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Congruences for sixth order mock theta functions λ(q) and ρ(q)

  • Received: 01 June 2021 Revised: 01 September 2021 Published: 26 October 2021
  • Primary: 11P83, 05A17

  • Ramanujan introduced sixth order mock theta functions λ(q) and ρ(q) defined as:

    λ(q)=n=0(1)nqn(q;q2)n(q;q)n,ρ(q)=n=0qn(n+1)/2(q;q)n(q;q2)n+1,

    listed in the Lost Notebook. In this paper, we present some Ramanujan-like congruences and also find their infinite families modulo 12 for the coefficients of mock theta functions mentioned above.

    Citation: Harman Kaur, Meenakshi Rana. Congruences for sixth order mock theta functions λ(q) and ρ(q)[J]. Electronic Research Archive, 2021, 29(6): 4257-4268. doi: 10.3934/era.2021084

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  • Ramanujan introduced sixth order mock theta functions λ(q) and ρ(q) defined as:

    λ(q)=n=0(1)nqn(q;q2)n(q;q)n,ρ(q)=n=0qn(n+1)/2(q;q)n(q;q2)n+1,

    listed in the Lost Notebook. In this paper, we present some Ramanujan-like congruences and also find their infinite families modulo 12 for the coefficients of mock theta functions mentioned above.



    In 1920, Ramanujan introduced 17 mock theta functions of odd order in his last letter to Hardy. In addition Ramanujan also gave the mock theta functions of order 6 that are listed in the Lost Notebook. In the study of the arithmetic properties of mock theta functions, many authors have found some congruence properties for their coefficients. For instance, Andrews et al. [1] found several congruences for the partition functions pω(n) and pν(n) corresponding to the mock theta functions ω(q) and ν(q) respectively, defined as:

    ω(q)=n=0q2(n2+n)(q;q2)2n+1,
    ν(q)=n=0qn(n+1)(q;q2)n+1,

    where

    (a;q)n=(1a)(1aq)(1aq2)(1aqn1), (1)

    for some positive integer n and

    (a;q)=j=0(1aqj) (2)

    with |q|<1.They proved congruences for modulo 2 and some infinite families of congruences for pω(n) and pν(n). In 2017, Fathima and Pore [4] obtained a number of congruences for pω(n) and pν(n) modulo 20 and some infinite families of congruences modulo 2. In the sequel, Baruah and Begum [2] in 2019 established many congruences for the same partition functions modulo powers of 5.

    Zhang in 2018 proved some congruences for the sixth order mock theta function β(q) shown below and also gave some conjectures in [8].

    β(q)=n=0q3n2+3n+1(q;q3)n+1(q2;q3)n+1.

    Brietzke, Silva, and Sellers [3] in 2019 found many arithmetic properties satisfied by the coefficients of the eighth order mock theta function V0(q) given as:

    V0(q)=1+2n=0(q;q2)nqn2(q;q2)n.

    Silva and Sellers in [7] proved some congruence relations for the third order mock theta function ξ(q) given by Gordon and McIntosh given below:

    ξ(q)=1+2n=1q6n26n+1(q;q6)n(q5;q6)n.

    The main purpose of this paper is to study the arithmetic properties of the sixth order mock theta functions λ(q) and ρ(q) given by Ramanujan where the two mock theta functions are defined as:

    λ(q)=n=0(1)nqn(q;q2)n(q;q)n=n=0pλ(n)qn, (3)
    ρ(q)=n=0qn(n+1)/2(q;q)n(q;q2)n+1=n=0pρ(n)qn. (4)

    Ramanujan also listed linear relations connecting the sixth order mock theta functions with each other as:

    2q1ψ6(q2)+λ(q)=(q;q2)2f(q,q5), (5)
    q1ψ6(q2)+ρ(q)=(q;q2)2f(q,q5), (6)

    where ψ6(q) is the sixth order mock theta function

    ψ6(q)=n=0(1)nq(n+1)2(q;q2)n(q;q)2n+1.

    The proof technique of all the congruences for mock theta functions involves applying identities on the coefficients in arithmetic progressions, we use the same idea to prove infinite family of congruences modulo certain numbers of the form 2α3β for pλ(n) and pρ(n). The main results are found in Theorem 3.2–3.7 given in Section 3. Before proceeding towards the main theorems, we need some preliminary results given in Section 2 for proving the results in Section 3.

