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Research article Special Issues

Existence of global solution to 3D density-dependent incompressible Navier-Stokes equations

  • Received: 09 November 2023 Revised: 06 January 2024 Accepted: 15 January 2024 Published: 23 February 2024
  • MSC : 35Q30, 76D05

  • In this article, we are committed to studying the three-dimensional incompressible Navier-Stokes equations, where the viscosity depends on density according to a power law. We investigate the Cauchy problem by constructing an approximation system and bootstrap argument. Finally, we establish the existence of a global strong solution under the conditions of small initial data and the compatibility condition. Meanwhile, the algebraic decay-in-time rates for the solution are also obtained. It is worth pointing out that the degradation of viscosity is allowed.

    Citation: Jianxia He, Ming Li. Existence of global solution to 3D density-dependent incompressible Navier-Stokes equations[J]. AIMS Mathematics, 2024, 9(3): 7728-7750. doi: 10.3934/math.2024375

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  • In this article, we are committed to studying the three-dimensional incompressible Navier-Stokes equations, where the viscosity depends on density according to a power law. We investigate the Cauchy problem by constructing an approximation system and bootstrap argument. Finally, we establish the existence of a global strong solution under the conditions of small initial data and the compatibility condition. Meanwhile, the algebraic decay-in-time rates for the solution are also obtained. It is worth pointing out that the degradation of viscosity is allowed.



    Let b2 and n1 be integers. We call n a palindrome in base b (or b-adic palindrome) if the b-adic expansion of n=(akak1a0)b with ak0 has the symmetric property aki=ai for 0ik. As usual, if we write a number without specifying the base, then it is always in base 10, and if we write n=(akak1a0)b, then it means that n=ki=0aibi, ak0, and 0ai<b for all i=0,1,,k. So, for example, 9=(1001)2=(100)3 is a palindrome in bases 2 and 10 but not in base 3.

    For a long time ago, palindromes were considered only as a part of recreational mathematics, but in recent years, there has been an increasing interest in the importance of palindromes in mathematics [1,2,3,16,25], theoretical computer science [4,15,17,18], and theoretical physics [5,13,21]. For example, in 2016, Banks [6] showed that every positive integer can be written as the sum of at most 49 palindromes in base 10. Two years later Cilleruelo, Luca, and Baxter [11] improved it by showing that if b5 is fixed, then every positive integer is the sum of at most three b-adic palindromes. Then Rajasekaran, Shallit, and Smith [31] completed the study by proving that the theorem of Cilleruelo, Luca, and Baxter [11] also holds when b{3,4}, and if b=2, then we need four summands to write every positive integer as a sum of b-adic palindromes. Nevertheless, we still have many other interesting open problems concerning palindromes. In particular, on Mathematics Stack Exchange, Vepir asked: which number base contains the most palindromic numbers?

    Pongsriiam and Subwattanachai [30] started the investigation by deriving an exact formula for the number of b-adic palindromes not exceeding n, denoted by Ab(n), but the formula is difficult to analyze, and so it does not give an answer to Vepir's question. Then Phunphayap and Pongsriiam [28] calculated the reciprocal sum of all b-adic palindromes implying that if b>b1 and if we use the logarithmic measure, then there are more b-adic palindromes than b1-adic palindromes. Nevertheless, this does not answer Vepir's question according to the counting measure.

    In this article, we obtain extremal orders of Ab(n) and show that Ab(n)Ab1(n) has infinitely many sign changes. We also obtain other related results and use them to solve Vepir's problem.

    For more information on the palindromes, we refer the reader to Banks, Hart, and Sakata [7] and Banks and Shparlinski [8] for some multiplicative properties of palindromes, Bašić [9,10], Di Scala and Sombra [14], Goins [19], Luca and Togbé [26] for the study of palindromes in different bases, Cilleruelo, Luca, and Tesoro [12] for palindromes in linear recurrence sequences, Harminc and Soták [20] for b-adic palindromes in arithmetic progressions, Korec [23] for nonpalindromic numbers having palindromic squares, and Pongsriiam [29] for the longest arithmetic progressions of palindromes.

