Legendre pairs of lengths ℓ ≡ 0 (mod 5)

Ilias S. Kotsireas; Christoph Koutschan; Dursun A. Bulutoglu; David M. Arquette; Jonathan S. Turner; Kenneth J. Ryan

doi:10.1515/spma-2023-0105

Article Open Access

Legendre pairs of lengths ℓ ≡ 0 (mod 5)

Ilias S. Kotsireas , Christoph Koutschan , Dursun A. Bulutoglu , David M. Arquette , Jonathan S. Turner and Kenneth J. Ryan

Published/Copyright: November 10, 2023

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From the journal Special Matrices Volume 11 Issue 1

Abstract

By assuming a type of balance for length ℓ = 87 and nontrivial subgroups of multiplier groups of Legendre pairs (LPs) for length ℓ = 85 , we find LPs of these lengths. We then study the power spectral density (PSD) values of m compressions of LPs of length 5 m . We also formulate a conjecture for LPs of lengths ℓ ≡ 0 (mod 5) and demonstrate how it can be used to decrease the search space and storage requirements for finding such LPs. The newly found LPs decrease the number of integers in the range ≤ 200 for which the existence question of LPs remains unsolved from 12 to 10.

Keywords: compressed vector; discrete Fourier transform; Hadamard matrix; periodic autocorrelation function; power spectral density; multiplier group

MSC 2010: 05B10; 05B20; 15B34

1 Introduction

The periodic autocorrelation function (PAF) of a (row) vector a ∈ C ℓ indexed by Z ℓ is defined as PAF a ( j ) = ∑ i = 0 ℓ − 1 a i a ¯ i − j , where a ¯ i is the complex conjugate of a i . By using the PAF, we define the concept of a Legendre pair (LP) first studied in [4].

Definition 1

Let a and b be { − 1 , 1 } row vectors indexed by Z ℓ . Then, ( a , b ) is an LP if

(1) PAF a ( j ) + PAF b ( j ) = − 2 ∀ j ∈ Z ℓ \ { 0 } , ∑ i = 0 ℓ − 1 a i = ∑ i = 0 ℓ − 1 b i .

An LP ( a , b ) must satisfy ∑ i = 0 ℓ − 1 a i = ∑ i = 0 ℓ − 1 b i = 1 or ∑ i = 0 ℓ − 1 a i = ∑ i = 0 ℓ − 1 b i = − 1 ; see [1].

A circulant shift of a vector a ∈ C ℓ by j ∈ Z ℓ , denoted c j ( a ) , is a transformation such that ( c j ( a ) ) i = a i − j for each i ∈ Z ℓ . Let C a be the circulant matrix obtained from a row vector, a , where the ( j + 1 ) th row of C a is c j ( a ) .

Let a , b ∈ { − 1 , 1 } ℓ be an LP such that ∑ i = 0 ℓ − 1 a i = ∑ i = 0 ℓ − 1 b i = 1 , and 1 be the length ℓ row vector of all 1s. For a complex number z = x + i y , let z ¯ = x − i y be the complex conjugate of z . Then,

(2) H = − 1 − 1 1 1 − 1 1 1 − 1 1 ⊤ 1 ⊤ C b C a 1 ⊤ − 1 ⊤ C a ¯ ⊤ − C b ¯ ⊤

is a ( 2 ℓ + 2 ) × ( 2 ℓ + 2 ) Hadamard matrix, a matrix with entries in { ± 1 } with orthogonal columns, where for a matrix A = ( a i j ) , A ¯ = ( a i j ¯ ) , which will soon become relevant. It is well known that a Hadamard matrix of order n does not exist if n is not divisible by 4 and n > 2 . Then, 4 ∣ ( 2 ℓ + 2 ) , implying that ℓ must be odd for an LP of length ℓ to exist. It is conjectured that an LP exists for all odd ℓ , or equivalently, there is an n × n Hadamard matrix constructed via LPs whenever n is a multiple of 4 [1].

An n × n matrix with entries from { − 1 , 1 , − i , i } with orthogonal columns with respect to the complex dot product is called a quaternary (complex) Hadamard matrix. The order of a quaternary Hadamard matrix must be even [6]. For even ℓ , if a , b ∈ { − 1 , 1 , − i , i } ℓ satisfy equation (1) and equations

∑ i = 0 ℓ − 1 a i = 1 + i , ∑ i = 0 ℓ − 1 b i = 0 ,

then H in equation (2) with the upper left 2 × 2 submatrix replaced by

− 1 i − i 1

is a quaternary Hadamard matrix, and such ( a , b ) is called a quaternary Legendre pair [6]. It is conjectured that a quaternary LP exists for every even ℓ [6]. Since if a quaternary Hadamard matrix of order n exists, then a Hadamard matrix of order 2 n must also exist [6], proving this conjecture would also prove the Hadamard conjecture, i.e., a Hadamard matrix of order n exists for every n divisible by 4.

For two groups N and H , and an action of H on N , let N ⋊ H be the semidirect product of N and H as defined by Rotman [8]. Let Z ℓ × be the group of units in Z ℓ . Let ( j , k ) i = k i + j for ( j , k ) ∈ Z ℓ ⋊ Z ℓ × and i ∈ Z ℓ . Then, the group Z ℓ ⋊ Z ℓ × acts on each vector a in C ℓ via ( j , k ) a i = a ( j , k ) − 1 i . The action of ( 0 , k ) ≔ d k for k ∈ Z ℓ × on a is called a decimation. Decimations and cyclic shifts do not commute. In fact, d r c i ( d r ) − 1 = d r c i d r − 1 = c i r . Equivalently,

c i d r = d r c i r − 1 or d r c k = c r k d r .

The action of the group Z ℓ ⋊ Z ℓ × on a ∈ C ℓ is used to define an action of ( Z ℓ × Z ℓ ) ⋊ Z ℓ × on LPs (quaternary LPs) ( a , b ) by ( ( j 1 , j 2 ) , k ) ( a , b ) = ( ( j 1 , k ) a , ( j 2 , k ) b ) , for ( j 1 , k ) ∈ Z ℓ ⋊ Z ℓ × and ( j 2 , k ) ∈ Z ℓ ⋊ Z ℓ × , see [1,6]. Two pairs of { − 1 , 1 } or { 0 , 1 } vectors ( a , b ) , ( a ′ , b ′ ) are equivalent if ( a , b ) is in the same orbit as ( a ′ , b ′ ) or ( b ′ , a ′ ) under the action of ( Z ℓ × Z ℓ ) ⋊ Z ℓ × . Two pairs of { − 1 , 1 , − i , i } vectors ( a , b ) , ( a ′ , b ′ ) are equivalent if ( a , b ) is in the same orbit as ( a ′ , b ′ ) under the action of ( Z ℓ × Z ℓ ) ⋊ Z ℓ × . Clearly, if ( a , b ) is an LP (quaternary LP), and ( a ′ , b ′ ) is equivalent to ( a , b ) , then ( a ′ , b ′ ) is also an LP (quaternary LP). The following definition characterizes the inherent symmetries of a vector a ∈ C ℓ .

Definition 2

For a vector a ∈ C ℓ , the group

G a = { j ∈ Z ℓ × ∣ ( i , j ) ( a ) = a for some i ∈ Z ℓ }

is called the multiplier group of a , and each element of G a is called a multiplier of a .

For g ∈ Z ℓ ⋊ Z ℓ × written as a product of cyclic shifts and decimations and q ∈ C ℓ , the following theorem shows the connection between G q and G g q .