    To shorten the notations, we define

    fl=(ql;ql),

    for some positive integer l.

    Now, we define Ramanujan's general theta function

    f(a,b)=n=an(n+1)/2bn(n1)/2,for|ab|<1, (7)

    Jacobi's triple product identity is defined as:

    f(a,b)=(a;ab)(b;ab)(ab;ab).

    The special cases for f(a,b) are:

    φ(q)=f(q,q)=n=qn2=f52f21f24, (8)
    ψ(q)=f(q,q3)=n=0qn(n+1)/2=f22f1, (9)
    φ(q)=f21f2, (10)
    ψ(q)=f1f4f2. (11)

    In some of the proofs, we make use of the following identities:

    f31=n=0(1)n(2n+1)qn(n+1)/2,(Jacobi's identity)f1=n=(1)nqn(3n1)/2,(Euler's Pentagonal number theorem).

    The following lemma exhibits the 3-dissection of ψ(q) and 1/φ(q).

    Lemma 2.1. We have

    ψ(q)=f6f29f3f18+qf218f9, (12)
    1φ(q)=f46f69f83f318+2qf36f39f73+4q2f26f318f63. (13)

    Proof. Identity (12) is equation (14.3.3) of [5]. Identity (13) comes from [6] shown in equation (2).

    Lemma 2.2. We have

    f31f3=f34f123qf22f312f4f26, (14)
    f3f31=f64f36f92f212+3qf24f6f212f72. (15)

    Proof. The above identity (14) is (22.1.13) in [5]. And (15) follows from (14) by replacing q by q and using

    (q;q)=f32f1f4.

    From the binomial theorem, for any positive integer l and prime p, we have

    (q;q)pl(qp;qp)pl1(modpl).

    We first prove the 2-dissection of λ(q).

    Theorem 3.1. We have

    n=0pλ(2n)qn=f32f23f31f6, (16)
    n=0pλ(2n+1)qn=3f36f1f2+2q1ψ6(q). (17)

    Proof. To prove the above dissections, consider (5) and replacing q by q in (5), we have

    λ(q)=(q;q2)2f(q,q5)+2q1ψ6(q2),=f31f26f32f3+2q1ψ6(q2),=f26f32(f34f123qf22f312f4f26)+2q1ψ6(q2),

    where the last equality follows from (14). Extracting even and odd terms from above equation, we get

    n=0pλ(2n)q2n=f26f34f32f12, (18)
    n=0pλ(2n+1)q2n+1=3qf312f2f4+2q1ψ6(q2). (19)

    Replace q2 by q in (18) to arrive at (16). Divide (19) by q and replace q2 by q to obtain (17).

    Theorem 3.2. We have

    pλ(2n){1(mod2)ifnis triangular number,0(mod2)otherwise.

    Proof. According to (16), we have

    pλ(2n)=f23f32f31f6f6f32f1f2f6=f22f1=ψ(q)=n=0qn(n+1)/2(mod2). (20)

    Therefore, we complete the proof.

    Theorem 3.3. We have

    pλ(2n){(1)k(mod3)ifn=3k(3k1)2,0(mod3)otherwise.

    Proof. According to (16), we have

    n=0pλ(2n)qn=f23f32f31f6f23f6f3f6=f3=n=(1)nq3n(3n1)/2(mod3). (21)

    Therefore, we complete the proof.

    Corollary 1. Let p>3 be a prime and r an integer such that 8r+1 is a quadratic non-residue modulo p. Then for all n0,

    pλ(2(pn+r))0(mod6).

    Proof. As

    pn+r=k(k+1)2rk(k+1)2(modp).

    Thus, 2rk2+k(modp) or 8r+1(2k+1)2(modp) which contradicts the fact that 8r+1 is a quadratic non-residue modulo p. Therefore, from Theorem 3.2,

    pλ(2(pn+r))0(mod2). (22)

    Similarly,

    pn+r=3k(3k1)2r3k(3k1)2(modp).

    then 2r9k23k(modp) or 8r+1(6k1)2(modp) which contradicts the fact that 8r+1 is a quadratic non-residue modulo p. Therefore, from Theorem 3.3,

    pλ(2(pn+r))0(mod3). (23)

    From (22) and (23), we readily arrive at the main result.

