Let b and b1 be distinct positive integers larger than 1, and let Ab(n) and Ab1(n) be the number of palindromes in bases b and b1 that are less than or equal to n, respectively. In this article, we finish the comparative study of the functions Ab(n) and Ab1(n). As a result, we present the full picture of the asymptotic behavior of their difference.
Citation: Phakhinkon Napp Phunphayap, Prapanpong Pongsriiam. A complete comparison for the number of palindromes in different bases[J]. AIMS Mathematics, 2023, 8(4): 9924-9932. doi: 10.3934/math.2023502
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Let b and b1 be distinct positive integers larger than 1, and let Ab(n) and Ab1(n) be the number of palindromes in bases b and b1 that are less than or equal to n, respectively. In this article, we finish the comparative study of the functions Ab(n) and Ab1(n). As a result, we present the full picture of the asymptotic behavior of their difference.
Let b≥2 and n≥1 be integers. We call n a palindrome in base b (or b-adic palindrome) if the b-adic expansion of n=(akak−1⋯a0)b with ak≠0 has the symmetric property ak−i=ai for 0≤i≤k. As usual, if we write a number without specifying the base, then it is always in base 10, and if we write n=(akak−1⋯a0)b, then it means that n=∑ki=0aibi, ak≠0, and 0≤ai<b for all i=0,1,…,k. Throughout this article, we let Ab(n) be the number of b-adic palindromes not exceeding n.
Previously, we [1] obtained an extremal order of Ab(n) and proved that if b>b1≥2 are integers, then
lim supn→∞(Ab(n)−Ab1(n))=+∞ and lim infn→∞(Ab(n)−Ab1(n))<0. |
In addition, if logblogb1 is irrational, then
lim infn→∞(Ab(n)−Ab1(n))=−∞. | (1.1) |
Therefore, it is interesting to determine the value of the left-hand side of (1.1) when b is a rational power of b1. In this article, we show in Theorem 3.1 that if logb/logb1 is an integer, then the left-hand side of (1.1) is −1, and we obtain in Theorem 3.2 that if logb/logb1 is rational but not integral, then the left-hand side of (1.1) is −∞. We also propose some possible research problems at the end of this article. We remark that the study on the ratio Ab(n)/Ab1(n) may be interesting too, but we were previously interested in the sign changes of Ab(n)−Ab1(n), and so we focus only on the difference not the ratio. Nevertheless, since Ab(n)−Ab1(n) has an infinite number of sign changes, if the limit of Ab(n)/Ab1(n) as n→∞ exists, then it must be one.
Perhaps, one of the popular problems in palindromes is to determine whether or not there are infinitely many palindromic primes. Although this problem is still open, Banks, Hart and Sakata [2] showed that almost all palindromes in any fixed base b≥2 are composite. Banks and Shparlinski [3] also obtained results on prime divisors of palindromes, and there are many other interesting articles on palindromes too. We refer the reader to Banks [4], Cilleruelo, Luca and Baxter [5], and Rajasekaran, Shallit and Smith [6] for additive properties of palindromes, Bašić [7,8], Di Scala and Sombra [9], Goins [10], Luca and Togbé [11] for the study of palindromes in different bases, Cilleruelo, Luca and Tesoro [12] for palindromes in linear recurrence sequences, Harminc and Soták [13] for b-adic palindromes in arithmetic progressions, and Pongsriiam [14] for the longest arithmetic progressions of palindromes.
In this section, we provide some results which are needed in the proof of the main theorems. Recall that for a real number x, ⌊x⌋ is the largest integer less than or equal to x, ⌈x⌉ is the smallest integer greater than or equal to x, and {x} is the fractional part of x given by {x}=x−⌊x⌋. It is also convenient to define Cb(n) as follows.
Definition 2.1. Let b≥2 and n=(akak−1⋯a1a0)b be positive integers. We define Cb(n)=(ckck−1⋯c1c0)b to be the b-adic palindrome satisfying ci=ai for k−⌊k/2⌋≤i≤k. In other words, Cb(n) is the b-adic palindrome having k+1 digits whose first half digits are the same as those of n in its b-adic expansion, that is, Cb(n)=(akak−1⋯ak−⌊k2⌋⋯ak−1ak)b.
In the following lemma, if P is a mathematical statement, then the Iverson notation [P] is defined by [P]=1 if P holds, and [P]=0 otherwise. Then the formula for Ab(n) is as follows.