Theorem 1

Let g ∈ Z ℓ ⋊ Z ℓ × and q ∈ C ℓ . Then, G g q = g G q g − 1 and ∣ G q ∣ = ∣ G g q ∣ .

Proof

Observe that r ∈ G q ⇔ d r ( q ) = c i ( q ) for some i ∈ Z ℓ ⇔ d r ( q ) = d r ( g − 1 ( g q ) ) = ( d r ( g − 1 ) ) ( g q ) = ( c i g − 1 ) ( g q ) = ( g − 1 c i ′ ) ( g q ) for some i ′ ∈ Z ℓ , where c i g − 1 = g − 1 c i ′ for some i ′ ∈ Z ℓ because Z ℓ ⊴ Z ℓ ⋊ Z ℓ × . Hence, r ∈ G q ⇔ g d r ( g − 1 ( g q ) ) = ( g d r g − 1 ) ( g q ) = c i ′ ( g q ) for some i ′ ∈ Z ℓ . This implies that g G q g − 1 = G g q . Then, ∣ G q ∣ = ∣ G g q ∣ as the map x → g x g − 1 is one-to-one.□

The following theorem appeared as Theorem 4 in the work of Turner et al. [9].

Theorem 2

Let ℓ be odd and b ∈ { 0 , 1 } ℓ such that ∑ i = 0 ℓ − 1 b i = δ with gcd ( δ , ℓ ) = 1 . Then, there is some i ∈ Z ℓ such that b ′ = c i ( b ) , and

G b ≤ G ^ b ′ = { j ∈ Z ℓ × ∣ d j ( b ′ ) = b ′ } .

The next lemma that follows from Theorems 1 and 2 shows that in Theorem 2, G b is equal to G ^ b ′ .

Lemma 1

Let ℓ be odd and b ∈ { 0 , 1 } ℓ such that ∑ i = 0 ℓ − 1 b i = δ with gcd ( δ , ℓ ) = 1 . Then, there is some i ∈ Z ℓ such that b ′ = c i ( b ) and

G b = G ^ b ′ = { j ∈ Z ℓ × ∣ d j ( b ′ ) = b ′ } .

The following lemma follows from Lemma 1.

Lemma 2

Let ℓ be odd and b ∈ { − 1 , 1 } ℓ such that ∑ i = 0 ℓ − 1 b i = δ ′ with gcd ( ( δ ′ + ℓ ) ⁄ 2 , ℓ ) = 1 . Then, there is some i ∈ Z ℓ such that b ′ = c i ( b ) , and

G b = G ^ b ′ = { j ∈ Z ℓ × ∣ d j ( b ′ ) = b ′ } .

Proof

Since ( i , j ) ( a ) = a for some i ∈ Z ℓ if and only if ( i , j ) ( a + 1 ) ⁄ 2 = ( a + 1 ) ⁄ 2 for some i ∈ Z ℓ ,

G b = G b + 1 2 .

Since, ∑ i = 0 ℓ − 1 ( b i + 1 ) ⁄ 2 = ( δ ′ + ℓ ) ⁄ 2 and gcd ( ( δ ′ + ℓ ) ⁄ 2 , ℓ ) = 1 , by Lemma 2, there is some i ∈ Z ℓ such that b ′ = c i ( ( b + 1 ) ⁄ 2 ) = ( c i ( b ) + 1 ) ⁄ 2 and

G b = G b + 1 2 = G ^ b ′ = G ^ c i ( b ) + 1 2 .

Now

G ^ c i ( b ) + 1 2 = G ^ c i ( b ) = G ^ b ′

follows from the definition of G ^ b ′ .□

The following corollary simplifies the search for an LP ( a , b ) with multiplier groups G a and G b .

Corollary 1

An LP ( a , b ) with multiplier groups G a and G b exists if and only if an equivalent LP ( a ′ , b ′ ) with multiplier groups G ^ a ′ = G a and G ^ b ′ = G b exists. Hence, a search for all nonequivalent LPs ( a , b ) with multiplier groups G a and G b can be implemented by searching for all nonequivalent LPs ( a ′ , b ′ ) with multiplier groups G ^ a ′ and G ^ b ′ .

Proof

For an LP ( a , b ) , ∑ i = 0 ℓ − 1 a i = ∑ i = 0 ℓ − 1 b i = δ ′ = ± 1 and gcd ( ( ℓ ± 1 ) ⁄ 2 , ℓ ) = 1 . Hence, by Lemma 2, there exits i 1 , i 2 ∈ Z ℓ such that G u = G ^ c i 1 ( a ) and G v = G ^ c i 2 ( b ) , and ( c i 1 ( a ) , c i 2 ( b ) ) is an LP equivalent to ( a , b ) with multiplier groups G ^ c i 2 ( a ) = G a , and G ^ c i 2 ( b ) = G b .□

Our repository of LPs provides many examples of LPs ( a , b ) with G u ≠ G v . However, for LPs ( a , b ) in our repository with nontrivial multiplier groups, more often than not G u = G v . On the other hand, for most LPs in our repository, G a = G b = { 1 } .

Throughout the rest of the article, let the row vectors a , b ∈ R ℓ , where ℓ be odd, and w = e 2 π i ⁄ ℓ . The discrete Fourier transform of a is DFT a ( j ) ≔ ∑ r = 0 ℓ − 1 w j r a r , and the power spectral density (PSD) of a is PSD a ( j ) ≔ ∣ DFT a ( j ) ∣ 2 for j ∈ Z ℓ . It is shown in the work of Fletcher et al. [4] that a pair of row vectors a , b ∈ { − 1 , 1 } ℓ form an LP of length ℓ if and only if

(3) ∑ i = 0 ℓ − 1 a i = ∑ i = 0 ℓ − 1 b i = ± 1 ,

(4) PSD a ( j ) + PSD b ( j ) = 2 ℓ + 2 ∀ j ∈ Z ℓ \ { 0 } .

If a ∈ { − 1 , 1 } ℓ , then ( a + 1 ) ⁄ 2 ∈ { 0 , 1 } ℓ . It is plain to see that

(5) PSD a + 1 2 ( j ) = PSD a ( j ) 4 ∀ j ∈ Z ℓ \ { 0 } ,

and

(6) PSD a + 1 2 ( 0 ) = ∣ DFT a + 1 2 ( 0 ) ∣ 2 = ∑ i = 0 ℓ − 1 a i + ℓ 2 2 = ℓ ± 1 2 2 .

The following lemma follows from equations (3)–(6).

Lemma 3

Let a , b ∈ { − 1 , 1 } ℓ form an LP of length ℓ . Then,

∑ i = 0 ℓ − 1 a i + 1 2 = ∑ i = 0 ℓ − 1 b i + 1 2 = ℓ ± 1 2 , PSD a + 1 2 ( j ) + PSD b + 1 2 ( j ) = ℓ + 1 2 ∀ j ∈ Z ℓ \ { 0 } .

The following theorem from the work of Fletcher et al. [4] determines the relation between PSD a ( j ) and PAF a ( j ) .

Theorem 3

(Wiener-Khinchin theorem) The PSD of a ∈ R ℓ is equal to the DFT of its PAF, i.e.,

PSD a ( k ) = ∑ j = 0 ℓ − 1 PAF a ( j ) w j k for k ∈ Z ℓ .

