For a fixed integer k ≥ 2, the k-generalized Fibonacci (k-nacci) recurrence is

with the conventional initial data F 1 ( k ) = 1 and F 0 ( k ) = F − 1 ( k ) = ⋯ = F 2 − k ( k ) = 0 . Thus k = 2 gives Fibonacci, k = 3 Tribonacci, k = 4 Tetranacci, and so forth. This terminology and recurrence are standard in the k-generalized Fibonacci literature [1]-[4].

The corresponding characteristic polynomial is x k − x k − 1 − ⋯ − x − 1 . It has a unique real root α k in (1, 2), the dominant k-nacci (multinacci) constant. Quasigeometric sequence (QGS) theory imposes an additional arithmetic condition: consecutive integer terms must also be related by nearest-integer multiplication by α k . In the golden-section case, this dual additive-quasigeometric framework was developed in earlier QGS work [5]-[7].

Recent work shows that higher-order Fibonacci recurrences remain an active subject across algebraic, combinatorial, matrix, convolution, and Diophantine directions. Özdemir and Karakaya developed matrix identities for k-generalized Fibonacci numbers [8], while Bérczes, Pink, and Young studied Cullen and Woodall numbers in k-generalized Fibonacci sequences [9]. Frontczak and Gryszka obtained general convolution sums whose specializations include Tribonacci, Tetranacci, and Pentanacci numbers [10]. Billore, Patel, and Makwana developed extended generalized Fibonacci and Tribonacci polynomials with Binet-type, generating-function, and classical identity formulas [11]. Recent Tribonacci-specific work includes the Fibonacci-product problem of Luca, Odjoumani, and Togbé [12], the generalized Wythoff/Trithoff construction of Chen et al. [13], and Gryszka’s self-convolution formula [14].

The higher-order literature has also continued to develop at the level of explicit recurrence structure. Gryszka treated sums of squares for Tribonacci and Tetranacci sequences with arbitrary initial conditions [15]; Bates et al. gave a new combinatorial interpretation of partial sums of m-step Fibonacci numbers [16]; Mane derived new expressions for k-generalized Fibonacci numbers using Fuss-Catalan numbers [17]; and Karakaya and Özdemir obtained further matrix identities for generalized Fibonacci and Tribonacci numbers [18]. Demirkol Ozkaya studied Tribonacci numbers represented as products of three Fibonacci numbers [19], while De la Sen analyzed boundedness, convergence, oscillation, and stability in generalized Fibonacci-type recurrences [20]. These recent developments complement the present QGS problem: rather than deriving further identities inside the additive recurrence alone, the present paper asks when the additive k-step law is compatible with nearest-integer multiplication by the dominant multinacci root.

The higher-order extension creates a compatibility problem. An orbit generated by nearest-integer multiplication need not satisfy the k-step additive recurrence. We call a positive integer admissible when the two rules agree over the first k steps and barren when they do not. The purpose of this paper is to develop an exact arithmetic description of this obstruction, distinguish primitive from inherited barrenness, and describe the resulting parent-number set.

Fix k ≥ 3. Let α k be the unique real root in (1, 2) of

equivalently,

Definition 2.1 (strict k-nacci QGS).

An integer sequence H = ( h 0 , h 1 , ⋯ ) is a strict k-nacci quasigeometric sequence when, at every applicable index n, it satisfies both

and

where rnd denotes nearest-integer rounding.

Lemma 2.2 (absence of rounding ties).

For every positive integer n, α k n is not a half-integer.

Proof

The polynomial in(3) is monic with integer coefficients. Its real root α k isirrational for k ≥ 2 .If α k n = m + 1 / 2 for integers m , n > 0 , then α k = ( 2 m + 1 ) / ( 2 n ) would be rational, a contradiction. Hence nearest-integer rounding is unambiguous. □

Define the rounded map

For n ∈ ℕ set x 0 = n and x j + 1 = R k ( x j ) . The local k-step additive discrepancy isgivenby Equation (7):

Definition 2.3 (admissible and barren).

A positive integer n is admissible when D k ( n ) = 0 and barren when D k ( n ) ≠ 0 . In parent-number terminology, the admissible integers form the genuine k-nacci parent-number set.

Lemma 3.1 (root identity).

For every k ≥ 2 ,

Proof

Multiply(2) by α k − 1 . The right-hand side telescopes, giving α k k + 1 − 2 α k k + 1 = 0 . Division by α k k yields α k − 2 + α k − k = 0 , which is (8). □

Set β k = α k − k = 2 − α k . Then

Because 2n is an integer and no half-integer tie occurs, rnd ( M − y ) = M − ⌊ y + 1 / 2 ⌋ for integer M. Hence

Define the integer-valued correction

Therefore the rounded map has the exact form

This establishes explicitly both the integer-valued nature of c_k and the exact passage from nearest-integer rounding to the correction representation.

Let x 0 = n and iterate

Lemma 4.1 (iteration formula).

For every j ≥ 1 , we state Equation (14):

Proof

The case j = 1is (13). If(14)holds for j,substitute it into x j + 1 = 2 x j − c k ( x j ) ;the powers and summation indices shift exactly to the stated formula for j + 1. □

Theorem 4.2 (correction-sum discrepancy formula).

For every positive integer n, we state Equation (15):

Proof

Substitute(14) into

The coefficient of n is

Fix i ∈ { 0 , ⋯ , k − 1 } . The correction c k ( x i ) occurs inx_kwith coefficient − 2 k − 1 − i and in

with total coefficient

Their sum is −1. Summing over i gives (15). □

Corollary 4.3 (finite admissibility test).

We state Equation (16):

A positive integer n is admissible if and only if Equation (16) holds, where x 0 = n and x j + 1 = 2 x j − ⌊ β k x j + 1 / 2 ⌋ .Thus, admissibility is decided by exactly k rounded iterations.

Lemma 5.1 (k-step recovery).

For every n ∈ ℕ , including n = 1,

Proof

Write x j + 1 = α k x j + ε j , where | ε j | < 1 / 2 by Lemma 2.2. Iteration gives Equation(18):

Dividing(2) by α k k gives

Therefore we have Equation (19):

Theinequalityis strict because every | ε j | < 1 / 2 . Hence x k / α k k lies strictly within one half-unit of n, so nearest-integer rounding gives n. □

Since R k k ( n ) = x k , Equation (17) may also be written rnd ( x k / α k k ) = n .

Theorem 5.2 (discrepancy conservation).

For every n ∈ ℕ ,

Proof

Let x j = R k j ( n ) . Then D k ( R k ( n ) ) − D k ( n ) = x k + 1 − 2 x k + x 0 . By (12), x k + 1 − 2 x k = − c k ( x k ) . From the recovery lemma and (10) applied backward to the recovered integer, c k ( x k ) = x 0 . Hence the difference is zero. Equivalently, the same identity follows directly by applying the correction-sum formula to the shifted orbit. □

Corollary 5.3 (orbit invariance).

For every j ≥ 0 , D k ( R k j ( n ) ) = D k ( n ) .Therefore admissibility, barrenness,discrepancy sign, and discrepancy magnitude are constant along a forward roundedorbit.

Theorem 6.1 (barren inheritance).

If n is barren,then R k j ( n ) is barren for every j ≥ 0 .

Proof

If D k ( n ) ≠ 0 ,Corollary5.3 gives D k ( R k j ( n ) ) = D k ( n ) ≠ 0 . □

For d ∈ ℤ define Equation (21):

Then R k ( ℬ k , d ) ⊆ ℬ k , d . The parent-number set is ℬ k , 0 ; every class with d ≠ 0 is barren. Thus forward-orbit inclusion follows directly from discrepancy conservation, not from a numerical test.

Lemma 7.1 (uniqueness of a positive predecessor).

For any y ∈ ℕ there is at most one x ∈ ℕ satisfying R k ( x ) = y .

Proof

R k ( x ) = y is equivalent to y − 1 / 2 < α k x < y + 1 / 2 .Thusx lies in an open interval of length 1 / α k < 1 . Any interval of length strictly less than one contains at most one integer. □

Lemma 7.2 (strict growth).

For every n ∈ ℕ and k ≥ 3 , R k ( n ) > n .

Proof

Since α k > 3 / 2 for k ≥ 3 , ( α k − 1 ) n > 1 / 2 for n ≥ 1 . Hence α k n > n + 1 / 2 ,and nearest-integer rounding gives R k ( n ) ≥ n + 1 . □

Definition 7.3.

A barren integer is inherited barren if it has a positive predecessor under R_k. Aprimitive barren integer is a barren integer with no positive predecessor. The phrase“no predecessor”is therefore not a third barren category; it is the defining dynamical property that distinguishes primitive from inherited barrenness.

If a barren y has a predecessor x, discrepancy conservation gives D k ( x ) = D k ( y ) ≠ 0 ; hence x is barren. Thus, for barren integers, “no barren predecessor” and “no positive predecessor” are equivalent without circularity.

The image of R_k is the shifted Beatty-type sequence

Let N k = ℕ \ A k . These are exactly the positive integers with no positive R_k-predecessor. The floor-function notation and standard integer-part identities used in the enumeration below follow the classical treatment in [21]. For y ≥ 1, the number of image values not exceeding y is

because ⌊ α k n + 1 / 2 ⌋ ≤ y is equivalent to n < ( y + 1 / 2 ) / α k , and the boundary cannot be integral by irrationality. Hence the complement counting function is

Theorem 8.1 (closed form for no-predecessor integers).

Let q k ( m ) be the m-thpositive integer omitted from A k . Then

Proof

Put γ = α k / ( α k − 1 ) and δ = − 1 / [ 2 ( α k − 1 ) ] . For q = ⌊ γ m + δ ⌋ ,the inequalities q ≤ γ m + δ < q + 1 are equivalent,after multiplication by α k − 1 ,to

Rearranging these inequalities yields N k ( q ) = m and N k ( q − 1 ) = m − 1 . Henceq is the least positive integer y for which N k ( y ) = m ,i.e. the m-th omitted integer. Irrationality excludes equality at the floor boundaries. □

Corollary 8.2.

Every primitive barren generator belongs to { q k ( m ) : m ≥ 1 } .

The term “Beatty-type” is used deliberately: (22) and (25) are inhomogeneous irrational floor sequences, but no claim of a classical homogeneous Beatty complementary pair is required.

For m ≥ 1 set y m , 0 = q k ( m ) , and recursively

Define

Theorem 9.1 (primitive barren generator classification).

The complete primitive barren generator set is

(29)

Proof

By Theorem 8.1,every q k ( m ) has no positive predecessor. By Corollary4.3, I k ( m ) = 1 is exactly equivalent to D k ( q k ( m ) ) ≠ 0 ,i.e. barrenness. Thus(29) selects precisely the barren integers with no predecessor. □

The index set { m : I k ( m ) = 1 } is determined by a finite correction itinerary. This motivates, but does not prove, possible connections with mechanical words or other symbolic-dynamical codings.

Theorem 10.1 (Barren Orbit Exhaustion Theorem).

Every barren integer belongs to a unique forward orbit generated by a primitive barren integer given by Equation (30):

(30)

Proof

Let n ∈ ℬ k . If n has no predecessor,then

.Otherwiseits unique predecessor n − 1 exists and is barren by discrepancy conservation. By Lemma 7.2, n = R k ( n − 1 ) > n − 1 ,so every backward step strictly decreases the positive integer. The process therefore terminates at a barren integer p with no predecessor,and n = R k j ( p ) for some j ≥ 0 . If two primitive forward orbits intersected,uniqueness of predecessors (Lemma 7.1) wouldforcetheir backward chains to coincide all the way to the same primitive root. Hence the union is disjoint. □

Corollary 10.2 (true parent-number set).

The genuine k-nacci parent-number set is

All numerical examples below were recomputed from the defining polynomial and the exact integer formulas (6), (7), and (16). High-precision arithmetic is used only to evaluate α_k safely away from rounding boundaries; the classification itself is integer-valued. The terminology Tribonacci, Tetranacci, and Pentanacci follows the standard m-step hierarchy used in recent recurrence literature [10][13][15]-[18].

For k = 3, the no-predecessor sequence begins

The first primitive barren generators include 8 and 16, with D 3 ( 8 ) = + 1 and D 3 ( 16 ) = − 1 . The forward orbit of 8 is

Direct verification uses R 3 ( n ) = ⌊ α 3 n + 1 / 2 ⌋ : R 3 ( 8 ) = 15 , R 3 ( 15 ) = 28 , R 3 ( 28 ) = 52 , and R 3 ( 52 ) = 96 . Thus 15 is barren but not primitive. For k = 4 the first primitive barren values are 11, 16, 24, 36, 45, …; for k=5 they begin 23, 27, 32, 34, 36, …

Identity (8) gives

Hence

The correction c k ( n ) vanishes exactly when n α k − k < 1 / 2 , equivalently

The first nonzero correction therefore occurs at the least integer n satisfying n > α k k / 2 ; equality cannot occur because α k is irrational. This threshold concerns the correction map only and does not by itself determine the first barren integer, because admissibility depends on the complete k-step correction sum.

The no-predecessor candidates form the explicit inhomogeneous floor sequence (25), while actual primitive barrenness is selected by the finite discrepancy criterion (28). This places the construction near classical work on complementary floor sequences and codings of irrational maps [22]-[26]. The present results require only the explicit counting and complement formulas proved above. They do not require the stronger assertion that the primitive-generator indices form a classical Beatty sequence or a Sturmian word.

Possible substitution, morphic, or mechanical-word descriptions of {m:I_k(m) = 1} are therefore proposed directions for further study. They are logically downstream of, and are not used in, the classification, recovery, conservation, or exhaustion theorems.

Three notions are now sharply separated. A no-predecessor integer is a gap in the image of R_k. A primitive barren integer is a no-predecessor integer with nonzero k-step discrepancy. An inherited barren integer is a forward image of a primitive barren generator. Discrepancy conservation explains why barrenness persists along the forward orbit.

The correction representation also yields a practical finite classification procedure. Parent-number admissibility is determined intrinsically from k rounded iterations, while Theorem 8.1 restricts the search for primitive generators to an explicitly enumerated no-predecessor sequence. The symbolic-dynamical remarks concern possible additional organization of the already-defined indicator sequence and do not enter any proof of the preceding theorems.

For every fixed k≥ 3, the dominant k-nacci root converts nearest-integer multiplication into an exact almost-doubling map with an integer correction. This identity yields a finite discrepancy formula, k-step recovery, discrepancy conservation, inheritance of barrenness, explicit enumeration of all integers without predecessors, classification of primitive barren generators, and a decomposition of the barren set into disjoint forward orbits. These are analytic statements for arbitrary fixed k≥ 3. The numerical lists for k = 3 , ⋯ , 7 are illustrations. Possible Beatty-complementarity refinements and symbolic-dynamical descriptions are not assumed by the proofs and remain separate structural questions.

1) Fix k ≥ 3 and compute α k as the dominant real root of x k − x k − 1 − ⋯ − x − 1 = 0 to sufficient precision.

2) Set β k = 2 − α k = α k − k .

3) Enumerate no-predecessor candidates q k ( m ) from (25).

4) For each candidate n = q k ( m ) , set x 0 = n and iterate x j + 1 = 2 x j − ⌊ β k x j + 1 / 2 ⌋ for j = 0 , ⋯ , k − 1 .

5) Compute

6) Record n as primitive barren if D k ( n ) ≠ 0 . Its entire inherited branch is { R k j ( n ) : j ≥ 0 } .

7) Numerical precision must be increased whenever β k x j + 1 / 2 lies close to an integer. Lemma 2.2 guarantees that exact ties do not occur.