Lemma 2.1. [15] Let b≥2, n≥1, and n=(akak−1⋯a1a0)b be integers. Then the number of b-adic palindromes less than or equal to n is given by
Ab(n)=b⌈k2⌉+∑0≤i≤⌊k2⌋ak−ib⌊k2⌋−i+[n≥Cb(n)]−2. |
Lemma 2.2. Let a,r,s≥2 be integers and (r,s)=1. If ars is an integer, then there exists an integer c≥2 such that a=cs.
Proof. Suppose ars=m is an integer. Then ar=ms, and so a and m have the same set of prime divisors. Let a=∏ki=1paii and m=∏ki=1pmii. Then air=mis for all i. Since s∣air and (s,r)=1, s∣ai for all i. Let c=∏ki=1pai/si. Then c is an integer, c≥2, and a=cs. So the proof is complete.
Theorem 3.1. Let b>b1≥2 and ℓ≥2 be integers. Suppose that b=bℓ1. Then, the following statements hold.
(i) Ab(n)−Ab1(n)≥−1 for all n∈N.
(ii) Ab(n)−Ab1(n)=−1 for infinitely many n∈N.
(iii) lim infn→∞(Ab(n)−Ab1(n))=−1.
Proof. We first prove (i). Let n≥1 and write
n=(akak−1⋯a0)b1=(crcr−1⋯c0)b. |
Since br≤crbr≤n<br+1, we see that r=⌊lognlogb⌋. Similarly, we have k=⌊lognlogb1⌋, and so r=⌊k/ℓ⌋. By the uniqueness of the b-adic and b1-adic representations, we can write c0, c1, c2,…, cr in terms of b1 and the aj as follows:
Considering n modulo b, we obtain
c0≡a0+a1b1+a2b21+⋯+aℓ−1bℓ−11(modb), |
and both sides of the congruence are nonnegative integers less than b, and so they are equal. Similarly, after reducing n modulo b2,b3,…,br+1, respectively, we obtain c1,c2,…cr. Thus
cj=ℓ−1∑i=0ajℓ+ibi1 for every j=0,1,2,…,r, |
where am=0 if m>k. By Lemma 2.1, we have
Ab1(n)=b⌈k2⌉1+∑0≤i≤⌊k2⌋ak−ib⌊k2⌋−i1+[n≥Cb1(n)]−2, | (3.1) |
Ab(n)=b⌈r2⌉+∑0≤j≤⌊r2⌋cr−jb⌊r2⌋−j+[n≥Cb(n)]−2=bℓ⌈⌊kℓ⌋2⌉1+∑0≤j≤⌊k2ℓ⌋(ℓ−1∑i=0a(r−j)ℓ+ibi1)bℓ(⌊k2ℓ⌋−j)1+[n≥Cb(n)]−2. | (3.2) |
It is useful to recall that if k∈Z and x∈R, then ⌊k+x⌋=k+⌊x⌋, and if k∈N and x∈R, then ⌊⌊x⌋k⌋=⌊xk⌋. We also let s=kmodℓ be the least nonnegative residue of k modulo ℓ, that is, k≡s(modℓ) and 0≤s<ℓ. Then, from (3.1) and (3.2), we obtain
Ab1(n)={bk21+∑0≤i≤k2ak−ibk2−i1+[n≥Cb1(n)]−2,if k is even;bk+121+∑0≤i≤k−12ak−ibk−12−i1+[n≥Cb1(n)]−2,if k is odd, | (3.3) |
Ab(n)={bk−s21+∑0≤j≤k−s2ℓℓ−1∑i=0ak−s−ℓj+ibk−s2−ℓj+i1+[n≥Cb(n)]−2,if ⌊kℓ⌋ is even;bk−s+ℓ21+∑0≤j≤k−s−ℓ2ℓℓ−1∑i=0ak−s−ℓj+ibk−s−ℓ2−ℓj+i1+[n≥Cb(n)]−2,if ⌊kℓ⌋ is odd. | (3.4) |
Next, we will reduce the double sum in (3.4) into a sum. We see that if ⌊kℓ⌋ is even, then −ℓj+i runs through the integers from −k−s2 to ℓ−1 exactly once as j runs through 0,1,2,…,k−s2ℓ and i runs through 0 to ℓ−1. Similarly, if ⌊kℓ⌋ is odd, then −ℓj+i ranges over the integers from −k−s−ℓ2 to ℓ−1 exactly once as j ranges over 0,1,2,…,k−s−ℓ2ℓ and i ranges over 0 to ℓ−1. So if ⌊kℓ⌋ is even, the first double sum in (3.4) reduces to
∑−k−s2≤i≤ℓ−1ak−s+ibk−s2+i1. |
We replace the index i by i−k−s2 and recall that s≤ℓ−1, ak−s2+i=0 if i>k+s2, and ℓ−1+k−s2≥k+s2. So the first double sum in (3.4) further reduces to
∑0≤i≤k+s2ak−s2+ibi1. |
Similarly, if ⌊kℓ⌋ is odd, then the second double sum in (3.4) reduces to
∑−k−s−ℓ2≤i≤ℓ−1ak−s+ibk−s−ℓ2+i1=∑0≤i≤k+s−ℓ2ak−s+ℓ2+ibi1. |
From (3.4) and the above calculation, we obtain
Ab(n)={bk−s21+∑0≤i≤k+s2ak−s2+ibi1+[n≥Cb(n)]−2,if ⌊kℓ⌋ is even;bk−s+ℓ21+∑0≤i≤k+s−ℓ2ak−s+ℓ2+ibi1+[n≥Cb(n)]−2,if ⌊kℓ⌋ is odd. | (3.5) |
Next, we divide the proof into four cases according to the parity of k and ⌊kℓ⌋.