    In this section, we provide some definitions and lemmas 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}=xx. Furthermore, for a mathematical statement P, the Iverson notation [P] is defined by

    [P]={1,if P holds;0,otherwise.

    In the proof of our main results, we often use Pongsriiam and Subwattanachai's formula [30]. So it is convenient to define Cb(n) as follows.

    Definition 1. Let b2 and n1 be integers, and n=(akak1a1a0)b. Define Cb(n)=(ckck1c1c0)b to be the b-adic palindrome satisfying ci=ai for kk/2ik. In other words, Cb(n) is the b-adic palindrome having k+1 digits, the first half of which are the same as those of n in its b-adic expansion, that is, Cb(n)=(akak1akk2ak1ak)b.

    Example 2. If m=(247853)9 and n=(1327021)8, then C9(m)=(247742)9 and C8(n)=(1327231)8.

    Note that our definition of Cb(n) is slightly different from that in Pongsriiam and Subwattanachai's formula [30]. In addition, while we focus only on positive integers, they [30] also count zero. After a slight modification, their formula is as follows.

    Lemma 3 (Pongsriiam and Subwattanachai [30]). Let b2 and n1 be integers, and n=(akak1a1a0)b. Then the number of b-adic palindromes less than or equal to n is given by

    Ab(n)=bk2+0ik2akibk2i+[nCb(n)]2,

    where [nCb(n)] is the Iverson notation.

    The next lemma gives an upper bound for Ab(n). By Theorem 9, we will see later that the inequality in Lemma 4 is sharp.

    Lemma 4. Let b2 and n1 be integers. Then

    Ab(n)+1(b+1b)n.

    Proof. We write n=(akak1a1a0)b and let

    y=0ik/2akibki  and  z=0ik/2akibi.

    We see that yn, y=bkz, and 1z(b1)i=0bi=b. By Lemma 3, we obtain

    Ab(n)+1bk2+0ik2akibk2i=bk2+bk2z.

    Therefore,

    Ab(n)+1nAb(n)+1ybk2+bk2zbk2z.

    We divide the consideration into two cases according to the parity of k.

    Case 1. k is even. Then

    Ab(n)+1nbk2+bk2zbk2z=z+1z.

    Since the function xx+1x is increasing on [1,] and 1zb, we have

    Ab(n)+1nz+1zb+1b.

    Case 2. k is odd. This case is similar to Case 1. We have

    Ab(n)+1nbk+12+bk12zbk2z=bz+zb=bz+1b/z.

    Since 1zb, we obtain 1b/zb and therefore

    Ab(n)+1nbz+zbb+1b.

    In any case, we obtain the desired result.

    Recall that a sequence (xn)n1 of real numbers is said to be uniformly distributed modulo 1 if for any 0a<b1,

    limx(1xnxa{xn}<b1)=ba.

    A well-known criterion for uniform distribution modulo 1 is as follows [24,page 7].

    Lemma 5. (Weyl's Criterion) The sequence (xn)n1 is uniformly distributed modulo 1 if and only if

    limN1NNn=1e2πihxn=0for all integers h0.

    Lemma 6. Let (xn)n1 be an arithmetic progression of real numbers, d the common difference, dZ{0}, and αR an irrational number. Then (αxn)n1 is uniformly distributed modulo 1.

    Proof. Since α is irrational, d0, and dZ, we have |e2πihαd1|0 for all integers h0. Recall that xn=x1+(n1)d for all n1. So if h is a nonzero integer, then

    |1NNn=1e2πihαxn|=|e2πihαx1||e2πihαNd1|N|e2πihαd1|=|e2πihαNd1|N|e2πihαd1|2N|e2πihαd1|,

    which converges to 0 as N. By Weyl's criterion, the desired result is verified.

    The next result is an important tool for obtaining Theorem 11.