Moreover, the PAF of a is equal to the inverse DFT of a ’s PSD, i.e.,

PAF a ( j ) = 1 ℓ ∑ k = 0 ℓ − 1 PSD a ( k ) w − j k for j ∈ Z ℓ .

The following corollary follows directly from Theorem 3 and equation (6).

Corollary 2

Let a , b ∈ { − 1 , 1 } ℓ form an LP of length ℓ . Then,

1 = PSD a ( 0 ) = ∑ j = 0 ℓ − 1 PAF a ( j ) ,

and

ℓ ± 1 2 2 = PSD a + 1 2 ( 0 ) = ∑ j = 0 ℓ − 1 PAF a + 1 2 ( j ) .

Throughout the article, WLOG, we assume that ∑ i = 0 ℓ − 1 a i = ∑ i = 0 ℓ − 1 b i = 1 . In addition, let e 2 ( a ) = ∑ i < j a i a j be the elementary symmetric polynomial of degree 2 and p 2 ( a ) = ∑ i a i 2 be the power sum symmetric polynomial of degree 2.

For a = [ a 0 , … , a ℓ − 1 ] and ℓ = n m for some positive integers n and m , define A j = ∑ i = 0 m − 1 a n i + j for j = 0 , … , n − 1 . The vector A m = [ A 0 , … , A n − 1 ] is called the m-compression of a [2]. Throughout the article, let A m = [ A 0 , … , A n − 1 ] and ℬ m = [ B 0 , … , B n − 1 ] be the m -compressions of a and b .

The following theorem from the work of Turner et al. [10] shows how the DFT and PSD of the m -compression of a vector a are related to the DFT and PSD of a .

Theorem 4

Let a be a vector of length ℓ = n m , and A m be the m -compression of a . Then, DFT A m ( j ) = DFT a ( m j ) and PSD A m ( j ) = PSD a ( m j ) for j ∈ { 0 , … , n − 1 } .

Compression of complementary vectors has proved to be a valuable tool for finding several previously unknown complementary vectors (vectors whose PAF vectors’ j th entries sum to a constant for j ≠ 0 ) in the past decade [2,5, 6,10]. Compression-based search algorithms are based on a two-step process. In the first step, several candidate compression vectors are computed. The second step involves searching for decompressions of the candidate compressed vectors from the first step.

In Section 2, we find restrictions on the PSD values of LP vectors computed using only the fifth primitive roots of unity. We then corroborate our theoretical restrictions by computing all PSD values, which are only based on the fifth primitive roots of unity for all LP vectors that we possess. Then, based on these restrictions, we develop Conjecture 1, which applies to ℓ ≡ 0 (mod 5) cases. We then show how this conjecture can be used to prune the search space. Moreover, we provide evidence for Conjecture 1 by confirming that it is valid for all ℓ ≤ 85 such that ℓ ≡ 0 (mod 5) (with the sole exception of ℓ = 75 ).

In Section 3, we first discuss the limitations of the methods used by Turner et al. [10] for finding LPs of length ℓ > 77 and how to overcome those limitations. Then, by assuming that the sought after LP ( a , b ) satisfies { 1 , 69 } ≤ G ^ a and { 1 , 69 } ≤ G ^ b and using the method in Section 4.1.1 of the work of Kotsireas and Koutschan [5] with H 1 = { 1 , 69 } , we find the first known examples of LPs of length 85 that satisfy this property. The length ℓ = 85 has been the smallest open-length case. In addition, we show how to exploit a different concept of balance applied to LPs of composite order. As an application, we find the first examples of LPs of length 87.

In Section 4, we describe the partial searches that have been implemented for LPs of length 115.

In Section 5, we provide the current list of ten ℓ values less than 200 for which the LP existence problem remains unsolved.

2 LPs of lengths ℓ ≡ 0 ( mod 5 )

For a vector of length ℓ = n m , the following lemma expresses a sum of PAFs in terms of the PAF of the compressed vector A m .

Lemma 4

Let a be a vector of length ℓ = n m for positive odd integers n and m. Then,

∑ j = 0 m − 1 PAF a ( n j + k ) = PAF A m ( k ) .

Proof

□ ∑ j = 0 m − 1 PAF a ( n j + k ) = ∑ j = 0 m − 1 ∑ i = 0 ℓ − 1 a i a i − n j − k = ∑ j = 0 m − 1 ∑ i = 0 m − 1 ∑ r = 0 n − 1 a n i + r a n i − n j + r − k = ∑ r = 0 n − 1 ∑ i = 0 m − 1 ∑ j = 0 m − 1 a n i + r a n j + r − k = ∑ r = 0 n − 1 ∑ i = 0 m − 1 a n i + r ∑ j = 0 m − 1 a n j + r − k = ∑ r = 0 n − 1 A r A r − k = PAF A m ( k ) .

For a vector of length ℓ = 5 m , the following lemma determines PSD a ( r m ) in terms of PAF a ( j ) for j ∈ Z ℓ .

Lemma 5

Let m ∈ Z ≥ 1 , r ∈ { 1 , … , 4 } , and a be a vector of length ℓ = 5 m . Then,

PSD a ( r m ) = ∑ j = 0 m − 1 PAF a ( 5 j ) − ∑ k = 1 2 1 + ( − 1 ) k + r 2 5 2 ∑ j = 0 m − 1 PAF a ( 5 j + k ) .

Proof

Let w = e 2 π i ⁄ ℓ and w ′ = w m . By Theorem 3,

PSD a ( r m ) = ∑ j = 0 ℓ − 1 PAF a ( j ) w r m j = ∑ j = 0 ℓ − 1 PAF a ( j ) ( w ′ ) r j = ∑ k = 0 4 ∑ j = 0 m − 1 PAF a ( 5 j + k ) ( w ′ ) r k = ∑ j = 0 m − 1 PAF a ( 5 j ) + ∑ k = 1 2 2 cos 2 π r k 5 ∑ j = 0 m − 1 PAF a ( 5 j + k ) = ∑ j = 0 m − 1 PAF a ( 5 j ) + ∑ k = 1 2 − 1 + ( − 1 ) ⌊ ( r k mod 5 ) ⁄ 2 ⌋ 5 2 ∑ j = 0 m − 1 PAF a ( 5 j + k ) ,

from which the assertion follows directly.□

Corollary 3

For a pair of vectors ( a , b ) of length ℓ = 5 m , the LP constraints

PSD a ( r m ) + PSD b ( r m ) = 2 ℓ + 2

are satisfied for r ∈ { 1 , … , 4 } if and only if

∑ j = 0 m − 1 ( PAF a ( 5 j ) + PAF b ( 5 j ) ) − ∑ j = 0 m − 1 ∑ k = 1 2 PAF a ( 5 j + k ) + PAF b ( 5 j + k ) 2 = 2 ℓ + 2

and

∑ j = 0 m − 1 ∑ k = 1 2 ( − 1 ) k ( PAF a ( 5 j + k ) + PAF b ( 5 j + k ) ) = 0 .

The following corollary follows from Lemmas 4 and 5.

Corollary 4

Let a be a vector of length 5 m for some m ∈ Z ≥ 1 . Let the m-compression of a be A m = [ A 0 , A 1 , A 2 , A 3 , A 4 ] . Then,

(7) PSD a ( r m ) = p 2 ( A m ) − 1 2 e 2 ( A m ) + ( − 1 ) r 2 5 PAF A m ( 1 ) − PAF A m ( 2 ) 2

for r = 1 , … , 4 .

