Algebraic independence results on the generating Lambert series of the powers of a fixed integer

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1 Algebrac ndependence results on the generatng Lambert seres of the powers of a fxed nteger Peter Bundschuh, Kejo Väänänen To cte ths verson: Peter Bundschuh, Kejo Väänänen. Algebrac ndependence results on the generatng Lambert seres of the powers of a fxed nteger. Hardy-Ramanujan Journal, Hardy-Ramanujan Socety, 2015, 38, pp <hal > HAL Id: hal Submtted on 11 Jan 2016 HAL s a mult-dscplnary open access archve for the depost and dssemnaton of scentfc research documents, whether they are publshed or not. The documents may come from teachng and research nsttutons n France or abroad, or from publc or prvate research centers. L archve ouverte plurdscplnare HAL, est destnée au dépôt et à la dffuson de documents scentfques de nveau recherche, publés ou non, émanant des établssements d ensegnement et de recherche franças ou étrangers, des laboratores publcs ou prvés.

2 Hardy-Ramanujan Journal 38 (2015), submtted 26/02/2015, accepted 07/07/2015, revsed 15/07/2015 Algebrac ndependence results on the generatng Lambert seres of the powers of a fxed nteger Peter Bundschuh and Kejo Väänänen Abstract. In ths paper, the algebrac ndependence of values of the functong d (z) := h 0 zdh /(1 z dh ), d>1 a fxed nteger, at non-zero algebrac ponts n the unt dsk s studed. Whereas the case of multplcatvely ndependent ponts has been resolved some tme ago, a partcularly nterestng case of multplcatvely dependent ponts s consdered here, and smlar results are obtaned for more general functons. The man tool s Mahler s method reducng the nvestgaton of the algebrac ndependence of numbers (over Q) to the one of functons (over the ratonal functon feld) f these satsfy certan types of functonal equatons. Keywords. Algebrac ndependence of numbers, Mahler s method, algebrac ndependence of functons Mathematcs Subject Classfcaton. 11J91; 11J81, 39B32 1. Introducton and man results A seres of type k=1 λ k z k 1 z k wth (λ k ) C N s called a Lambert seres. Denotng by d the nteger n the ttle, supposng always d 2, and takng λ k to be 1 or 0 dependng on whether k s a power of d or not, our above seres reduces to z dh G d (z) :=. (1.1) dh 1 z Ths seres converges exactly on the open unt dsk D and defnes there a holomorphc functon. The smlar-lookng seres z dh F d (z) := (1.2) 1 + z dh has the same analytc propertes and, ndeed, we have F d (z) = G d ( z) n D f d s odd. Thus, n ths case, both functons F d, G d are very closely related. The am of the present paper s to study the algebrac ndependence of the values of the functons G d (z) and F d (z) at certan multplcatvely dependent ponts α 1,..., α n D. The arthmetcal nature of the values of these functons, beng typcal examples of Mahler functons, has been studed n several works (see [BV15c], [Coo12], [Coo13], [Mah69], [Sch67]). In partcular, t s known that G d (α 1 ),..., G d (α n ) are algebracally ndependent (over Q) f α 1,..., α n Q D are multplcatvely ndependent, and the same holds for F d nstead of G d. Here and n the sequel, Q denotes the feld of all complex algebrac numbers. In ths work, we suppose always α := α m ( = 1,..., n) wth α Q D and all m N. The followng result was establshed n [BV15c, Theorem 3.1]. We thank epscences.org for provdng open access hostng of the electronc journal Hardy-Ramanujan Journal