    Since gcd(6, p) = 1, among the p1 residues modulo p, there are (p1)/2 residues r for which 8r+1 is a quadratic non-residue modulo p. So the above result leads us to a number of congruences for different primes p>3 as shown below:

    pλ(10n+i)0(mod6),i{4,8},pλ(14n+i)0(mod6),i{4,8,10}.

    Theorem 3.4. We have

    pλ(6n+2)0(mod3), (24)
    pλ(6n+4)0(mod6), (25)
    pλ(18n+8)0(mod18), (26)
    pλ(18n+14)0(mod72). (27)

    Proof. From (16), we get

    n=0pλ(2n)qn=f23f6f22f1f2f21. (28)

    Using (9) and (10), we have

    n=0pλ(2n)qn=φ(q3)ψ(q)φ(q).

    Using Lemma 2.1, we obtain

    n=0pλ(2n)qn=φ(q3)(f6f29f3f18+qf218f9)(f46f69f83f318+2qf36f39f73+4q2f26f318f63).

    Extracting the terms involving q3n,q3n+1,q3n+2 from above equation, we have

    n=0pλ(6n)q3n=φ(q3)(f56f89f93f418+4q3f26f518f63f9), (29)
    n=0pλ(6n+2)q3n+1=φ(q3)(2qf46f59f83f18+qf46f59f83f18), (30)
    n=0pλ(6n+4)q3n+2=φ(q3)(4q2f36f29f218f73+2q2f36f29f218f73). (31)

    To prove (25), dividing (31) by q2 and replacing q3 by q, we have

    n=0pλ(6n+4)qn=6φ(q)f32f23f26f71

    which can also be written as

    n=0pλ(6n+4)qn=6f22f23f26f51.

    The above equation readily implies (25). Consider (30), dividing by q and replacing q3 by q, we have

    n=0pλ(6n+2)qn=φ(q)(2f42f53f81f6+f42f53f81f6),=3f53f6(f2f21)3, (32)

    which proves (24). Now using (13) in above equation,

    n=0pλ(6n+2)qn=3f53f6(f46f69f83f318+2qf36f39f73+4q2f26f318f63)3. (33)

    Extracting the terms involving q3n,q3n+1,q3n+2 from (33), we have

    n=0pλ(18n+2)q3n=3f53f6(f126f189f243f918+56q3f96f99f213+64q6f66f918f183), (34)
    n=0pλ(18n+8)q3n+1=3f53f6(6qf116f159f233f618+96q4f86f69f318f203), (35)
    n=0pλ(18n+14)q3n+2=3f53f6(24q2f106f129f153f318+96q5f76f39f618f193). (36)

    Dividing (35) and (36) by q and q2 respectively, replacing q3 by q, we arrive at

    n=0pλ(18n+8)qn=3f51f2(6f112f153f231f66+96qf82f63f36f201),n=0pλ(18n+14)qn=3f51f2(24f102f123f221f36+96qf72f33f66f191).

    From the above equations, we get (26) and (27).

    Corollary 2. We have

    n=0pλ(18n+8)qn18f21f33(mod72). (37)

    Now we present the infinite families of congruences modulo 12.

    Theorem 3.5. For prime p5, we have

    pλ(6p2n+6pi+p214)0(mod12) (38)

    where i=1,2,,p1.

    Proof. From (29), replacing q3 by q, we have

    n=0pλ(6n)qn=φ(q)(f52f83f91f46+4qf22f56f61f3)

    Reducing modulo 4,

    n=0pλ(6n)qnf21f2f52f83f91f46(mod4)f42f1f81(mod4)f1(mod4). (39)

    By (21), we have

    n=0pλ(2n)qnf3(mod3).

    Extracting the terms involving q3n and replacing q3 by q, we have

    n=0pλ(6n)qnf1(mod3). (40)

    From (39) and (40), we have

    n=0pλ(6n)qnf1=n=(1)nqn(3n1)/2(mod12).

    This implies that pλ(6k)0(mod12) unless k is a pentagonal number, or equivalently, unless 24k+1 is a square. Now letting k=p2n+pi+(p21)/24 where i=1,2,,(p1) and p is a prime, we have that 24k+1=24p2n+24pi+p2, and this is evidently not a square since p2 divides the first and third terms but not the middle term. Thus pλ(6k)=pλ(6p2n+6pi+(p21)/4)0(mod12).

    Theorem 3.6. For m1, we have

    n=0pλ(34m32n+22m2j=032j)qn3f2f3f6f21(mod12), (41)
    n=0pλ(34m12n+22m1j=032j)qn9f2f3f6f21(mod12), (42)
    n=0pλ(34m22n+22m2j=032j)qn3f32f23f31f6(mod12), (43)
    n=0pλ(34m2n+22m1j=032j)qn9f32f23f31f6(mod12), (44)
    n=0pλ(32m+12n+32m22+2m1j=032j)qn6f36f1(mod12), (45)
    n=0pλ(32m2n+32m12+2m1j=032j)qn6f2f33(mod12), (46)
    n=0pλ(34m12n+22m2j=032j)qn3f1(mod12), (47)
    n=0pλ(34m+12n+22m1j=032j)qn9f1(mod12). (48)

    Theorem 3.7. For m1, we have

    pλ(32m2n+32m122+2m1j=032j)0(mod12). (49)

    Proof of Theorems 3.6 and 3.7. The proof for the above theorems follows by induction. Let us first prove the first step of induction, for m=1.

    From (32), we have

    n=0pλ(6n+2)qn=3f32f53f61f6.

    Reducing modulo 12, we have

    n=0pλ(6n+2)qn3f3f6f2f21(mod12) (50)

    which proves (41) for m=1. Using (13), we get

    n=0pλ(6n+2)qn3f3f6(f46f69f83f318+2qf36f39f73+4q2f26f318f63)(mod12)

    or

    n=0pλ(6n+2)qn3f56f69f73f318+6qf46f39f63(mod12).

    Extracting the terms involving q3n,q3n+1,q3n+2 from above, we have

    n=0pλ(18n+2)q3n3f56f69f73f318(mod12), (51)
    n=0pλ(18n+8)q3n+16qf46f39f63(mod12), (52)
    n=0pλ(18n+14)q3n+20(mod12). (53)

    Dividing (52) and (53) by q and q2 respectively then replacing q3 by q, we have

    n=0pλ(18n+8)qn6f42f33f616f2f33(mod12), (54)
    n=0pλ(18n+14)qn0(mod12). (55)

    Here (54) and (55) proves (46) and (49), respectively for m=1. Replacing q3 by q in (51), we have

    n=0pλ(18n+2)qn3f52f63f71f36(mod12)3f32f23f31f6(mod12)=3f22f1f2f21f23f6(mod12).

    The above equation proves (43) for m=1 and it can also be written as:

    n=0pλ(18n+2)qn3φ(q3)ψ(q)φ(q)(mod12). (56)

    Using Lemma 2.1, we have

    n=0pλ(18n+2)qn3φ(q3)(f6f29f3f18+qf218f9)(f46f69f83f318+2qf36f39f73+4q2f26f318f63)(mod12).

    Extracting the terms involving q3n,q3n+1,q3n+2 from above, we get

    n=0pλ(54n+2)q3n3φ(q3)f56f89f93f418(mod12), (57)
    n=0pλ(54n+20)q3n+19qφ(q3)f46f59f83f18(mod12), (58)
    n=0pλ(54n+38)q3n+26q2φ(q3)f36f29f218f73(mod12). (59)

    Replacing q3 by q in (57), we have

    n=0pλ(54n+2)qn3f42f83f71f463f1(mod12)

    which proves (47) for m=1. Now, dividing (58) and (59) by q and q2 respectively then replacing q3 by q,

    n=0pλ(54n+20)qn9f32f53f61f69f2f3f6f21(mod12),
    n=0pλ(54n+38)qn6f22f23f26f516f36f1(mod12).

    The above congruences imply (42) and (45) for m=1. Consider

    n=0pλ(54n+20)qn9f3f6(f46f69f83f318+2qf36f39f73+4q2f26f318f63)(mod12)

    or

    n=0pλ(54n+20)qn9f3f6(f46f69f83f318+2qf36f39f73)(mod12).