Proposition 1

Let ( a , b ) be an LP of length ℓ = 5 m with m odd. Then, there exist integers n 1 , n 2 ∈ Z ≥ 0 with n 1 + n 2 = 2 ℓ + 2 and x ∈ Z such that

(8) PSD a ( r m ) = n 1 + ( − 1 ) r 2 5 x , PSD b ( r m ) = n 2 − ( − 1 ) r 2 5 x

for r = 1 , … , 4 , where

(9) x = PAF A m ( 1 ) − PAF A m ( 2 ) 2 = − PAF ℬ m ( 1 ) − PAF ℬ m ( 2 ) 2 .

Proof

Equation (8) directly follows from equation (7), and the fact that an LP ( a , b ) of length ℓ = 5 m satisfies PSD a ( r m ) + PSD b ( r m ) = 2 ℓ + 2 for r = 1 , … , 4 . Since, by our general assumption, the vectors a and b satisfy ∑ i = 0 ℓ − 1 a i = ∑ i = 0 ℓ − 1 b i = 1 , the same holds true for their m -compressions, i.e., ∑ i = 0 4 A i = ∑ i = 0 4 B i = 1 . Consequently, ∑ i = 0 4 A i 2 = p 2 ( A m ) + 2 e 2 ( A m ) = 1 , and analogously for ℬ m . Then, by Corollary 4,

(10) n 1 = p 2 ( A m ) − 1 2 e 2 ( A m ) = 1 4 ( 5 p 2 ( A m ) − 1 ) , n 2 = p 2 ( ℬ m ) − 1 2 e 2 ( ℬ m ) = 1 4 ( 5 p 2 ( ℬ m ) − 1 ) .

This shows that n 1 and n 2 are integers, because p 2 ( A m ) is the sum of five odd squares and hence equals 1 mod 4 . To see that x is an integer, note that an odd m implies that PAF A m ( k ) and PAF ℬ m ( k ) for k = 1 , 2 are all odd. Equation (9) follows from Corollary 4 and equation (8).

We will next prove that 0 ≤ n 1 . By symmetry, this will also give us 0 ≤ n 2 . By equations (7) and (10)

4 n 1 25 = p 2 ( A m ) 5 − 1 25 = p 2 ( A m ) 5 − ∑ i = 0 4 A i 5 2 .

Hence, 4 n 1 ⁄ 25 is the population variance of the numbers in { A 0 , A 1 , A 2 , A 3 , A 4 } implying that 0 ≤ n 1 .□

Experimental evidence gathered for ℓ = 15 , 25 , 35 , 45 , and 55 indicates that there are LPs of these orders such that x = 0 and

(11) n 1 = 1 4 ( 5 p 2 ( A m ) − 1 ) = ℓ + 1 = 5 m + 1 , n 2 = 1 4 ( 5 p 2 ( ℬ m ) − 1 ) = ℓ + 1 = 5 m + 1 .

Solving equation (11) for p 2 ( A m ) and p 2 ( ℬ m ) , we obtain that equation (11) is equivalent to

p 2 ( A m ) = p 2 ( ℬ m ) = 4 m + 1 .

The next two lemmas determine restrictions on n 1 , n 2 , and x in Proposition 1.

Lemma 6

Each of n 1 , n 2 , and x in Proposition 1 must be even.

Proof

By symmetry, it suffices to prove the result for n 1 and x . Let a be as in Proposition 1. By replacing a with ( a + 1 ) ⁄ 2 in Lemma 5 and by equation (5), we obtain that

PSD a + 1 2 ( m ) = α 1 2 + α 2 2 5 = n 1 + 5 x 4

for some α 1 , α 2 ∈ Z . Hence, by the linear independence of 1 and 5 in the field extension Q [ 5 ] , we have α 1 = n 1 ⁄ 2 and α 2 = x ⁄ 2 , implying that both n 1 and x must be even.□

Lemma 7

Let n 1 , n 2 , and x be as in Proposition 1. Then, n i + x ≡ 0 ( mod 4 ) for i = 1 , 2 .

Proof

By symmetry, it suffices to prove the result for n 1 and x . Let w ′ = e 2 π i ⁄ 5 and a be as in Proposition 1. Then, by the proof of Lemma 5,

PSD a ( m ) = ∑ k = 0 4 ∑ j = 0 m − 1 PAF a ( 5 j + k ) ( w ′ ) k = n 1 + 5 x .

Now, by equation (5),

PSD a + 1 2 ( m ) = ∑ k = 0 4 ∑ j = 0 m − 1 PAF a + 1 2 ( 5 j + k ) ( w ′ ) k = n 1 + 5 x 4 .

Since PAF ( a + 1 ) ⁄ 2 ( 5 j + k ) = PAF ( a + 1 ) ⁄ 2 ( − 5 j − k ) for k ∈ { 1 , 2 } ,

PSD a + 1 2 ( m ) = S 0 + S 1 ( w ′ + w ′ ¯ ) + S 2 ( ( w ′ ) 2 + ( w ′ ¯ ) 2 ) ,

where S 0 = ∑ j = 0 m − 1 PAF ( a + 1 ) ⁄ 2 ( 5 j ) and S i = ∑ j = 0 m − 1 PAF ( a + 1 ) ⁄ 2 ( 5 j + i ) for i = 1 , 2 . Observe that S 0 , S 1 , S 2 ∈ Z ≥ 1 . Then, since ( w ′ ) 2 + ( w ′ ¯ ) 2 = ( w ′ + w ′ ¯ ) 2 − 2 ∈ Z [ w ′ + w ′ ¯ ] ,

PSD a + 1 2 ( m ) = S 0 + S 1 ( w ′ + w ′ ¯ ) + S 2 ( ( w ′ ) 2 + ( w ′ ¯ ) 2 ) ∈ Z [ w ′ + w ′ ¯ ] .

Then, since 5 = 2 w ′ + 2 w ′ ¯ + 1 ,

PSD a + 1 2 ( m ) = n 1 + 5 x 4 = n 1 + x + 2 x ( w ′ + w ′ ¯ ) 4 ∈ Z [ w ′ + w ′ ¯ ] .

Hence, n 1 + x ≡ 0 ( mod 4 ) .□

The following general lemma will be used to find another restriction on n 1 and n 2 in Proposition 1.

Lemma 8

Let a be a vector of length ℓ = m n for some positive integers m and n . Then,

∑ k = 0 n − 1 PSD a ( k m ) = n ∑ x = 0 m − 1 PAF a ( x n ) = n PAF A m ( 0 ) .

Proof

Since PSD a ( k m ) = PSD A m ( k ) , by applying the inverse Fourier transform to [ PSD A m ( 0 ) … PSD A m ( n − 1 ) ] , we obtain [ PAF A m ( 0 ) , … , PAF A m ( n − 1 ) ] , which at 0 is what the lemma claims (cf. Theorem 3).□

The following lemma provides an additional restriction on the values of n 1 and n 2 in Proposition 1.

Lemma 9

Let a , b , n 1 , and n 2 be as in Proposition 1. Then, n i ≡ 6 ( mod 10 ) for i = 1 , 2 .

Proof

By symmetry, it suffices to prove the result for i = 1 . By Lemma 8 and Proposition 1,

∑ k = 0 4 PSD a ( k m ) = 1 + 4 n 1 = 5 ∑ x = 0 m − 1 PAF a ( 5 x ) ≡ 0 ( mod 5 ) .