3 Bundschuh and Väänänen, Algebrac ndependence results 37 Theorem 1. Let m 1,..., m n be n 2 postve ntegers, and let α Q D. Then G d (α 1 ),..., G d (α n ) are algebracally ndependent f and only f holds for any par (, j) wth j. m j m / d Z (1.3) To state our frst new result, we defne, for any functon f, the notaton f(±β) to mean ether f(β) or f( β). Theorem 2. Let m 1,..., m n be n 2 postve ntegers satsfyng the condton (1.3), and let α Q D. Then, for each choce of n sgns, the values G d (±α 1 ),..., G d (±α n ) are algebracally ndependent, and the same holds for F d (±α 1 ),..., F d (±α n ). Note that, as an mmedate corollary of ths result, we obtan the analogue of Theorem 1 for F d. To state our next result, we defne G d (z) := a h z dh 1 z dh, F d(z) := where (a h ), (b h ) are non-zero perodc sequences of algebrac numbers. Theorem 3. Let m 1,..., m n be n 2 postve ntegers such that b h z dh, (1.4) dh 1 + z m j m / N (1.5) holds for any par (, j) wth j, and let α Q D. Then, for each choce of n sgns, the values G r (±α ) ( = 1,..., n; r N \{1}) are algebracally ndependent. In partcular, the numbers G r (±α ) ( = 1,..., n; r N \{1}) are algebracally ndependent. The same holds f G r, G r are replaced by F r, F r. In the remanng results, both functons G d, F d are studed smultaneously. But here the case d = 2 has to be excluded n a natural way snce, usng (1.1) and (1.2), we fnd for z D G 2 (z) + F 2 (z) = 2 z 2h 1 z 2h+1 = 2 z (1+2k)2h = 2 z n = k=0 n=1 2z 1 z. (1.6) Theorem 4. Suppose d 3, and let m 1,..., m n be n 2 postve ntegers satsfyng (1.3) and m j m / 2d Z (1.7) for any par (, j) wth j. If α Q D, then the numbers G d (α 1 ),..., G d (α n ), F d (α 1 ),..., F d (α n ) are algebracally ndependent. It may be of some nterest to see a typcal example of an applcaton of Theorem 4 nvolvng recprocal sums of the usual Fbonacc numbers Φ n (or Lucas numbers Λ n, respectvely). To ths purpose, we frst deduce from (1.1) and (1.2) that G d (β m ) + F d (β m ), up to an algebrac summand,

4 38 2. The man lemma equals to (2/ 5) h>0 1/Φ md h for any d, m N wth 2 d, where we put β := (1 5)/2. Assumng even 4 d, Theorem 4 tells us the algebrac ndependence of all numbers 1 Φ h 0 md h (m 2N 1). Smlarly we establsh, for odd d > 2, the algebrac ndependence of all h 0 dvsble by d. Theorem 5. Let m 1,..., m n be n 2 postve ntegers such that 1 Λ md h wth odd m not 2m j m / N (1.8) holds for any par (, j) wth j. If α Q D, then the numbers G r (α ), F r (α ) ( = 1,..., n; r N \ {1}, r / 2 2N 1 ) are algebracally ndependent. In partcular, the numbers G r (α ), F r (α ) ( = 1,..., n; r N \{1}, r / 2 2N 1 ) are algebracally ndependent. Remark. Assumng that (d h ) N N 0 satsfes a lnear recurrence d h+t = c 1 d h+t c t d h wth certan condtons on t and (c 1,..., c t ) N t 0 \ {0} excludng, n partcular, the case of (d h) beng a geometrc progresson, Tanaka [Tan05] settled the algebrac ndependence problem for the values of the Lambert seres h 0 zd h/(1 z d h) at dstnct ponts α 1,..., α n Q D. Thus, our nvestgatons on the Lambert seres G d (z) just concern the mportant remanng case, where (d h ) reduces to the geometrc progresson (d h ). 