    Lemma 7. Let (xn)n1, d1, and α satisfy the same assumption as in Lemma 6, xnZ for every n, and let c(0,1). Then there are infinitely many integers k1 such that

    {αxk}<c and αxk0 (mod 2). (2.1)

    Proof. Suppose for a contradiction that there is only a finite number of integers k1 such that xk satisfies (2.1). Let k be the largest such an integer if it exists and let k=1 otherwise. Since d1, we see that (xn)n1 is strictly increasing, and so if n>k, then xn does not satisfy (2.1). We will get a contradiction by constructing an integer x=xn satisfying (2.1) and n>k. By Lemma 6, (αxn)n1 is uniformly distributed modulo 1. From this point on, we apply the above fact repeatedly without reference. Let k1,k2,,kd be integers satisfying

    kd>kd1>>k1>k  and  {αxki}<cd+1  for all i=1,2,,d.

    Then αxki1 (mod 2) for all i=1,2,,d. We divide the consideration into two cases.

    Case 1. d is even. Let kd+1 be an integer such that

    kd+1>kd  and  {αxkd+1}>1di=1{αxki}. (2.2)

    Case 1.1 αxkd+11 (mod 2). We see that

    αd+1i=1xki=d+1i=1αxki+d+1i=1{αxki},  and1<d+1i=1{αxki}<1+di=1{αxki}<1+c.

    This implies that

    {αd+1i=1xki}<c  and  αd+1i=1xki=1+d+1i=1αxki1+d+10 (mod 2). (2.3)

    Let A=d+1i=1xki. Then we have

    A>xk  and  Ad+1i=1(x1+(ki1)d)(d+1)x1x1 (mod d).

    Since A>xk and Ax1 (mod d), we see that A=xn for some n>k. But by (2.3), A=xn satisfies (2.1), contradicting the fact that k is the largest integer such that xk satisfies (2.1).

    Case 1.2 αxkd+10 (mod 2). Let M be the maximum value of {αxki} for i=1,2,,d. Since d is even, we must have d2. Then

    M<cd+1  and  0<1d<1d+cdM<1.

    Therefore there are infinitely many tN such that

    1d<{αxt}<1d+cdM. (2.4)

    Then one of the two sets

    X={tN | xt satisfies (2.4) and αxt0 (mod 2)}  andY={tN | xt satisfies (2.4) and αxt1 (mod 2)}

    is infinite, say Y. Then we can choose integers td>td1>>t1>k in Y so that

    1d<{αxti}<1d+cdM  for all i=1,2,,d, (2.5)

    and

    di=1αxti1+1++1d0 (mod 2).

    (If X is infinite, then we can choose such t1,t2,,tdX too.) Let

    B=d+1i=1xki+di=1xti.

    Then

    αB=d+1i=1αxki+di=1αxti+d+1i=1{αxki}+di=1{αxti}.

    By (2.2) and (2.5), we obtain

    2=1+di=11d<d+1i=1{αxki}+di=1{αxti}<{αxkd+1}+di=1({αxki}M)+di=1(1d+cd){αxkd+1}+1+c<2+c,

    which implies that

    {αB}<c  and  αB=2+d+1i=1αxki+di=1αxti0 (mod 2).

    Similar to Case 1.1, we have

    B>xk  and  Bx1 (mod d).

    Therefore B=xn for some n>k and xn satisfies (2.1), a contradiction.

    Case 2. d is odd. Let kd+1 be an integer satisfying

    kd+1>kd  and  {αxkd+1}<cd+1.

    This case is similar to Case 1. Let D=d+1i=1xki. Then D>xk, Dx1 (mod d), and

    αD=d+1i=1αxki=d+1i=1αxki+d+1i=1{αxki}. (2.6)

    Since kd+1>k and {αxkd+1}<c, we see that αxkd+11 (mod 2). In addition, we have

    d+1i=1{αxki}<d+1i=1cd+1=c, (2.7)
    d+1i=1αxki1+1++1d+10 (mod 2). (2.8)

    From (2.6), (2.7), and (2.8), we obtain

    {αD}=d+1i=1{αxki}<c  and  αD=d+1i=1αxki0 (mod 2).

    Therefore D=xn for some n>k and xn satisfies (2.1), a contradiction.

    In any case, we have a contradiction. So the proof is complete.