Case 1. Assume that k and ⌊kℓ⌋ are even. Then, s is even and
∑0≤i≤k+s2ak−s2+ibi1≥bs21∑s2≤i≤k+s2ak−s2+ibi−s21=bs21∑0≤i≤k2ak−ibk2−i1. |
By (3.3), (3.5) and the above inequality, we obtain that Ab(n)−Ab1(n) is at least
bk−s21−bk21+(bs21−1)∑0≤i≤k2ak−ibk2−i1−1≥bk−s21−bk21+akbk21(bs21−1)−1=bk21(akbs21+b−s21−ak−1)−1. |
Since the function x↦akx+x−1 is increasing on [1,∞), the number in the above parenthesis is nonnegative, and so Ab(n)−Ab1(n)≥−1.
Case 2. Assume that k is odd and ⌊kℓ⌋ is even. Then s is odd. Similar to Case 1, we obtain
∑0≤i≤k+s2ak−s2+ibi1≥bs+121∑s+12≤i≤k+s2ak−s2+ibi−s+121≥bs+121∑0≤i≤k−12ak−ibk−12−i1. |
By (3.3), (3.5) and the above inequality, we see that Ab(n)−Ab1(n) is at least
bk−s21−bk+121+akbk−121(bs+121−1)−1=bk21(akbs21+b−s21−akb−121−b121)−1. |
Since x↦akx+x−1 is increasing on [1,∞) and bs21≥b121≥1, we have akbs21+b−s21≥akb121+b−121, and
Ab(n)−Ab1(n)≥bk21(akb121+b−121−akb−121−b121)−1=bk21(ak−1)(b121−b−121)−1≥−1. |
Case 3. Assume that k is even and ⌊kℓ⌋ is odd. Then ℓ≡k−s≡s(mod2). So ℓ−s2 is an integer and ℓ−s2>0. So ℓ−s2≥1. Considering the first sum in (3.3) and changing the index from i to k+s−ℓ2−i, we see that
∑0≤i≤k2ak−ibk2−i1=∑−ℓ−s2≤i≤k+s−ℓ2ak−s+ℓ2+ibi+ℓ−s21=bℓ−s21∑0≤i≤k+s−ℓ2ak−s+ℓ2+ibi1+∑−ℓ−s2≤i<0ak−s+ℓ2+ibi+ℓ−s21=bℓ−s21∑0≤i≤k+s−ℓ2ak−s+ℓ2+ibi1+∑0≤i<ℓ−s2ak2+ibi1≤bℓ−s21∑0≤i≤k+s−ℓ2ak−s+ℓ2+ibi1+(bℓ−s21−1). |
By (3.3), (3.5) and the above inequality, we obtain
Ab(n)−Ab1(n)≥bk−s+ℓ21−bk21+(1−bℓ−s21)∑0≤i≤k+s−ℓ2ak−s+ℓ2+ibi1−(bℓ−s21−1)−1. | (3.6) |
The sum on the right-hand side of (3.6) is less than or equal to bk+s−ℓ2+11−1 and 1−bℓ−s21 is negative. Therefore, Ab(n)−Ab1(n) is at least
bk−s+ℓ21−bk21+(1−bℓ−s21)(bk+s−ℓ2+11−1)+(1−bℓ−s21)−1=bk21(bℓ−s21+b1−ℓ−s21−b1−1)−1. |
Since the function x↦bx1+b1−x1 is increasing on [1,∞) and ℓ−s2≥1, the number in the above parenthesis is nonnegative, and so Ab(n)−Ab1(n)≥−1.