Alternatively, we can employ p 2 ( A m ) = 4 s + 1 (see the proof of Proposition 1) to deduce that n 1 = ( 5 p 2 ( A m ) − 1 ) ⁄ 4 = 5 s + 1 for some integer s . Now the result follows from the Chinese remainder theorem since n 1 must be even by Lemma 6.□

The following corollary provides upper and lower bounds on n 1 and n 2 in Proposition 1.

Corollary 5

Let ℓ = 5 m , and let a , b , n 1 , and n 2 be as in Proposition 1. Then, 6 ≤ n i ≤ 2 ℓ − 4 for i = 1 , 2 .

Proof

By Proposition 1, we obtain 0 ≤ n i ≤ 2 ℓ + 2 = 10 m + 2 for i = 1 , 2 . By Lemma 9, we obtain 6 ≤ n i ≤ 2 ℓ − 4 = 10 m − 4 for i = 1 , 2 .□

The lower bound for n i in Corollary 5 is achieved by taking A m = [ 1 , 1 , 1 , − 1 , − 1 ] .

The following corollary now follows from Proposition 1, Corollary 5, and Lemma 7.

Corollary 6

Let ℓ = 5 m , and let a , b , n 1 , n 2 , and x be as in Proposition 1. Then, there exist nonnegative integers k 1 and k 2 such that n i = 10 k i + 6 for i = 1 , 2 , k 1 + k 2 = m − 1 , and 2 k i + 2 ≡ x (mod 4).

Corollary 6 summarizes all the constraints obtained in this section regarding n 1 , n 2 , and x in equation (8). In fact, all the constraints regarding n 1 , n 2 , and x that appear before Corollary 6 are implied by Corollary 6.

Exhaustive searches for ℓ = 5 , 15 , 25 have revealed that most LPs of these lengths have n 1 = n 2 = ℓ + 1 , where n 1 and n 2 are as in Proposition 1. The case n 1 = n 2 = ℓ + 1 corresponds to taking k 1 = k 2 = ( m − 1 ) ⁄ 2 in Corollary 6. Non-exhaustive searches for larger odd values of ℓ , which are multiples of 5, have revealed the same pattern, i.e., the standard relationship PSD a ( m ) + PSD b ( m ) = 2 ℓ + 2 is materialized (often but not always) in the same manner. These observations give rise to the following conjecture.

Conjecture 1

For every odd positive integer ℓ ≡ 0 ( mod 5 ) , there exists at least one LP ( a , b ) of length ℓ = 5 m such that

(12) PSD a ( r m ) = ℓ + 1 + ( − 1 ) r 2 5 x PSD b ( r m ) = ℓ + 1 − ( − 1 ) r 2 5 x

for r = 1 , … , 4 , and for some nonnegative integer x ≡ − ( ℓ + 1 ) ≡ ℓ + 1 ( mod 4 ) .

By formula (8), Conjecture 1 holds for one r if and only if n 1 = n 2 . Again, by formula (8), if Conjecture 1 holds for one r , then it also holds for the other r s. Conjecture 1 states that for each odd positive ℓ ≡ 0 ( mod 5 ) , there exists a length ℓ LP ( a , b ) such that n 1 = n 2 = ℓ + 1 in Proposition 1. That is, the constant 2 ℓ + 2 can be distributed in a “balanced” manner between PSD a ( r m ) and PSD b ( r m ) for r ∈ { 1 , … , 4 } . On the other hand, there exist LPs ( a , b ) of length ℓ = 5 m , which do not satisfy n 1 = n 2 . Table 1 provides computational evidence for Conjecture 1. Cases m = 1 , 3 , 5 are based on complete – and the remaining cases are based on partial – classifications of all nonequivalent LPs, where equivalent LPs satisfying Conjecture 1 have the same x value. This is because

PSD ( j 1 , k ) a ( r m ) = PSD a ( k r m )

for each ( j 1 , k ) ∈ Z ℓ ⋊ Z ℓ × and r ∈ { 1 , … , 4 } .

Table 1

Computationally verified cases for Conjecture 1 with their corresponding x value(s)

m	ℓ = 5 m	x
1	5	2
3	15	0 , 4
5	25	2 , 6 , 10
7	35	0 , 8 , 16
9	45	2 , 6 , 10 , 14 , 18
11	55	4 , 8 , 12 , 20 , 24
13	65	14
17	85	18

The following proposition provides necessary and sufficient constraints on the m -compressed vectors ( A m , ℬ m ) of an LP ( a , b ) of length ℓ = 5 m to satisfy equation (12).

Proposition 2

The m-compressed vectors A m = [ A 0 , … , A 4 ] and ℬ m = [ B 0 , … , B 4 ] of an LP ( a , b ) of length ℓ = 5 m satisfy the conditions in Conjecture 1 if and only if

(13) p 2 ( A m ) = PAF A m ( 0 ) = p 2 ( ℬ m ) = PAF ℬ m ( 0 ) = 4 m + 1 .

Proof

This result follows from Proposition 1 and equation (10).□

Equation (13) in Proposition 2 drastically reduces the search space for an LP ( a , b ) satisfying the conditions in Conjecture 1. For such an LP, we are only interested in all-odd solutions to equation (13) since A i and B i are odd numbers for i = 0 , … , 4 . Another consequence of Proposition 2 is that the alphabet of the possible m -compressed vectors ( A m , ℬ m ) is also significantly truncated. More specifically, while the full alphabet for candidate m -compressed vectors is

{ − m , − ( m − 2 ) , … , − 1 , + 1 , … , ( m − 2 ) , m } ,

by Proposition 2, A i is an odd integer with ∣ A i ∣ ≤ 4 m − 3 < 2 m for i ∈ Z ℓ .

Next, we show that Conjecture 1 cannot be ruled out by the necessary constraints [equation (13)] and ∑ i = 0 4 A i = ∑ i = 0 4 B i = 1 for LPs of lengths ℓ = 5 , 15 , 25 , 35 , 45 , 55 , 65 , 75 , 85 , 95 , 105 , and 115. In Table 2, we summarize the all-odd solutions to equation (13) for odd ℓ such that ℓ = 5 m and m ∈ { 1 , … , 23 } . In the last column, we record the all-odd solutions (up to sign changes and permutations) of the five sums-of-squares Diophantine equations of the fourth column. Since the linear equations ∑ i = 0 4 A i = ∑ i = 0 4 B i = 1 must also be satisfied, some of these all-odd solutions are ruled out, and this is indicated by boldface. Conjecture 1 has practical value as it can be used to prune the search space when ℓ ≡ 0 (mod 5) by only decompressing m -compressions, which are compatible with the all-odd solutions in Table 2.