    Lemma 8. Suppose a and b are positive rational numbers and a,b1. Then logba is rational if and only if there exist integers m and n such that am=bn.

    Proof. This is well-known. For more details, see for example, pages 24–25, Chapter 2 in the book of Niven [27].

    We are now ready to prove our main theorems. We begin with maximal and minimal orders of Ab(n).

    Theorem 9. Let b2 be an integer. Then

    lim supnAb(n)n=b+1b.

    In particular, a maximal order of Ab(n) is (b+1b)n.

    Proof. Let ε>0. By Lemma 4, it remains to show that Ab(n)/nb+1bε for infinitely many nN. To prove this, we construct a strictly increasing sequence (nk)k1 of positive integers such that

    Ab(nk)nkb+1b  as  k.

    For each kN, let nk=b2k+1+1. By Lemma 3, we obtain that Ab(nk)=bk+1+bk1. Therefore,

    Ab(nk)nk=bk+1+bk1bk+121+b2k1=(b+1b1bk+12)11+b2k1b+1b  as k,

    as desired.

    Theorem 10. Let b2 be an integer. Then

    lim infnAb(n)n=2.

    In particular, a minimal order of Ab(n) is 2n.

    Proof. Let ε>0. We first show that Ab(n)/n2ε for all large n. Let N be a large positive integer to be determined later and let nbN. Since [bN,)=k=N[bk,bk+1), we see that bkn<bk+1 for some kN. Then the number of digits of n in its b-adic expansion is k+1. Let

    n=(akak1a1a0)b,  y=0ik/2akibki,  and  z=0ik/2akibi.

    So y=(akak1akk2000)b, ny, and Ab(n)Ab(y). We divide the calculation into two cases according to the parity of k.

    Case 1. k is even. Then by Lemma 3,

    Ab(y)=bk2+0ik2akibk2i2=bk2(1+z)2.

    Observe that

    n0ik2akibki+(b1)k2<ikbki=0ik2akibki+bk21=bkz+bk21.

    Then nbk/2z+δ, where δ=bk/2bk>0. Therefore

    Ab(n)nAb(y)nbk2(1+z)bk2z+δ2n=1+zz+δ22n+2=(z+δ1)2z+δδz+δ2n+222nδz+δ.

    As N, we see that n, k, and δ/z+δδ=bk2bk0. Therefore we can choose N large enough so that

    Ab(n)n22nδz+δ2ε,  for all nbN.

    Case 2. k is odd. Some calculations in this part are similar to those in Case 1, so we skip some details. Let δ=b(k1)/2bk>0. By Lemma 3,

    Ab(y)=bk+12+0ik12akibk12i2=bk2(b+zb)2.

    In addition,

    nbkz+bk+121=bk2z+δ,  and
    Ab(n)nAb(y)nbk2(b+zb)bk2z+δ2n=b+zb(z+δ)22n+2=(z+δb)2b(z+δ)δb(z+δ)2n+222nδb(z+δ)2ε  when N is large enough.

    In any case, we see that if N is large and nbN, then Ab(n)/n2ε. So it remains to show that Ab(n)/n2+ε for infinitely many nN. For each kN, let n=nk=b2k2. Then n=(a2k1a2k2a1a0)b where a0=b2 and ai=b1 for all i=1,2,,2k1, and n<b2k1=Cb(n). By Lemma 3, we have

    Ab(n)=bk+(b1)0ik1bk1i2=2bk3.

    This implies

    Ab(n)n=2bkb2k23b2k22  as k.

    Since k is arbitrary, there are infinitely many nN such that Ab(n)/n2+ε. This completes the proof.

    We used a computer to compare the values of Ab(n) when b=2,3,5,10 and n=10k for k=1,2,,20. The data are shown in Table 1 at the end of this article. We see that for distinct b,b1{2,3,5,10}, there is an integer n1 such that Ab(n)>Ab1(n) and there is an integer m1 such that Ab(m)<Ab1(m). For example, A2(1020)>A10(1020) while A2(1019)<A10(1019). In general, we have the following theorem, which is the main result of this paper.