Case 4. Assume that k and ⌊kℓ⌋ are odd. Changing the index from i to k−12−ℓ−s−12−i, the second sum in (3.3) is
∑0≤i≤k−12ak−ibk−12−i1=∑−ℓ−s−12≤i≤k+s−ℓ2ak−s+ℓ2+ibi+ℓ−s−121=bℓ−s−121∑0≤i≤k+s−ℓ2ak−s+ℓ2+ibi1+∑−ℓ−s−12≤i<0ak−s+ℓ2+ibi+ℓ−s−121≤bℓ−s−121∑0≤i≤k+s−ℓ2ak−s+ℓ2+ibi1+(bℓ−s−121−1). |
By (3.3), (3.5) and the above inequality, we obtain that Ab(n)−Ab1(n) is at least
bk−s+ℓ21−bk+121+(1−bℓ−s−121)∑0≤i≤k+s−ℓ2ak−s+ℓ2+ibi1−(bℓ−s−121−1)−1≥bk−s+ℓ21−bk+121+(1−bℓ−s−121)(bk+s−ℓ2+11−1)+(1−bℓ−s−121)−1=bk+121(bℓ−s−121+b−ℓ−s−121−2)−1. |
Since x+x−1≥2 for all x>0, we see that Ab(n)−Ab1(n)≥−1.
In any case, we obtain Ab(n)−Ab1(n)≥−1, as desired. This proves (i).
Next, we prove (ii). For each k∈N, let n=nk=b2ℓk−11+1. Then n=bℓ−11b2k−1+1. By Lemma 2.1, we obtain
Ab1(n)=b⌈2ℓk−12⌉1+b⌊2ℓk−12⌋1−1=bkℓ1+bkℓ−11−1, |
Ab(n)=b⌈2k−12⌉+bℓ−11b⌊2k−12⌋−2=bk+bℓ−11bk−1−2=bkℓ1+bkℓ−11−2. |
Therefore Ab(n)−Ab1(n)=−1. Since n=nk and k is arbitrary, this shows that there are infinitely many n∈N such that Ab(n)−Ab1(n)=−1. This proves (ii). Then (iii) follows immediately from (i) and (ii). So the proof is complete.
Since we have already got the answers to the cases when logb/logb1 is irrational and when it is integral, it remains to consider the case when logb/logb1 is a rational number and is not an integer.
Theorem 3.2. Let b>b1≥2 be integers and b=brs1 where r,s∈N, r>s>1, and (r,s)=1. Then
lim infn→∞(Ab(n)−Ab1(n))=−∞. |
Proof. To prove this theorem, it is enough to find a sequence (nk)k≥1 of positive integers such that nk→+∞ and Ab1(nk)−Ab(nk)→+∞ as k→+∞. We divide the calculation into three cases according to parity of r and s, and adjust the exponents so that Ab1(n) is large and Ab(n) is small.