Table 2

All-odd solutions to equations (13)

ℓ	m	PSD a ( r m ) PSD b ( r m )	∑ i = 0 4 A i 2 ∑ i = 0 4 B i 2	All-odd solutions
5	1	5 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 1 + 1 = 5	[1, 1, 1, 1, 1]
15	3	15 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 3 + 1 = 13	[1, 1, 1, 1, 3]
25	5	25 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 5 + 1 = 21	[1, 1, 1, 3, 3]
35	7	35 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 7 + 1 = 29	[1, 1, 1, 1, 5], [1, 1, 3, 3, 3]
45	9	45 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 9 + 1 = 37	[1, 1, 1, 3, 5], [1, 3, 3, 3, 3]
55	11	55 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 11 + 1 = 45	[1, 1, 3, 3, 5], [3, 3, 3, 3, 3]
65	13	65 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 13 + 1 = 53	[1, 1, 1, 1, 7], [1, 1, 1, 5, 5], [1, 3, 3, 3, 5]
75	15	75 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 15 + 1 = 61	[1, 1, 1, 3, 7], [1, 1, 3, 5, 5], [3, 3, 3, 3, 5]
85	17	85 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 17 + 1 = 69	[1, 1, 3, 3, 7], [1, 3, 3, 5, 5]
95	19	95 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 19 + 1 = 77	[1, 1, 1, 5, 7], [1, 1, 5, 5, 5], [1, 3, 3, 3, 7], [3, 3, 3, 5, 5]
105	21	105 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 21 + 1 = 85	[1, 1, 1, 1, 9], [1, 1, 3, 5, 7], [1, 3, 5, 5, 5], [3, 3, 3, 3, 7]
115	23	115 + 1 ± ( − 1 ) r 2 5 x	4 ⋅ 23 + 1 = 93	[1, 1, 1, 3, 9], [1, 3, 3, 5, 7], [3, 3, 5, 5, 5]

3 Finding LPs of lengths ℓ = 85 and ℓ = 87

The orbit of a vector b under circulant shifts and decimations is called the decimation class of b [9]. Since ( Z ℓ × Z ℓ ) ⋊ Z ℓ × acts on LPs, the search space for LPs is drastically reduced by searching only across decimation class representatives [4]. In fact, to classify { 0 , 1 } LPs up to equivalence, in light of equation (3), Fletcher et al. [4] exhaustively generated a set of all decimation class representatives b ∈ { 0 , 1 } ℓ for odd ℓ ≤ 47 such that ∑ i = 0 ℓ − 1 b i = ( ℓ + 1 ) ⁄ 2 . By equation (3), they then deleted the class representatives b such that PSD b ( j ) > ( ℓ + 1 ) ⁄ 2 for some j ∈ Z ℓ \ { 0 } . Then, among all decimations of a given vector b , they selected the decimation d k ( b ) with ∣ PSD d k ( b ) ( 1 ) − ( ℓ + 1 ) ⁄ 4 ∣ of greatest magnitude to be the representative of its decimation class, where they required a vector b to have two different decimation class representatives if

max k ∈ Z × PSD d k ( b ) ( 1 ) − ℓ + 1 4 = − min k ∈ Z × PSD d k ( b ) ( 1 ) − ℓ + 1 4 ≠ 0 .

In the second step, they sorted the list of class representative b s according to

PSD b ( 1 ) − ℓ + 1 4 .

In the last step, they located pairs of class representatives a and b that satisfied

(14) PSD a ( 1 ) − ℓ + 1 4 + PSD b ( 1 ) − ℓ + 1 4 = 0 .

To confirm that the LP property was satisfied by all the resulting pairs, they also deleted pairs ( a , b ) satisfying equation (14) but not equation (1). In the end, Fletcher et al. [4] classified all LPs of length ≤ 47 up to equivalence using this method. A similar method that additionally exploited simultaneous decompressions of candidate compressed vectors was implemented in the work of Turner et al. [10]. This method found an LP of length 77 for the first time.

For lengths ℓ = 55 and 77, Table 3 reports the decimation class counts for the number of decimation classes of length ℓ , { 0 , 1 } vectors with ( ℓ + 1 ) ⁄ 2 ones reported in [9], and the solution times for the simultaneous decompression-based searches reported in [10].

Table 3

Decimation class counts in the work of Turner et al. [9] and solution times for simultaneous decompression-based searches in the work of Turner et al. [10]

Length ℓ	Number of decimation classes	Exhaustive search (CPU hours)	One solution (CPU hours)
55	1.738341231644 × 1 0 12	101 , 542.4	—
77	2.945564382817 × 1 0 18	—	182 , 280

Assuming that the number of compressed pairs and the time required for decompression scales linearly on average with the number of decimation classes, we expect the Table 4 time requirements for an exhaustive search or partial search to first solution if the method in [10] is used for lengths ℓ = 85 and ℓ = 87 .

Table 4

Expected time needed for an exhaustive search and partial search to first solution using the method in the work of Turner et al. [10]

Length ℓ	Number of decimation classes	Exhaustive search (CPU hours)	One solution (CPU hours)
85	6.100692175209 × 1 0 20	3.563621 × 1 0 13	37 , 752 , 839
87	2.693812140345 × 1 0 21	1.573547 × 1 0 14	166 , 700 , 847

These time estimates are optimistic as experiments suggest that the growth rate is superlinear in the number of decimation classes. This growing magnitude of complexity precluded Turner et al. [10] from investigating larger open LP problems. Partial searches for LPs using the compression method for ℓ > 77 only remain viable by reducing the number of complementary compressed vectors or rather by selecting the complementary compressions with a higher likelihood of producing an LP. Alternatively, by Corollary 1, we can search for an LP ( a , b ) that has multiplier groups G ^ a and G ^ b , both containing some nontrivial elements of Z n × . This would reduce the search space drastically and can bring ℓ > 77 cases within computational reach if such an LP exists. For ℓ = 3 m , m ∈ { 39 , 43 , 49 } , Kotsireas and Koutschan [5] used a method that exploits such a property along with a method that reduces the possible values of PSD a ( m ) and keeps only the vectors whose PSD a ( m ) are within the reduced set and found the first examples of LPs of lengths 117, 129, and 147. Kotsireas and Koutschan [5] also found an LP of length 133 by solely restricting the search to LPs ( a , b ) whose multiplier groups G ^ a and G ^ b that both contained the group G = { 1 , 11 , 121 } < Z 133 × .

3.1 LPs of length ℓ = 85

We used the method in Section 4.1.1 in the work of Kotsireas and Koutschan [5] with H 1 = { 1 , 69 } and Corollary 1 to restrict the search space for LPs of length ℓ = 85 by assuming that the sought-after LP ( a , b ) has multiplier groups G ^ a and G ^ b satisfying H 1 = { 1 , 69 } ≤ G ^ a and H 1 = { 1 , 69 } ≤ G ^ b . This method found the first known examples of LPs of length ℓ = 85 . This had been the smallest previously unknown length case for LPs. The subgroup H 1 = { 1 , 69 } of Z 85 × acts on Z 85 and yields 16 orbits of size 1 and 34 orbits of size 2. We searched for an LP of length ℓ = 85 , which could be obtained by combining the orbits of the subgroup H 1 . This restriction has the benefit of reducing the search space, provided that such an LP exists. We chose 12 orbits of size 1 and 15 orbits of size 2 to make blocks of size 12 ⋅ 1 + 15 ⋅ 2 = 42 . Here, each block of size 42 consists of positions of − 1 s determining the vectors a and b . Therefore, the size of the search space was 16 12 ⋅ 34 15 = 1,820 ⋅ 1,855,967,520 = 3,377,860,886,400 . This search was not exhaustive and was interrupted after traversing 2.6% of the search space. This took about 100 h of CPU time. This search was done on 64 compute nodes, each with two 8-core Intel Haswell CPUs (Xeon E5-2630v3, 2.4Ghz) and 128 GB RAM. For this search, we implemented the same PSD test in Section 4 of the work of Kotsireas and Koutschan [5] without checking if the PSD value at ℓ ⁄ 3 is from a finite list of candidates, and used the same method as in Section 4 of the work of Kotsireas and Koutschan [5] to identify LPs among vectors that passed the PSD test. However, this search did not take advantage of Conjecture 1 by only keeping the vectors a that satisfied PAF A 17 ( 0 ) = 69 for the 17 compression A 17 of a as Conjecture 1 had not yet been formulated at the time of the search. Our search yielded four equivalent (one nonequivalent) LPs of length ℓ = 85 made out of six different vectors. Their lexicographic rank encodings as subsets of size 12 out of 16 and 15 out of 34 are

( { 12 , 1321116338 } , { 42 , 1275934280 } ) , ( { 12 , 1843909851 } , { 42 , 606586783 } ) , ( { 42 , 1275934280 } , { 9 , 1555522731 } ) , ( { 42 , 606586783 } , { 9 , 788215097 } ) .