    Theorem 11. Let b>b12 be integers. Then Ab(n)Ab1(n) has infinitely many sign changes as n. More precisely, the following statements hold.

    (i) There are infinitely many nN such that Ab1(n)>Ab(n).

    (ii) There are infinitely many nN such that Ab(n)>Ab1(n).

    Proof. Throughout the proof, we apply Lemma 3 repeatedly without reference and separate the proof into two parts. In the first part, we show that (i) holds.

    Case 1(i). b is not a rational power of b1. By Lemma 8, we see that logbb1 is irrational. Applying Lemma 7 to the sequence of odd positive integers (1,3,5,7,) with d=2, α=logbb1, and c=logb(1+1b2), we obtain that there are infinitely many integers k such that

    k1 (mod 2),  k5logb1b,  {klogbb1}<logb(1+1b2), and  klogbb10 (mod 2). (3.1)

    Let k be one of those integers. Since k1 (mod 2) and bk1<bk1+1=Cb1(bk1), we obtain

    Ab1(bk1)=bk21+bk212=bk+121+bk1212=(b1+1b1)bk212. (3.2)

    Next, we write bk1=(arar1a1a0)b. Since brarbrbk1<br+1, we see that r is the largest integer such that brbk1. Therefore r=klogbb15logb1blogbb1=5. In addition, arbrbk1<(ar+1)br, so ar is the largest integer such that arbrbk1. Therefore

    ar=bk1br=bk1bklogbb1{klogbb1}=b{klogbb1}.

    Since {klogbb1}<logb(1+1/b2)<logb2, we see that b{klogbb1}<blogb2=2. Thus b{klogbb1}<2, which implies ar<2. So ar=1. Next, we calculate ar1. We see that

    ar1=bk1brbr1=b1+{klogbb1}b. (3.3)

    Since {klogbb1}<logb(1+1/b2)logb(1+1/b), we have

    b1+{klogbb1}<b1+logb(1+1b)=b(1+1b)=b+1.

    We obtain from (3.3) that ar1<b+1b=1, which implies ar1=0. Similarly, we have

    ar2=bk1brbr2=b2+{klogbb1}b2<b2(1+1b2)b2=1.

    This implies ar2=0. So we have ar=1 and ar1=ar2=0, and thus bk1br+br2. Recall that r=klogbb10 (mod 2), r5, and brbk1. Since br+br2<br+br2+b2+1=Cb(br+br2), we obtain

    Ab(bk1)Ab(br+br2)=br2+br2+br222=2br2+br2222bk21+bk21b22=(2+b2)bk212.

    From this and (3.2), we obtain

    Ab1(bk1)Ab(bk1)(b1+1b1)bk21(2+1b2)bk21=(b1+1b121b2)bk21. (3.4)

    Since b>b12 and the function xx+1x is increasing on [1,),

    b1+1b121b22+122132>0.

    Therefore Ab1(bk1)Ab(bk1)>0. Since Ab1(bk1)Ab(bk1)>0 holds for any k satisfying (3.1), we can choose n=bk1 and obtain that Ab1(n)Ab(n)>0 for infinitely many n, as required.

    Case 2(i). b is a rational power of b1. Let bs=bt1 for some s,tN with gcd(s,t)=1. Let mN and k=2mt+1. Then k is odd. Since bk1+1=Cb1(bk1+1), we obtain

    Ab1(bk1+1)=bk21+bk211=bk+121+bk1211=(b1+1)bk1211. (3.5)

    Since bs=bt1 and k=2mt+1, we obtain bk1=b2mt+11=b1b2ms. Therefore bk1+1=(arar1a0)b where r=2ms, ar=b1, a0=1, and ai=0 for i=1,2,,r1. So

    Ab(bk1+1)=br2+arbr22=(b1+1)bms2=(b1+1)bmt12=(b1+1)bk1212.

    From this and (3.5), we obtain Ab1(bk1+1)Ab(bk1+1)=1. Since m is arbitrary, we can choose k=2mt+1 and n=bk1+1 so that Ab1(n)Ab(n)>0 for infinitely many n.