Case 1. Assume that s is even. Let k be a positive integer and n=nk=bs(2k+1)+1. Since (r,s)=1, s is even, and br1=bs, we see that r is odd and n=br(2k+1)1+1. By Lemma 2.1, we obtain Ab(n)=2bsk+s2−1 and
Ab1(n)=b⌈r(2k+1)2⌉1+b⌊r(2k+1)2⌋1−1=bkr+r+121+bkr+r−121−1=b121bsk+s2+b−121bsk+s2−1. |
Since x+x−1>2 for all x>1, we see that b121+b−121−2 is a positive constant. Therefore,
Ab1(n)−Ab(n)=bsk+s2(b121+b−121−2)→+∞ as k→+∞. |
Case 2. Assume that s and r are odd. Since brs1=b is an integer, we obtain by Lemma 2.2 an integer b2≥2 such that b1=bs2, and so b=br2. Since (2s,r)=1, there exists a negative integer x such that 2sx≡1(modr). Since the proof of Case 1 is finished, we will define a new sequence (nk)k≥1 using the same notation. Let k be a positive integer and let n=nk=b2x(1−s)+2rk+11+1. For convenience, let ℓ=x(1−s)+rk. So ℓ∈N, ℓ≥5, and n=b2ℓ+11+1. By Lemma 2.1 and the fact that b1=bs2, we obtain
Ab1(n)=bℓ+11+bℓ1−1=bsℓ+s2+bsℓ2−1. | (3.7) |
Next, let y=s(2ℓ+1)−1r. By the definition of ℓ and x, we have
s(2ℓ+1)=2sx(1−s)+2srk+s≡1(modr). |
Therefore y is a positive integer. Since s(2ℓ+1)−1 is even and r is odd, we see that y is even. To calculate Ab(n), we write
n=b2ℓ+11+1=bs(2ℓ+1)2+1=bry+12+1=b2⋅by+1. |
So b2 is the leading digit in the b-adic representation of n. Since y is even and b=br2, we obtain by Lemma 2.1 that
Ab(n)=bry22(1+b2)−2=bsℓ+s−122(1+b2)−2. | (3.8) |
From (3.7) and (3.8), we obtain
Ab1(n)−Ab(n)=bsℓ+s22B+1, where B=bs22+b−s22−b−122−b122. |
Since the function x↦x+x−1 is strictly increasing on (1,∞) and bs22>b122>1, we see that B is a positive constant. Therefore, Ab1(n)−Ab(n)=bsℓ+s22B+1→+∞ as ℓ→+∞. Since ℓ=x(1−s)+rk≥k, we see that Ab1(n)−Ab(n)→+∞ as k→+∞. So the proof of Case 2 is complete.
Case 3. Assume that s is odd and r is even. This case is similar to Case 2 and we only need to modify some calculations. Again, we have b1=bs2 and b=br2 for some integer b2≥2. Since (2r,2s)=2 and 2∣1−s, there exists a positive integer k such that
2sk≡1−s(mod2r). | (3.9) |
In fact, there are infinitely many positive integers k satisfying (3.9), so we can choose k to be arbitrarily large. Let n=nk=b2k+11+1. By Lemma 2.1 and the fact that b1=bs2, we obtain
Ab1(n)=bk+11+bk1−1=bsk+s2+bsk2−1. | (3.10) |
Next, let y=s(2k+1)−1r. By (3.9), we have s(2k+1)=2sk+s≡1(mod2r). Therefore 2r divides s(2k+1)−1, and thus y=2(s(2k+1)−12r) is an even positive integer. To calculate Ab(n), we write
n=b2k+11+1=bs(2k+1)2+1=bry+12+1=b2⋅by+1. |
By Lemma 2.1, we obtain
Ab(n)=bry22(1+b2)−2=bks+s−122(1+b2)−2. | (3.11) |
From (3.10), (3.11) and a similar reason as in Case 2, we obtain
Ab1(n)−Ab(n)=bks+s22(bs22+b−s22−b−122−b122)+1→+∞ as k→+∞. |
This completes the proof.
Let us record all related results that we obtained as follows.
Summary of the results:
Let b>b1≥2 be integers. Then, the following statements hold.
(i) If logblogb1 is not an integer, then,
lim supn→∞(Ab(n)−Ab1(n))=+∞ and lim infn→∞(Ab(n)−Ab1(n))=−∞. |
(ii) If logblogb1 is an integer, then,
lim supn→∞(Ab(n)−Ab1(n))=+∞ and lim infn→∞(Ab(n)−Ab1(n))=−1. |
(iii) Ab(n)−Ab1(n) has an infinite number of sign changes as n→∞.
(iv) If sb and sb1 are the sums of the reciprocal of palindromes in bases b and b1, respectively, then sb and sb1 are finite and sb>sb1.
The statements (i)–(iii) come directly from [1, Theorems 11 and 12], and Theorems 3.1 and 3.2 of this article. The finiteness of sb and sb1 in (iv) is given by Shallit and Klauser [16] and the inequality sb>sb1 is obtained from [17, Theorem 3].
Now that for all cases we have obtained results for the comparison between the number of palindromes in two bases, it is natural to extend this to more than two bases. So let k≥2 and b1>b2>⋯>bk≥2 be integers. Let c1,c2,…,ck be any permutation of b1,b2,…,bk.
Question 4.1. Does the inequality Ac1(n)<Ac2(n)<⋯<Ack(n) hold for infinitely many n∈N? We conjecture that if logbi/logbj is irrational for every distinct i,j=1,2,…,k, then the inequality holds for infinitely many n∈N. What are the answers when one of the following situations occur?