For lexicographic ranking and unranking algorithms for subsets of size k , we refer the readers to the work of Kreher and Stinson [7]. For the first LP ( a , b ) of length ℓ = 85 shown above, its lexicographic rank encoding ( { 12 , 13211163 38 } , { 42 , 1275934280 } ) is decoded as follows:

In the space of 16 12 = 1,820 , 12 subsets of { 1 , … , 16 } , decode
12 as a ones = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 14 , 15 } , 42 as b ones = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 10 , 11 , 14 , 15 } .
In the space of 34 15 = 1 , 855 , 967 , 520 15 subsets of { 1 , … , 34 } , decode
1321116338 as a twos = { 3 , 4 , 5 , 7 , 10 , 11 , 22 , 24 , 25 , 27 , 28 , 29 , 30 , 31 , 34 } , 1275934280 as b twos = { 2 , 8 , 10 , 11 , 12 , 15 , 19 , 21 , 23 , 25 , 26 , 28 , 29 , 33 , 34 } .
Enumerate the 16 orbits of size 1 in increasing order as follows:
O 1 = { { 5 } , { 10 } , { 15 } , { 20 } , { 25 } , { 30 } , { 35 } , { 40 } , { 45 } , { 50 } , { 55 } , { 60 } , { 65 } , { 70 } , { 75 } , { 80 } } .
Enumerate the 34 orbits of size 2 in increasing order of their smallest element as O 2 = { { 1 , 69 } , { 2 , 53 } , { 3 , 37 } , { 4 , 21 } , { 6 , 74 } , { 7 , 58 } , { 8 , 42 } , { 9 , 26 } , { 11 , 79 } , { 12 , 63 } , { 13 , 47 } , { 14 , 31 } , { 16 , 84 } , { 17 , 68 } , { 18 , 52 } , { 19 , 36 ] , { 22 , 73 ] , {23, 57} , {24, 41} , {27, 78} , {28, 62} , {29, 46} , {32, 83} , {33, 67} , {34, 51} , {38, 72} , {39, 56} , { 43 , 77 } , { 44 , 61 } , { 48 , 82 } , { 49 , 66 } , { 54 , 71 } , { 59 , 76 } , { 64 , 81 } } .
Make a block of size 42 of the indices of the positions of the − 1 elements in a , by combining 12 elements of O 1 whose indices are given by a ones and 15 elements of O 2 whose indices are given by a twos . This yields the following a block of size 42: { 5 , 10, 15, 20, 25, 30, 35, 40, 45, 50, 70, 75, 3, 37, 4, 21, 6, 74, 8, 42, 12, 63, 13, 47, 29, 46, 33, 67, 34, 51, 39, 56, 43, 77, 44, 61, 48, 82, 49, 66, 64, 81 } .
Make a block of size 42 of the indices of the positions of the − 1 elements in b by combining 12 elements of O 1 whose indices are given by b ones and 15 elements of O 2 whose indices are given by b twos . This yields the following b block of size 42: { 5 , 10, 15, 20, 25, 30, 35, 40, 50, 55, 70, 75, 2, 53, 9, 26, 12, 63, 13, 47, 14, 31, 18, 52, 24, 41, 28, 62, 32, 83, 34, 51, 38, 72, 43, 77, 44, 61, 59, 76, 64, 81 } .
The above a and b blocks (both of size 42) yield an LP for ℓ = 85 .

For the first LP ( a , b ) of length ℓ = 85 shown above, the 17 compressions are

A 17 = [ 1 , 3 , 3 , 1 , − 7 ] , ℬ 17 = [ 3 , 1 , 1 , 3 , − 7 ] .

These possess the properties required by Proposition 2. That is,

PAF A 17 ( 0 ) = PAF ℬ 17 ( 0 ) = 4 ⋅ 17 + 1 = 69 .

As a consequence, the coefficients of 5 in PSD a ( 17 r ) and PSD b ( 17 r ) cancel out for r = 1 , … , 4 . Specifically,

PSD a ( 17 r ) = 85 + 1 + ( − 1 ) r 2 5 ⋅ 18 PSD b ( 17 r ) = 85 + 1 − ( − 1 ) r 2 5 ⋅ 18 .

Remark 1

We did not use Conjecture 1 or Corollary 6 in our search for an LP of length 85 because neither Conjecture 1 nor Corollary 6 had been formulated when we implemented our search. However, the LP we found happened to satisfy equation (12) in Conjecture 1.

3.2 LPs of length ℓ = 87

Next, we describe our search method for LPs ( a , b ) of length ℓ = 87 . First, we need the following theorem, which is a special case of Theorem 3 in [2]:

Theorem 5

Let ( a , b ) be an LP of length ℓ = d m . Then, the m-compressed vectors ( A m , ℬ m ) satisfy

PAF A m ( 0 ) + PAF ℬ m ( 0 ) = 2 ℓ − 2 ℓ d − 1 = 2 ℓ + 2 − 2 ℓ d , PAF A m ( j ) + PAF ℬ m ( j ) = − 2 ℓ d , ∀ j ∈ Z d \ { 0 } .

By Theorem 5, the three-compressed vectors ( A 3 , ℬ 3 ) of an LP ( a , b ) of length ℓ = 87 must contain 14 elements with an absolute value equal to 3 and 2 ⋅ 29 − 14 = 44 elements with an absolute value equal to 1. Experimental evidence from the study of other lengths that are divisible by 3 indicates that A 3 and ℬ 3 , which have an equal number of elements with an absolute value equal to 3, are more likely to yield LPs. Hence, we computed approximately 6,000 candidate 3-compressions satisfying the following constraints:

The vectors A 3 , ℬ 3 ∈ { − 3 , − 1 , + 1 , + 3 } 29 contain 14 elements with an absolute value equal to 3 and 44 elements with an absolute value equal to 1.
PAF A 3 ( s ) + PAF ℬ 3 ( s ) = ( − 2 ) ⋅ 3 = − 6 for s = 1 , … , 28 .
PSD A 3 ( s ) + PSD ℬ 3 ( s ) = 2 ℓ + 2 = 2 ⋅ 87 + 2 = 176 for s = 1 , … , 28 .
∑ i = 0 28 A i = ∑ i = 0 28 B i = 1 .
Each of A 3 and ℬ 3 contains 7 elements with an absolute value equal to 3 and 22 elements with an absolute value equal to 1.

Among the above constraints, only constraint 5 is not necessary for a length 87 LP to exist. However, imposing constraint 5 was essential in greatly reducing the search space to a part where solutions are most likely to exist. Subsequently, we ran our C 3-uncompression code for approximately 2,000 candidate 3-compressions and discovered the following two LPs of order ℓ = 87 :

a = [ − 1 , − 1 , − 1 , − 1 , 1 , − 1 , − 1 , − 1 , 1 , 1 , − 1 , 1 , − 1 , 1 , − 1 , 1 , 1 , − 1 , − 1 , 1 , 1 , − 1 , − 1 , 1 , − 1 , 1 , − 1 , 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , − 1 , − 1 , − 1 , − 1 , − 1 , 1 , − 1 , − 1 , 1 , − 1 , − 1 , − 1 , 1 , 1 , 1 , 1 , − 1 , − 1 , − 1 , 1 , − 1 , 1 , 1 , 1 , − 1 , 1 , 1 , − 1 , − 1 , − 1 , − 1 , 1 , 1 , 1 , 1 , 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , − 1 , 1 , 1 , 1 , 1 , 1 , − 1 , 1 , − 1 , 1 , − 1 ] , b = [ − 1 , − 1 , − 1 , − 1 , − 1 , − 1 , − 1 , − 1 , 1 , 1 , 1 , 1 , − 1 , 1 , 1 , − 1 , 1 , 1 , − 1 , 1 , − 1 , 1 , 1 , 1 , 1 , − 1 , − 1 , 1 , 1 , − 1 , 1 , 1 , − 1 , 1 , − 1 , − 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , − 1 , 1 , − 1 , 1 , 1 , − 1 , − 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , 1 , 1 , − 1 , − 1 , − 1 , 1 , 1 , 1 , − 1 , − 1 , 1 , − 1 , 1 , − 1 , − 1 , 1 , − 1 , 1 , − 1 , − 1 , − 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , − 1 , 1 , − 1 , 1 , − 1 ] , a = [ − 1 , − 1 , − 1 , − 1 , 1 , − 1 , − 1 , − 1 , 1 , 1 , 1 , 1 , 1 , 1 , − 1 , 1 , − 1 , − 1 , − 1 , 1 , 1 , 1 , − 1 , 1 , − 1 , − 1 , − 1 , 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , − 1 , − 1 , − 1 , 1 , − 1 , − 1 , 1 , − 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , − 1 , 1 , 1 , − 1 , 1 , 1 , − 1 , − 1 , − 1 , − 1 , 1 , − 1 , 1 , 1 , − 1 , − 1 , − 1 , − 1 , 1 , 1 , 1 , − 1 , 1 , 1 , − 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , − 1 ] , b = [ − 1 , 1 , − 1 , 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , 1 , 1 , 1 , − 1 , 1 , − 1 , 1 , 1 , 1 , − 1 , − 1 , 1 , − 1 , 1 , − 1 , 1 , − 1 , 1 , − 1 , − 1 , − 1 , 1 , − 1 , − 1 , − 1 , − 1 , − 1 , − 1 , − 1 , 1 , − 1 , 1 , 1 , 1 , − 1 , 1 , − 1 , − 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , − 1 , − 1 , 1 , 1 , − 1 , − 1 , 1 , − 1 , 1 , − 1 , − 1 , 1 , − 1 , 1 , − 1 , 1 , − 1 , − 1 , 1 , − 1 , − 1 , 1 , − 1 , − 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , − 1 ] .

Both of the above LPs of length ℓ = 87 3-compress to

A 3 = [ − 3 , − 1 , 1 , − 1 , 1 , − 3 , − 3 , − 1 , 1 , 1 , 1 , 1 , − 1 , 1 , − 3 , 1 , 1 , 1 , − 1 , 3 , 3 , − 1 , − 1 , 1 , − 1 , 1 , − 1 , 3 , 1 ] , ℬ 3 = [ − 3 , − 1 , 1 , − 1 , 1 , − 3 , − 3 , 1 , − 1 , 1 , 1 , 1 , 1 , − 1 , 3 , − 3 , 1 , 1 , − 1 , − 1 , − 1 , 1 , 1 , 3 , 1 , 1 , − 1 , 3 , − 1 ] .

There are seven ± 3 s in each of the two vectors A 3 and ℬ 3 , and this “balanced” configuration yields LPs of length ℓ = 87 . Based on the analysis at the beginning of this section, we can make the following claim.

Claim 1

Without imposing constraint 5, a successful outcome of our search for a length 87 LP would not have been possible.

4 Searching for LPs of length ℓ = 115

By using Corollary 1, two nonexhaustive searches for an LP ( a , b ) of length 115 with multiplier groups G ^ a and G ^ b both containing the subgroup { 1 , 91 } were performed. The first search was done before Conjecture 1 was formulated, hence it did not use it. For the first search, the combinations of 0 cosets of size 1 and 29 cosets of size 2 were considered, yielding a search space of size 4 0 ⋅ 55 29 = 3,560,597,348,629,860 . The computation was aborted after almost 4% of the search space was traversed. This took about 4,359 days of CPU time, and the output files took up 76 GB of disk space. The second search was done with the implementation of Conjecture 1. This time, combinations of four cosets of size 1 and 27 cosets of size 2 were considered, yielding a search space of size 4 4 ⋅ 55 27 = 3,824,345,300,380,220 . This computation was aborted after 3,436 CPU days, when slightly more than 10% of the search space was traversed. The output files occupied about 64 GB of disk space. For both of the searches, the same PSD test in Section 4 of the work of Kotsireas and Koutschan [5], without checking if the PSD value at ℓ ⁄ 3 is from a finite list of candidates, was implemented. Both searches used the same method as in Section 4 of the work of Kotsireas and Koutschan [5] to identify LPs among vectors that passed the PSD test. The comparison of the two partial searches shows the significant gain (both time-wise and space-wise) that is obtained from implementing Conjecture 1. Neither the first, nor the second search yielded an LP of length 115. Both of the searches were done on 64 compute nodes, each with two 8-core Intel Haswell CPUs (Xeon E5-2630v3, 2.4Ghz) and 128 GB RAM.

5 Conclusion and future research

Recently, LPs of length 77 were found in [10], and lengths 117, 129, 133, and 147 were found in the work of Kotsireas and Koutschan [5] for the first time. In this article, we find LPs of (the previously open) lengths 85 and 87. This reduces the list of integers < 200 for which the existence of LP problem remains open to the following ten values:

115 , 145 , 159 , 161 , 169 , 175 , 177 , 185 , 187 , 195 .

An LP ( a , b ) as defined in this article corresponds to a difference family in Z ℓ [1]. In the work of Djokovic and Kotsireas [3], previously known theory to search for difference families in Z ℓ was generalized to search for difference families in finite abelian groups. A possible direction for future research is generalizing the theory in this article to difference families in finite abelian groups, i.e., by the structure theorem of finite abelian groups, groups of the form Z d 1 × Z d 2 × ⋯ × Z d m , where d i divides d i + 1 for i = 1 , … , m − 1 , and d m ≡ 0 ( mod 5 ) .

Acknowledgments

The views expressed in this article are those of the authors and do not reflect the official policy or position of the United States Air Force, Department of Defense, or the US Government. The authors thank three anonymous for carefully reading the article and providing comments that significantly improved the article.

Funding information: Christoph Koutschan was supported by the Austrian Science Fund (FWF): F5011-N15.
Conflict of interest: Authors state no conflict of interest.
Data availability statement: The datasets generated during and/or analysed during the current study are available from the corresponding author on reasonable request.

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Received: 2023-05-01

Revised: 2023-08-21

Accepted: 2023-09-07

Published Online: 2023-11-10

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/spma-2023-0105

Keywords for this article

compressed vector; discrete Fourier transform; Hadamard matrix; periodic autocorrelation function; power spectral density; multiplier group

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