    Case 1(i) and Case 2(i) give a proof of (i). The proof of (ii) is quite similar to that of (i). So we omit some details. We divide the consideration into two cases.

    Case 1(ii). b is not a rational power of b1. Then logb1b is irrational. Similar to Case 1(i), we apply Lemma 7 and let k be an integer such that

    k1 (mod 2),  k5,  {klogb1b}<logb1(1+1b21),  and  klogb1b0 (mod 2). (3.6)

    Then

    Ab(bk)=bk+12+bk122=(b+1b)bk22. (3.7)

    We write bk=(arar1a1a0)b1. Then r=klogb1bk5 and

    ar=bkbr1=bkbklogb1b{klogb1b}1=b{klogb1b}1<1+1b21<2,

    which implies ar=1. Similarly,

    ar1=bkbr1br11=b1+{klogb1b}1b1<b1(1+1b21)b1<1,

    which implies ar1=0. Then

    ar2=bkbr1br21=b2+{klogb1b}1b21<b21(1+1b21)b21=1,

    which implies ar2=0. So bkbr1+br21<Cb1(br1+br21). Recall that r=klogb1b0 (mod 2) and r5. Then

    Ab1(bk)Ab1(br1+br21)=br21+br21+br2212=2br21+br21b2122bk2+bk2b212=(2+b21)bk22.

    From this and (3.7), we obtain

    Ab(bk)Ab1(bk)(b+1b21b21)bk2. (3.8)

    Since b>b12 and the function xx+1x is increasing on [1,), we have

    b+1b21b213+132122>0.

    Therefore Ab(bk)Ab1(bk)>0. By Lemma 7, there are infinitely many integers k satisfying (3.6). So we can choose n=bk so that Ab(n)Ab1(n)>0 for infinitely many n.

    Case 2(ii). bs=bt1 for some s,tN with gcd(s,t)=1. Let mN and k=2ms+1. Then k is odd and

    Ab(bk)=bk+12+bk122=(b+1b)bk22. (3.9)

    Since bs=bt1, we obtain bk=b2ms+1=b2mt1bts1=b{ts}1b2mt+ts1. Since {ts}=js for some j=0,1,,s1, we obtain b{ts}1=bjs1. Recall that for any positive integers x and y, x1/y is either an irrational number or an integer. Then b{ts}1=(bj1)1s is either an irrational number or an integer. If b{ts}1 is irrational, then b=bts1=b{ts}1bts1 is irrational, a contradiction. So b{ts}1N and bk=(arar1a0)b1 where r=2mt+ts, ar=b{ts}1=bjs1, and ai=0 for all i=0,1,,r1. Therefore

    Ab1(bk)=br21+arbr212.

    In addition, bk2=bms+12=bmt+t2s1. Suppose first that ts is even. Then

    Ab1(bk)=bmt+12ts1+bjs1bmt+12ts12=(1+bjs1)bmt+t2s1b12{ts}12=(1+bjs1)bk2bj2s12=(bjs1+1bjs)bk22. (3.10)

    Since b>bjs11 and the function xx+1x is strictly increasing on [1,), we obtain from (3.9) and (3.10) that

    Ab(bk)Ab1(bk)=(b+1bbjs11bjs1)bk2>0. (3.11)

    Suppose ts is odd. Then similar to (3.10), we obtain

    Ab1(bk)=bmt+12ts+121+b{ts}1bmt+12ts1212=bmt+t2s+1212{ts}1+bmt+t2s12+12{ts}12=bk2(b12(1{ts})1+b12(1{ts})1)2=(b1+1b1)bk22,

    where =1{ts}. From this and (3.9), we obtain

    Ab(bk)Ab1(bk)=(b+1bb11b1)bk2>0. (3.12)

    Since m is arbitrary, we can choose k=2ms+1 and n=bk so that Ab(n)Ab1(n)>0 for infinitely many n.

    Case 1(ii) and Case 2(ii) give a proof of (ii). Therefore the proof of this theorem is complete.

    Observing the proof of Theorem 11 carefully, we can state some parts of Theorem 11 in a stronger form as follows.