2. The man lemma The man tool n the proof of [BV15c, Theorem 3.1] was the followng auxlary result. Lemma 1. Let m 1,..., m n be n 2 postve ntegers satsfyng condton (1.3). Then the functons G d (z m 1 ),..., G d (z mn ) are lnearly ndependent over C modulo C(z). Combnng the proof of ths lemma wth some new deas, we are now able to generalze Lemma 1. To state ths generalzaton, we ntroduce, for fxed a C, the functons G d (a, z) := a h z dh 1 z dh, F d(a, z) := a h z dh. (2.9) dh 1 + z Lemma 2. Let m 1,..., m n be n 2 postve ntegers satsfyng condton (1.3). Assume that I 1 and I 2 are (possbly empty) dsjont sets of postve ntegers satsfyng I 1 I 2 = {1,..., n}. Then, for any root of unty ζ, the functons G d (ζ, z m ) ( I 1 ), F d (ζ, z m ) ( I 2 ) are lnearly ndependent over C modulo C(z). Proof. We frst note that the functons satsfy the functonal equatons g (z) := G d (ζ, z m ), f (z) := F d (ζ, z m ) ( = 1,..., n) (2.10) ζg (z d ) = g (z) + zm z m, ζf (z d ) = f (z) zm 1 z m ( = 1,..., n). (2.11) + 1

5 Bundschuh and Väänänen, Algebrac ndependence results 39 Assume now, contrary to Lemma 2, that there exsts some c := (c 1,..., c n ) C n \ {0} such that r(z) := I 1 c g (z) + I 2 c f (z) s a ratonal functon. For {1,..., n}, we now wrte m = d t() k wth ntegers t() 0, k > 0 such that d k. Then condton (1.3) s equvalent to the dstnctness of k 1,..., k n. By usng a sutable permutaton of {1,..., n}, we may assume wthout loss of generalty that, for some m {1,..., n}, the condtons c 1 c m 0, c m+1 = = c n = 0 hold and, moreover, k 1 > > k m. If, under ths permutaton, { I j : c 0} changes to J j (j = 1, 2), then J 1 J 2 = {1,..., m} and we may wrte r(z) := J 1 c g (z) + J 2 c f (z). From (2.11) we see z m z m ζr(z d ) = r(z) + c z m c 1 z m. + 1 J 1 J 2 Wth c and r as above, we defne the ratonal functon s by s(z) := r(z) t() 1 J 1 Ths new functon satsfes τ=0 c z dτ k ζ t() τ (z dτ k 1) + J 2 c z k ζs(z d ) = s(z) + ζ t() (z k J 1) 1 J 2 = s(z) + J 1 c ζ t() J 2 c c t() 1 τ=0 ζ t() + ζ t() (z k + J 1) 1 J 2 c z dτ k ζ t() τ (z dτ k + 1). c z k ζ t() (z k + 1) (2.12) c ζ t() (z k + 1). Snce all polynomals z k 1, z k + 1 ( = 1,..., m) dvde z L 1, where L := 2 lcm(k 1,..., k m ), t follows from [Ns97, Lemma 1] that s must be of the form s(z) = a(z) z L 1 wth some a C[z]. By consderng poles on the rght-hand sde of the frst lne of (2.12), one easly concludes a 0. Moreover, consderng the same lne near, we obtan deg a L, whence σ := s( ) C and s 1 (z) := s(z) σ tends to 0 as z. All n all, we conclude from the second lne of (2.12) ζs 1 (z d ) = s 1 (z) + c ζ t() (z k + c J 1) ζ t() (z k 1 J + 1) 2 = m c s 1 (z) + ζ t() (z k 2c =1 1) ζ t() (z 2k. J 1) 2 Let now d 2k exactly for {(1),..., (p)} ( J 2, clearly such can exst only f d s even), and wrte 2k = dl for these. Then d l and the l s are dstnct. Assume that j(1),..., j(q) are those values J 2 wth d 2k. Snce an equaton l (u) = 2k j(v) leads to the contradcton d k (u), the ntersecton {l (1),..., l (p) } {2k j(1),..., 2k j(q) } s empty. It follows that the ratonal functon S(z) := s 1 (z) + p u=1 2c (u) ζ t((u))+1 (z l (u) 1)