    Extracting the terms involving q3n,q3n+1,q3n+2 from above, we arrive at

    n=0pλ(162n+20)q3n9f56f69f73f318(mod12), (60)
    n=0pλ(162n+74)q3n+16qf46f39f63(mod12), (61)
    n=0pλ(162n+128)q3n+20(mod12). (62)

    Replacing q3 by q in (60), we obtain

    n=0pλ(162n+20)qn9f52f63f71f36(mod12)9f32f23f31f6(mod12)

    which proves (44) for m=1.

    n=0pλ(162n+20)qn9φ(q3)ψ(q)φ(q)(mod12)

    Similar to (56), extracting the terms involving q3n,q3n+1,q3n+2, we ultimately get

    n=0pλ(1623n+20)qn9f1(mod12), (63)
    n=0pλ(162(3n+1)+20)qn3f2f3f6f21(mod12), (64)
    n=0pλ(162(3n+2)+20)qn6f36f1(mod12). (65)

    Here (63) is the case when m=1 in (48). For the next step of induction, let us suppose that (41)-(49) holds true for m=k. Then, for m=(k+1), we prove the relations in similar manner mentioned above starting from (50) (taking m=k) and obtain (64) (m=k+1). Same process follows for other parts.

    Corollary 3. For m1, we have

    pλ(34m12n+22m1j=032j)3pλ(34m32n+22m2j=032j)(mod12), (66)
    pλ(34m2n+22m1j=032j)3pλ(34m22n+22m2j=032j)(mod12), (67)
    pλ(34m+12n+22m1j=032j)3pλ(34m12n+22m2j=032j)(mod12). (68)

    Now, we prove the 2-dissection of ρ(q).

    Theorem 3.8. We have

    n=0pρ(2n)qn=f32f23f31f6, (69)
    n=0pρ(2n+1)qn=3f36f1f2q1ψ6(q2). (70)

    Proof. From (6), we have

    ρ(q)=f62f12f34f6.f3f31q1ψ6(q2).

    Substituting the value from (15), we have

    ρ(q)=f62f12f34f6(f64f36f92f212+3qf24f6f212f72)q1ψ6(q2).

    Extracting even and odd terms from above equation, we easily arrive at (69) and (70).

    Finally on comparing (16) and (69), we arrive at the following theorem.

    Theorem 3.9. We have

    pλ(2n)=pρ(2n). (71)

    The above theorem yields that all the results shown above for λ(q) also hold for ρ(q).

    We have provided elementary proofs of numerous infinite family of congruences satisfied by pλ(n) and pρ(n). We did not carry out a computer search for congruences, and so we are unaware whether other congruences hold beyond the ones we prove in this paper, but certainly there is a possibility to explore more in this direction.

    The first author is supported by UGC, under grant Ref No. 971/ (CSIR-UGC NET JUNE 2018) and the second author is supported by SERB-MATRICS grant MTR/2019/000123. We would like to thank the two referees for carefully reading our paper and offering corrections and helpful suggestions.



    [1] Congruences related to the Ramanujan/Watson mock theta functions ω(q) and ν(q). Ramanujan J. (2017) 43: 347-357.
    [2] Generating functions and congruences for some partition functions related to mock theta functions. Int. J. Number Theory (2020) 16: 423-446.
    [3] Congruences related to an eighth order mock theta function of Gordon and McIntosh. J. Math. Anal. Appl. (2019) 479: 62-89.
    [4] Some congruences for partition functions related to mock theta functions ω(q) and ν(q). New Zealand J. Math. (2017) 47: 161-168.
    [5] M. D. Hirschhorn, The Power of q, Developments in Mathematics, 49. Springer, Cham, 2017. doi: 10.1007/978-3-319-57762-3
    [6] Arithmetic relations for overpartitions. J. Combin. Math. Combin. Comput. (2005) 53: 65-73.
    [7] R. da Silva and J. A. Sellers, Congruences for the coefficients of the Gordon and McIntosh mock theta function ξ(q), Ramanujan J., (2021), 1–20. doi: 10.1007/s11139-021-00479-8
    [8] Congruences for the coefficients of the mock theta function β(q). Ramanujan J. (2019) 49: 257-267.
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