    Theorem 12. Let b>b12 be integers. Then the following statements hold.

    (i) There are infinitely many kN and a constant c>0 depending at most on b and b1 and not on k such that

    Ab(bk)Ab1(bk)cbk2.

    Consequently, lim supn(Ab(n)Ab1(n))=+.

    (ii) Suppose b is not a rational power of b1. Then there are infinitely many kN and a constant d>0 which depends at most on b and b1 and not on k such that

    Ab1(bk1)Ab(bk1)dbk21.

    Consequently, lim infn(Ab(n)Ab1(n))=

    Proof. For (i), we consider Case 1(ii) and Case 2(ii) in the proof of Theorem 11. In Case 1(ii), we see from (3.8) that we can take c=b+1b21b21 so that Ab(bk)Ab1(bk)cbk2. Next, we consider (3.11) and (3.12) in Case 2(ii). Let 0<α<1. Since b1>bα11 and the function xx+1x is increasing on [1,), we obtain

    b+1bbα11bα1b+1bb11b1>0.

    Setting α=js in (3.11) and let α= in (3.12), we see that we can take

    c=b+1bb11b1>0

    so that

    Ab(bk)Ab1(bk)cbk2.

    If we would like to obtain c that works in (3.8), (3.11), and (3.12), then we can choose

    c=min{b+1b21b21,b+1bb11b1}.

    This proves (i). For (ii), we consider (3.4) in Case 1(i), take d=b1+1b121b2, and obtain that

    Ab1(bk1)Ab(bk1)dbk21.

    This proves (ii). So the proof is complete.

    We are now ready to give a complete answer to Vepir's question [32] posted on Mathematics Stack Exchange. The title of Vepir's post is as follows:

     which number base contains the most palindromic numbers? 

    In the comment, Vepir also says that he is only interested in the palindromes having more than one digit. Therefore for each integers b2 and n1, we let

    fb(n)=Ab(n)(b1).

    So fb(n) is the number of b-adic palindromes which have more than one digit and are less than or equal to n. We have the following corollary.

    Corollary 13. Let b>b12 be integers. Then fb(n)fb1(n) changes sign infinitely often as n. In other words, if we use counting measure, then the races between palindromes in any two different bases have infinitely many wins and infinitely many losses.

    Proof. By Theorem 11, there are infinitely many nN such that Ab1(n)>Ab(n). So if n is such an integer, then

    fb1(n)fb(n)=Ab1(n)Ab(n)+bb1>0.

    By Theorem 12, there are infinitely many mN such that Ab(m)Ab1(m)b. So if m is such an integer, then

    fb(m)fb1(m)=Ab(m)Ab1(m)b+b1b1>0.

    This completes the proof.

    Corollary 14. Let b>b12 be integers. Then Ab(n)Ab1(n)=0 for infinitely many nN.

    Proof. For each nN, let g(n)=Ab(n)Ab1(n). We know that, for any nN, both Ab(n+1)Ab(n) and Ab1(n+1)Ab1(n) are either 0 or 1. So g(n+1)g(n) is 1,0, or 1, that is, the difference of any two consecutive terms of the sequence (Ab(n)Ab1(n))n1 is one of 1, 0, or 1. Therefore if Ab(r)Ab1(r)<0 and Ab(m)Ab1(m)>0, then there exists an integer n lying between r and m such that Ab(n)Ab1(n)=0. By Theorem 11, there are infinitely many nN such that Ab(n)Ab1(n)=0, as required.

    We obtain extremal orders of the palindromes counting function Ab(n) and show that if b>b12, then Ab(n)Ab1(n) has infinitely many sign changes as n. Moreover, we obtain that

    lim supn(Ab(n)Ab1(n))=+,

    and if b is not a rational power of b1, then

    lim infn(Ab(n)Ab1(n))=.

    Problem 1. Suppose b>b1 and b is a rational power of b1. Then, perhaps, lim infn(Ab(n)Ab1(n)) is either or 1. More precisely, it is 1 if and only if b is an integral power of b1, and it is if and only if bbm1 for any mN. We do not have a proof of this yet but we plan to do it in the future.