6 40 2. The man lemma satsfes ζs(z d ) S(z) = m =1 c ζ t() (z k 1) p u=1 2c (u) ζ t((u))+1 (z l (u) 1) q v=1 2c j(v) ζ t(j(v)) (z 2k j(v) 1), (2.13) where, wthout loss of generalty, we may assume l (1) > > l (p) and 2k j(1) > > 2k j(q). The possble poles on both sdes of (2.13) are roots of unty. If there ever s a pole, let a prmtve Nth root of unty be one of these on the left-hand sde wth maxmal N. Crucal to our fnal reasonng wll be the fact d N. Indeed, ths dvsblty property follows from the proof of [BV15c, Lemma 3.3], but, for the sake of completeness, we brefly explan here the reasonng. For ths purpose, we may wrte down the partal fracton decomposton of S as S(z) = δ L s δ (z) wth s δ (z) := δ 1 j=0 (j,δ)=1 where the s δ,j s are complex constants and ζ δ := e 2π/δ. From ths defnton of s δ for postve dvsors δ of L we obtan s δ,j z ζ j δ, s δ (z d ) = δ 1 j=0 (j,δ)=1 s δ,j z d ζ j δ = δ 1 j=0 (j,δ)=1 d 1 s δ,j κ=0 1 dζ (j+κδ)(d 1) dδ (z ζ j+κδ dδ ). (2.14) Suppose, from now on, p ν(1) 1... p ν(ω) ω to be the canoncal factorzaton of d. Assume that p 1,..., p σ are not dvsors of δ but p σ+1,..., p ω are, where we have to consder the cases σ = 0,..., ω. Then we have the followng equvalence (j, δ) = 1 (j + κδ, δ) = (j + κδ, dδ/ σ =1 pν() ) = 1 wth the usual conventon here and later that empty products (or sums) equal 1 (or 0, respectvely). Now, any postve dvsor D of p 1... p σ s relatvely prme to δ, whence there are precsely d D numbers κ {0,..., d 1} satsfyng D (j + κδ). Thus, by the well-known ncluson-excluson prncple, we can say that, for fxed coprme j, δ, the number of κ {0,..., d 1} such that j + κδ s prme to p 1... p σ (or equvalently to σ =1 pν() ) equals d σ =1 (1 1/p ). Therefore we can note that, for fxed coprme j, δ, there are exactly d σ =1 (1 1/p ) values κ {0,..., d 1} such that (j + κδ, dδ) = 1 holds. Hence we conclude s δ (z d ) = dδ 1 j=0 (j,dδ)=1 s δ,j [j/δ]δ dζ j(d 1) dδ (z ζ j dδ ) + Σ δ(z) (2.15) from the double sum n (2.14). The ratonal functon Σ δ n (2.15) vanshes dentcally n case σ = 0, whereas, n the cases 1 σ ω, t may have poles at certan prmtve ρth roots of unty but wth ρ < dδ only. Snce s δ 0 s equvalent to the fact that not all s δ,j [j/δ]δ, j {0,..., dδ 1} and prme to dδ, vansh, we conclude from (2.15) that, n ths case of δ, the dfference s δ (z d ) s δ (z) has poles at (dδ)th roots of unty. Thus, the number N defned after (2.13) must be of the form dδ, whence d N holds. If the set J 2 s empty (p + q = 0), then we obtan N = k 1 from the rght-hand sde of (2.13), hence d k 1, a contradcton, and t suffces to subsequently consder only the case p + q 1. In case d = 2, we have q = 0 and k = l for all J 2. Thus, the rght-hand sde of (2.13) s of the form m c ζ t() (z k 2c =1 1) ζ t()+1 (z k, J 1) 2

7 Bundschuh and Väänänen, Algebrac ndependence results 41 whence agan N = k 1, a contradcton. We next suppose d 3. If k 1 max{l (1),..., l (p), 2k j(1),..., 2k j(q) }, we are led to the same contradcton as before. Hence, let k 1 be equal to that maxmum. Ths mples k 1 = 2k j(1) observng that l (1) < k 1 follows from d 3. If c 1 /ζ t(1) 2c j(1) /ζ t(j(1)), we have our contradcton, whence c 1 = 2c j(1) ζ t(1) t(j(1)) must hold. We may now contnue n the same way and obtan a contradcton unless {k 1,..., k p+q } = {l (1),..., l (p), 2k j(1),..., 2k j(q) }. Moreover, for each {1,..., p + q}, there must exst a unque µ() {(1),..., (p), j(1),..., j(q)} such that c = 2c µ() ζ t() t(µ()) δ wth δ = 1 f µ() {(1),..., (p)} and δ = 0 otherwse. If