(i) logbilogbi+1 is integral for every i=1,2,…,k−1;
(ii) logbilogbj is rational but not integral for each distinct i, j;
(iii) The set {logbilogbj | 1≤i<j≤k} contains both an irrational number and a rational number.
Prapanpong Pongsriiam's research project is funded jointly by the Faculty of Science Silpakorn University and the National Research Council of Thailand (NRCT), grant number NRCT5-RSA63021-02. He is also supported by the Tosio Kato Fellowship given by the Mathematical Society of Japan during his visit at Nagoya University in July 2022 to July 2023.
The authors declare no conflicts of interest regarding the publication of this article.\newpage
[1] |
P. Phunphayap, P. Pongsriiam, Extremal orders and races between palindromes in different bases, AIMS Math., 7 (2022), 2237–2254. https://doi.org/10.3934/math.2022127 doi: 10.3934/math.2022127
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[2] |
W. D. Banks, D. N. Hart, M. Sakata, Almost all palindromes are composite, Math. Res. Lett., 11 (2004), 853–868. https://doi.org/10.4310/MRL.2004.v11.n6.a10 doi: 10.4310/MRL.2004.v11.n6.a10
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[3] |
W. D. Banks, I. E. Shparlinski, Prime divisors of palindromes, Period. Math. Hung., 51 (2005), 1–10. https://doi.org/10.1007/s10998-005-0016-6 doi: 10.1007/s10998-005-0016-6
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[4] | W. D. Banks, Every natural number is the sum of forty-nine palindromes, Integers, 16 (2016), 1–9. |
[5] |
J. Cilleruelo, F. Luca, L. Baxter, Every positive integer is a sum of three palindromes, Math. Comput., 87 (2018), 3023–3055. https://doi.org/10.1090/mcom/3221 doi: 10.1090/mcom/3221
![]() |
[6] | A. Rajasekaran, J. Shallit, T. Smith, Sums of palindromes: an approach via automata, In: 35th symposium on theoretical aspects of computer science (STACS 2018), 54: 1–54: 12. https://doi.org/10.4230/LIPIcs.STACS.2018.54 |
[7] |
B. Bašić, On d-digit palindromes in different bases: the number of bases is unbounded, Int. J. Number Theory, 8 (2012), 1387–1390. https://doi.org/10.1142/S1793042112500819 doi: 10.1142/S1793042112500819
![]() |
[8] |
B. Bašić, On " very palindromic" sequences, J. Korean Math. Soc., 52 (2015), 765–780. https://doi.org/10.4134/jkms.2015.52.4.765 doi: 10.4134/jkms.2015.52.4.765
![]() |
[9] | A. J. Di Scala, M. Sombra, Intrinsic palindromes, Fibonacci Quart., 42 (2004), 76–81. |
[10] |
E. H. Goins, Palindromes in different bases: a conjecture of J. Ernest Wilkins, Integers, 9 (2009), 725–734. https://doi.org/10.1515/INTEG.2009.059 doi: 10.1515/INTEG.2009.059
![]() |
[11] |
F. Luca, A. Togbé, On binary palindromes of the form 10n±1, C. R. Math., 346 (2008), 487–489. https://doi.org/10.1016/j.crma.2008.03.015 doi: 10.1016/j.crma.2008.03.015
![]() |
[12] |
J. Cilleruelo, R. Tesoro, F. Luca, Palindromes in linear recurrence sequences, Monatsh. Math., 171 (2013), 433–442. https://doi.org/10.1007/s00605-013-0477-2 doi: 10.1007/s00605-013-0477-2
![]() |
[13] | M. Harminc, R. Soták, Palindromic numbers in arithmetic progressions, Fibonacci Quart., 36 (1998), 259–261. |
[14] |
P. Pongsriiam, Longest arithmetic progressions of palindromes, J. Number Theory, 222 (2021), 362–375. https://doi.org/10.1016/j.jnt.2020.10.018 doi: 10.1016/j.jnt.2020.10.018
![]() |
[15] | P. Pongsriiam, K. Subwattanachai, Exact formulas for the number of palindromes up to a given positive integer, Int. J. Math. Comput. Sci., 14 (2019), 27–46. |
[16] | H. Klauser, J. Shallit, Sum of base-b palindrome reciprocals, Fibonacci Quart., 19 (1981), 469. |
[17] | P. Phunphayap, P. Pongsriiam, Reciprocal sum of palindromes, J. Integer Seq., 22 (2019), 1–13. |