    Problem 2. Suppose b1,b2,,bk are distinct integers larger than 1. We believe that our results can be extended to the string of inequalities

    Ab1(n)<Ab2(n)<<Abk(n)

    for infinitely many nN. Maybe, if δi{0,1,1} for every i, δ1+δ2++δk=0, and b1,b2,,bk satisfy some natural conditions such as logbi/logbj is not rational for any ij, then

    lim supn(ki=1δiAbi(n))=+andlim infn(ki=1δiAbi(n))=.

    Problem 3. For positive integers b2, n1, q2, and 1aq, let Ab(n,q,a) be the number of b-adic palindromes which are less than or equal to n and are congruent to a modulo q. Perhaps, we can find an asymptotic formula for Ab(n,q,a). Then the study of the race between palindromes in different congruence classes (in the same or different bases) may be interesting. For example, under some natural conditions on b and q, are there infinitely many sign changes in Ab(n,q,a1)Ab(n,q,a2) for distinct a1, a2? If kN is fixed, are there b, q, a1, a2 such that the number of sign changes in Ab(n,q,a1)Ab(n,q,a2) is exactly k?

    Problem 4. Can Lemma 7 be extended to any congruence classes? For example, suppose (xn) is an arithmetic progression, xnZ for all nN, d=x2x11, α is an irrational number, and 0c1<c21. Then for any q2 and 0a<q, we may be able to prove that there are infinitely many kN such that c1<{αxk}<c2 and αxka(modq). Furthermore, we may be able to replace the assumption that xnZ for all nN by a weaker condition.

    Problem 5. Considering Remark 3.19 in the article by Kawsumarng et al. [22], we see that there exists a palindromic pattern in the sumset B(α2)+B(α2) with respect to the Fibonacci numbers, where B(α2) is the upper Wythoff sequence. It may be interesting to see whether or not these kinds of palindromic patterns occur in the h-fold sumset hB(x) with respect to the members of linear recurrence sequence (an), where x is a particular root of the characteristic polynomial of the sequence (an) and h+1 is the smallest positive integer such that (h+1)B(x) is cofinite.

    We plan to solve some of these problems in the future but we do not mind if the readers solve them before us.

    We are grateful to the editor and reviewers for their comments and suggestions which improve the quality of this article. Prapanpong Pongsriiam's research project is supported jointly by the Faculty of Science Silpakorn University and the National Research Council of Thailand (NRCT), grant number NRCT5-RSA63021-02.

    The authors declare that there is no conflict of interests regarding the publication of this article.

    Table 1.  The value of Ab(n) when b=2,3,5,10.
    n A2(n) A3(n) A5(n) A10(n)
    10
    102
    103
    104
    105
    106
    107
    108
    109
    1010
    1011
    1012
    1013
    1014
    1015
    1016
    1017
    1018
    1019
    1020
    5
    19
    61
    204
    644
    1,999
    6,535
    20,397
    63,283
    207,364
    643,612
    2,002,248
    6,578,488
    20,309,535
    63,356,753
    208,723,532
    640,964,484
    2,005,064,397
    6,623,273,731
    20,231,466,772
    5
    19
    62
    202
    652
    2,099
    6,758
    21,801
    70,487
    228,398
    719,607
    2,221,547
    6,873,719
    21,318,077
    66,277,292
    206,575,404
    645,537,966
    2,022,653,063
    6,354,756,390
    20,020,259,837
    5
    23
    63
    203
    783
    2,223
    6,323
    22,023
    79,623
    206,123
    646,623
    2,465,123
    7,073,123
    20,005,623
    69,308,123
    253,628,123
    653,740,623
    2,039,903,123
    7,741,915,623
    22,487,515,623
    9
    18
    108
    198
    1,098
    1,998
    10,998
    19,998
    109,998
    199,998
    1,099,998
    1,999,998
    10,999,998
    19,999,998
    109,999,998
    199,999,998
    1,099,999,998
    1,999,999,998
    10,999,999,998
    19,999,999,998

     | Show Table
    DownLoad: CSV


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