p + q < m, then N = k p+q+1 holds under the precedng condtons, and we end up at our standard contradcton. Therefore, the only remanng possblty s that p + q = m, J 1 =. In ths case, the above condtons lead to c 1 = 2 j ζ ν c 1 wth some j {1,..., m}, ν Z, and ths contradcton completes the proof of Lemma 2. We next apply some deas ntroduced n [Ns02],[NTT99] (see also [BV14],[BV15b]) to the functons g,d (z) := G d (z m ), f,d (z) := F d (z m ) ( = 1,..., n), the G d, F d as defned n (1.4). We obtan from Lemma 2 and [BV14, Lemma 5] the followng Lemma 3. Under the assumptons of Lemma 2, the functons g,d j(z) = are algebracally ndependent over C(z). a h z m d jh 1 z m d jh ( I 1, j N), f,d j(z) = a h z m d jh 1 + z m d jh ( I 2, j N) Smlarly to the proof of [BV14, Theorem 3], and notng that condton (1.5) mples (1.3), ths lemma leads to the followng result. Theorem 6. Let m 1,..., m n be n 2 postve ntegers satsfyng (1.5), and let α Q D. Then the numbers G r (α ) wth I 1, r N \ {1}, and F r (α ) wth I 2, r N \ {1} are algebracally ndependent. In partcular, the numbers G r (α ) ( I 1, r N \ {1}), F r (α ) ( I 2, r N \ {1}) are algebracally ndependent. 3. Proof of Theorems 2 and 3 To ths end, we denote I 1 := { : α = α m }, I 2 := { : α = α m }. Clearly, these I j satsfy the condtons of Lemma 2. Further, let h (ζ, z) := G d (ζ, z m ) for I 1, h (ζ, z) := G d (ζ, z m ) for I 2, where G d (a, z) (and ts F -analogue) s defned n (2.9). If I 2, then h (ζ, z) = F d (ζ, z m ) for odd d, and h (ζ, z) = G d (ζ, z m ) + 2z m /(z 2m 1) for even d. Therefore, Lemma 2 mmedately gves Lemma 4. Let m 1,..., m n be n 2 postve ntegers satsfyng (1.3). Then, for any root of unty ζ, the functons h (ζ, z) ( = 1,..., n) are lnearly ndependent over C modulo C(z). By usng [Ns96, Theorem ], ths lemma provdes us drectly Theorem 2.

8 42 4. Proof of Theorems 4 and 5 Lemma 4 leads also to an analogue of Lemma 3 for the functons h,d (z) := g,d (z) f I 1, h,d (z) := a h ( z m ) dh 1 ( z m ) d h f I 2. Lemma 5. If the assumptons of Lemma 4 are satsfed, then the functons h,d j(z) ( = 1,..., n, j N) are algebracally ndependent over C(z). Analogously to Theorem 6, ths lemma mples Theorem Proof of Theorems 4 and 5 In the followng, we want to study algebrac ndependence of the functons G d and F d. By (1.6), t s natural to suppose d 3 for ths consderaton. Frst we note that, for any m N, the functonal equaton ζr(z d ) = r(z) + zm z m 1 2z2m z 2m 1 + zm z m + 1. has r(z) = 0 as a soluton. Thus, the functons G d (ζ, z m ), G d (ζ, z 2m ), and F d (ζ, z m ) are lnearly dependent over C. More generally, f m j d Z 2m holds, then the functons g j (z), g (z), and f (z) ntroduced n (2.10) are lnearly dependent over C modulo C(z). Indeed, f m j = d t 2m wth t 0, then the ratonal functon satsfes mplyng ζr 1 (z d ) = r 1 (z) Furthermore, f d t m j = 2m wth t > 0, then s a soluton of mplyng t 1 z dτ 2m r 1 (z) := 2 ζ t τ (z dτ 2m 1) τ=0 2zm j z m j 1 + z m ζ t (z m 1) + r 1 (z) = 2g j (z) + ζ t g (z) ζ t f (z). t 1 z dτ m j r 2 (z) := 2 ζ t τ (z dτ m j 1) τ=0 z m j z m ζ t (z m + 1) ζr 2 (z d ) = r 2 (z) 2ζ t z m j 1 + zm z m + zm 1 z m + 1 r 2 (z) = 2ζ t g j (z) + g (z) f (z). Ths makes t evdent that we have to suppose condton (1.7) for our lnear ndependence consderatons. Lemma 6. Let m 1,..., m n be n 2 postve ntegers, let d 3, and assume that condtons (1.3) and (1.7) are satsfed. Then, for any root of unty ζ, the functons G d (ζ, z m 1 ),..., G d (ζ, z mn ), F d (ζ, z m 1 ),

9 Bundschuh and Väänänen, Algebrac ndependence results 43..., F d (ζ, z mn ) are lnearly ndependent over C modulo C(z). Proof. Assume, contrary to Lemma 6, that there exsts a C := (b 1,..., b n, c 1,..., c n ) C 2n \ {0} such that r(z) := n =1 (b g (z) + c f (z)) s a ratonal functon satsfyng ζr(z d ) = r(z) + n =1 b z m z m 1 We may now argue as n the proof of Lemma 2 to obtan ζs(z d ) S(z) = n =1 b + c ζ t() (z k 1) p u=1 n =1 c 2c (u) ζ t((u))+1 (z l (u) 1) z m z m + 1. q v=1 2c j(v) ζ t(j(v)) (z 2k j(v) 1) (4.16) as an analogue of (2.13), where p+q = n and some of b, c may vansh. An equaton k = 2k j(v) s mpossble, by (1.7). Thus, k 2k j(v) for all pars (, j(v)), and smlarly k l (u) for all pars (, (u)). Further, {l (1),..., l (p) } {2k j(1),..., 2k j(q) } = was shown n the proof of Lemma 2. All n all, ths means that {k 1,..., k n, l (1),..., l (p), 2k j(1),..., 2k j(q) } s a set of 2n dstnct postve ntegers, each one not dvsble by d. Moreover, condton C 0 mples (b 1 + c 1,..., b n + c n, c 1,..., c n ) 0. Therefore, f we now defne N for the equaton (4.16) as we dd t for (2.13), then we get d N from the left-hand sde of (4.16), but clearly d N from ts rght-hand sde, a contradcton. To establsh Theorem 4, we proceed smlarly to our proof of Theorem 2 n the prevous secton. Moreover, the followng analogue of Lemma 5 holds. Lemma 7. Let m 1,..., m n be n 2 postve ntegers, let d 3, and assume that condtons (1.3) and (1.7) are satsfed. Then the functons g,d j(z), f,d j(z) ( = 1,..., n; j N) are algebracally ndependent over C(z). Outlne of Proof. By notng that (1.8) mples (1.3) and (1.7) for any d 3, the proof of Theorem 5 runs very much parallel to that of Theorem 6. Acknowledgement. The authors are grateful to the referees for suggestons mprovng the work. References [BV15a] [BV14] [BV15b] [BV15c] P. Bundschuh and K. Väänänen, Algebrac ndependence of certan Mahler functons and of ther values, J. Aust. Math. Soc. 98 (2015) P. Bundschuh and K. Väänänen, Algebrac ndependence of recprocal sums of certan Fbonacc-type numbers, arxv: v1 [math. NT] (2014). P. Bundschuh and K. Väänänen, Algebrac ndependence of recprocal sums of powers of certan Fbonacc-type numbers, Funct. Approx. Comment. Math. (to appear). P. Bundschuh and K. Väänänen, Guded by Schwarz functons: a walk through the garden of Mahler s transcendence method, submtted. [Coo12] M. Coons, Extenson of some theorems of W. Schwarz, Canad. Math. Bull. 55 (2012) [Coo13] M. Coons, On the ratonal approxmaton of the sum of the recprocals of the Fermat numbers, The Ramanujan Journal 30 no. 1 (2013) [Mah69] K. Mahler, Remarks on a paper by W. Schwarz, J. Number Theory 1 (1969) [Ns96] K. Nshoka, Mahler Functons and Transcendence LNM 1631 (Sprnger, Berln et al., 1996). [Ns97] K. Nshoka, Algebrac ndependence of recprocal sums of bnary recurrences, Monatsh. Math. 123 (1997) [Ns02] K. Nshoka, Algebrac ndependence of recprocal sums of bnary recurrences II, Monatsh. Math. 136 (2002) [NTT99] K. Nshoka, T. Tanaka, and T. Toshmtsu, Algebrac ndependence of sums of recprocals of the Fbonacc numbers, Math. Nachr. 202 (1999)

10 44 4. