
+
f2 f1


+

β
.
f2 f1

So the quantity
T = fn fn−1 + α

1
1+
fn fn−1+β
fn fn−1

must be conserved. (This equation defines an elliptic curve in the(fn , fn−1 ) plane.)
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Invariance of the first integral T2
fn+1 fn2 fn−1 − αfn = β = fn+2 fn+1
fn − αfn+1

αβαfn−1 fn −== fn+1 fn+2 −fn+1
fn fn+1
fn

fn−1 fn + fn fn+1 +Alexey Ustinov (Khabarovsk)/(fn fn+1 )+fn fn+1 + α

1
1+
fn fn+1αα= T = fn fn+1 + fn+1 fn+2 +fn
fn+1

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Invariance of the first integral T2
fn+1 fn2 fn−1 − αfn = β = fn+2 fn+1
fn − αfn+1

αβαfn−1 fn −== fn+1 fn+2 −fn+1
fn fn+1
fn

fn−1 fn + fn fn+1 +Alexey Ustinov (Khabarovsk)/(fn fn+1 )+fn fn+1 + α

1
1+
fn fn+1αα= T = fn fn+1 + fn+1 fn+2 +fn
fn+1

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Invariance of the first integral T2
fn+1 fn2 fn−1 − αfn = β = fn+2 fn+1
fn − αfn+1

αβαfn−1 fn −== fn+1 fn+2 −fn+1
fn fn+1
fn

fn−1 fn + fn fn+1 +Alexey Ustinov (Khabarovsk)/(fn fn+1 )+fn fn+1 + α

1
1+
fn fn+1αα= T = fn fn+1 + fn+1 fn+2 +fn
fn+1

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Invariance of the first integral T2
fn+1 fn2 fn−1 − αfn = β = fn+2 fn+1
fn − αfn+1

βα== fn+1 fn+2 −fn−1 fn −fn+1
fn fn+1fnαfn−1 fn + fn fn+1 +/(fn fn+1 )+fn fn+1 + α

1
1+
fn fn+1αα= T = fn fn+1 + fn+1 fn+2 +fn
fn+1

In particular this proof gives new representation for T
fn2 (fn−1 + fn+1 ) = Tfn − αTogether with the main equation
fn+1 fn2 fn−1 = αfn + β
it allows to express fn−1 + fn+1 and fn−1 fn+1 in terms of fn .
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Induction

Desired formulasn+k sn−k = αk sn+1 sn−1 − βk sn2
can be rewritten as

sn+k sn−k= αk fn − βk .
sn2

Induction step:
sn+k sn−k =(sn+k sn−k+2 )(sn−k sn+k−2 ).
sn+k−2 sn−k+2

So we can apply induction hypothesis

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Induction

Desired formulasn+k sn−k = αk sn+1 sn−1 − βk sn2
can be rewritten as

sn+k sn−k= αk fn − βk .
sn2

Induction step:
sn+k sn−k =(sn+k sn−k+2 )(sn−k sn+k−2 ).
sn+k−2 sn−k+2

So we can apply induction hypothesis

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Induction
Desired formulasn+k sn−k = αk sn+1 sn−1 − βk sn2
can be rewritten as

sn+k sn−k= αk fn − βk .
sn2

Induction step is equivalent to(αk+1 fn − βk+1 )(αk−1 fn − βk−1 ) = (αk fn+1 − βk )(αk fn−1 − βk )fn2
RHS as a symmetric function of fn+1 and fn−1 can be expressed as afunction of fn only
(αk+1 fn −βk+1 )(αk−1 fn −βk−1 ) = βk2 fn2 +αk (αk β−βk T )fn +αk (αk β+αβk )Comparing coefficients from both sides of this formula we getrecurrences for αk and βk .
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Induction
Desired formulasn+k sn−k = αk sn+1 sn−1 − βk sn2
can be rewritten as

sn+k sn−k= αk fn − βk .
sn2

Induction step is equivalent to(αk+1 fn − βk+1 )(αk−1 fn − βk−1 ) = (αk fn+1 − βk )(αk fn−1 − βk )fn2
RHS as a symmetric function of fn+1 and fn−1 can be expressed as afunction of fn only
(αk+1 fn −βk+1 )(αk−1 fn −βk−1 ) = βk2 fn2 +αk (αk β−βk T )fn +αk (αk β+αβk )Comparing coefficients from both sides of this formula we getrecurrences for αk and βk .
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Induction
Desired formulasn+k sn−k = αk sn+1 sn−1 − βk sn2
can be rewritten as

sn+k sn−k= αk fn − βk .
sn2

Induction step is equivalent to(αk+1 fn − βk+1 )(αk−1 fn − βk−1 ) = (αk fn+1 − βk )(αk fn−1 − βk )fn2
RHS as a symmetric function of fn+1 and fn−1 can be expressed as afunction of fn only
(αk+1 fn −βk+1 )(αk−1 fn −βk−1 ) = βk2 fn2 +αk (αk β−βk T )fn +αk (αk β+αβk )Comparing coefficients from both sides of this formula we getrecurrences for αk and βk .
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Induction
The coefficients of fn0 and fn2 give recurrences
αk+1 αk−1 = βk2 ,βk+1 βk−1 = αk (βαk + αβk ).

The coefficient of fn1 give extra condition on αk and βk . In terms ofvariables gk = βk /αk it can be written as


1
β
1
+
+
=T
gn gn−1 + α
gn gn−1
gn gn−1
But we know that this condition follows from recurrence
gk+1 gk−1 =1(αgk + β)
gk2

which is indeed the case.
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Induction
The coefficients of fn0 and fn2 give recurrences
αk+1 αk−1 = βk2 ,βk+1 βk−1 = αk (βαk + αβk ).

The coefficient of fn1 give extra condition on αk and βk . In terms ofvariables gk = βk /αk it can be written as


1
β
1
+
+
=T
gn gn−1 + α
gn gn−1
gn gn−1
But we know that this condition follows from recurrence
gk+1 gk−1 =1(αgk + β)
gk2

which is indeed the case.
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Induction
The coefficients of fn0 and fn2 give recurrences
αk+1 αk−1 = βk2 ,βk+1 βk−1 = αk (βαk + αβk ).

The coefficient of fn1 give extra condition on αk and βk . In terms ofvariables gk = βk /αk it can be written as


1
β
1
+
+
=T
gn gn−1 + α
gn gn−1
gn gn−1
But we know that this condition follows from recurrence
gk+1 gk−1 =1(αgk + β)
gk2

which is indeed the case.
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Elliptic divisibility sequences
We can associate with gk = βk /αk a Somos-4 sequence {Wk } such
W
Wk+1
. It is sufficient to take Wk−1 Wk+1 = βk andthat gk = k−1
W2kWk2 = αk (small ambiguousness can be resolved via initial conditions).
This sequence is known as Elliptic divisibility sequence.
Additional formula for general Somos-4
sn+k sn−k = αk sn+1 sn−1 − βk sn2 = Wk2 sn+1 sn−1 − Wk−1 Wk+1 sn2may be applied to sn = Wn yielding
Wn+k Wn−k = Wk2 Wn+1 Wn−1 − Wk−1 Wk+1 Wn2 .

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Elliptic divisibility sequences
We can associate with gk = βk /αk a Somos-4 sequence {Wk } such
W
Wk+1
. It is sufficient to take Wk−1 Wk+1 = βk andthat gk = k−1
W2kWk2 = αk (small ambiguousness can be resolved via initial conditions).
This sequence is known as Elliptic divisibility sequence.
Additional formula for general Somos-4
sn+k sn−k = αk sn+1 sn−1 − βk sn2 = Wk2 sn+1 sn−1 − Wk−1 Wk+1 sn2may be applied to sn = Wn yielding
Wn+k Wn−k = Wk2 Wn+1 Wn−1 − Wk−1 Wk+1 Wn2 .

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Elliptic divisibility sequences
We can associate with gk = βk /αk a Somos-4 sequence {Wk } such
W
Wk+1
. It is sufficient to take Wk−1 Wk+1 = βk andthat gk = k−1
W2kWk2 = αk (small ambiguousness can be resolved via initial conditions).
This sequence is known as Elliptic divisibility sequence.
Additional formula for general Somos-4
sn+k sn−k = αk sn+1 sn−1 − βk sn2 = Wk2 sn+1 sn−1 − Wk−1 Wk+1 sn2may be applied to sn = Wn yielding
Wn+k Wn−k = Wk2 Wn+1 Wn−1 − Wk−1 Wk+1 Wn2 .

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General additional formulasa+b sa−b wc+d wc−d + sa+d sa−d wb+c wb−c − sa+c sa−c wb+d wb−d = 0
is an equivalent analogue of forbidden Weierstrass three term identity
σ(a + b)σ(a − b)σ(c + d)σ(c − d)−−σ(a + c)σ(a − c)σ(b + d)σ(b − d)++σ(a + d)σ(a − d)σ(b + c)σ(b − c) = 0.
Equivalent form:0 sl+k wl−k
Pf 
sm+k wm−k
sn+k wn−kAlexey Ustinov (Khabarovsk)

sk+l wk−l0sm+l wm−l
sn+l wn−l
sk +m wk−m sk+n wk−nsl+m wl−msl+n wl−n  = 0.
0sm+n wm−n 
sn+m wn−m0

An elementary approach to Somos-4

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Conclusions

Main properties of Somos-4 sequences can be drived by purely
algebraic way without any help of special functions.
First integral and elliptic divisibility sequences arise naturally on
this way.
Hopefully this approach will be suitable for higher rank sequences.

Alexey Ustinov (Khabarovsk)

An elementary approach to Somos-4

25 / 27

Somos sequences: further directions
Different sides of integrability

Properties which seemed to be more or less equivalent for general
Somos sequences:
Integrality
Laurent phenomenon
Finite rank
Periodicity (mod N)
General formula in terms of theta-functions
A natural candidate for experiments is the Gale – Robinson sequence
generated by
sm+n sm = αsm+r sm+n−r + βsm+p sm+n−p + γsm+q sm+n−q ,where r + p + q = n.
Alexey Ustinov (Khabarovsk)

An elementary approach to Somos-4

26 / 27

Somos sequences: further directions
Different sides of integrability

Properties which seemed to be more or less equivalent for general
Somos sequences:
Integrality
Laurent phenomenon
Finite rank
Periodicity (mod N)
General formula in terms of theta-functions
A natural candidate is the Gale–Robinson sequence generated by
sm+n sm = αsm+r sm+n−r + βsm+p sm+n−p + γsm+q sm+n−q ,where r + p + q = n.

Alexey Ustinov (Khabarovsk)

An elementary approach to Somos-4

26 / 27

Questions?

Alexey Ustinov (Khabarovsk)

An elementary approach to Somos-4

27 / 27