On Diophantine Equations Related to Narayana’s Cows Sequence and Double Factorials or Repdigits

: In this paper, we determine all the Narayana’s cows numbers that are factorials or double factorials. We also show that 88 is the only repdigit (i


Introduction
In 1356, Indian mathematician Narayana Pandita proposed the problem of a herd of cows and calves in his famous book titled Ganita Kaumudi [1].It is a problem similar to Fibonacci's rabbit problem.One can see that the number of cows in each year forms a sequence with the first few terms: 1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, . . . .This sequence is named Narayana's cows sequence.It is also called the Fibonacci-Narayana sequence or Narayana sequence.However, there is another sequence that is also named the Narayana sequence (see [2]).The Narayana's cows sequence can be written as the following recurrence: In the literature, there are several results dealing with Diophantine equations involving factorials, repdigits, and recurrence sequences.In 1999, Luca [3] proved that F 12 = 2! 2 3! 2 = 3!4! and L 3 = (2!) 2 are the largest Fibonacci and Lucas numbers that can be represented as the products of factorials, respectively.In 2006, Luca and Stȃnicȃ [4] found all products of Fibonacci numbers that are products of factorials.Those results can be proven by applying the primitive divisor theorem.Meanwhile, a high-order recurrence version of the primitive divisor theorem seems to be out of reach.By characterizing the 2-adic valuation of tribonacci numbers, Marques and Lengyel [5] determined all the factorials in a tribonacci sequence.Using the same method, Irmak [6] identified the factorials in Perrin or Padovan sequences, and Guadalupe [7] found all the factorials in Narayana's cows numbers.
A repdigit is a positive integer with only one distinct digit in its decimal expansion.It has the form a(10 m − 1)/9 for some m ≥ 1 and 1 ≤ a ≤ 9.In 2000, Luca [8] showed that the largest repdigits in Fibonacci and Lucas sequences are F 10 = 55 and L 5 = 11.Since then, this result has been generalized and extended in various directions.For example, Faye and Luca [9] proved that P 3 = 5 and Q 2 = 6 are the largest repdigits in Pell and Pell-Lucas sequences, respectively.Bravo et al. [10] obtained all base b repdigits that are the sum of two Narayana numbers.Considering the consecutive product of the recurrence sequence, Marques and Togbé [11] showed that the product of consecutive Fibonacci numbers can never be a repdigit greater than 10.Irmak and Togbé [12] verified that the largest repdigit appearing as the product of consecutive Lucas numbers is 77.Rayaguru and Panda [13] studied repdigits as products of consecutive balancing and Lucas-balancing numbers.Bravo et al. [14] proved that 44 is the largest repdigit in the product of consecutive tribonacci numbers.Recently, Rihane and Togbé [15] dealt with repdigits that can be written as the products of consecutive Padovan or/and Perrin numbers.
Motivated by the results of [3][4][5][6][8][9][10][11][12][13][14][15], it is natural to ask what will happen if we consider Narayana's cows numbers.We denote the double factorial n!! as the product of the natural numbers less than or equal to n that have the same parity as n.In this paper, we investigate double factorials and repdigits in Narayana's cows sequence.We mainly solve the Diophantine equation G n = m!! and study repdigits that can be written as the product of consecutive Narayana's cows numbers.More precisely, we prove the following results.
From the above theorem, we have the following corollary.
Therefore, we obtain the following lower bound.
To prove Theorem 1, we need the following lemmas.
Lemma 4 (Guadalupe [7]).For n ≥ 1, we have where v p (r) is the exponent of prime p in the factorization of r.
Lemma 5 (Grossman,Luca [17]).For any prime p and positive integer n ≥ p, we have From the above lemma, we have a similar inequality for double factorials.
Lemma 6.For any odd prime p and positive integer n ≥ p, we have Proof.If n is even, then Thus,

Linear Forms in Logarithms
For any non-zero algebraic number γ of degree d over Q, whose minimal polynomial over Z is a d ∏ j=1 (X − γ (j) ), we denote by the usual absolute logarithmic height of γ.
To prove Theorem 2, we use lower bounds for linear forms in logarithms to bound the subscript n appearing in Equation ( 1).We quote the following result.
Lemma 7 (Bugeaud et al. [18], Matveev [19]).Let γ 1 , . . ., γ s be real algebraic numbers and let b 1 , . . ., b s be the non-zero rational integer numbers.Let D be the degree of the number field Q(γ 1 , . . ., γ s ) over Q and let A j be a positive real number satisfying The next step is to reduce the bound of n, which is generally too large.To this end, we present a variant of the reduction method of Baker and Davenport, which was introduced by de Weger [20].
Then, the solutions of ( 8) and (9) satisfy To apply Lemma 7 the following lemma will help us later to obtain inequality similar to (8).

Proof of Theorem 1
It is easy to check that the only solutions are the ones listed in Theorem 1 if m ≤ 3. Thus, we shall suppose that m ≥ 4. By using Lemma 6 (for p = 3) together with Lemma 4, we derive that Again, from Lemma 1, Thus, n < 2.65m log( m 2 ) + 2. Substituting this in (10), we arrive at This inequality yields m ≤ 503 and then n < 2.65 • 503 • log( 67 2 ) + 2 = 7369.805• • • .Now, we use a simple routine written in Mathematica that does not return any solution in the range 4 ≤ m ≤ 503 and 1 ≤ n ≤ 7369.The proof is complete.

Proof of Theorem 2 4.1. Absolute Bounds on Variables
In this section, we will use Baker's method and the p-adic valuation to completely prove Theorem 2.
First, we give an upper bound for .
Take the Galois automorphism σ := (αβ) and absolute values on both sides of the resulting equality, and we obtain It is a contradiction.Thus, H = 0. Now, we give estimates to A i for i = 1, 2, 3.By the properties of the absolute logarithmic height, we have 3 log 31 and thus h(γ 1 ) ≤ 2 log 9 + 2 log 31.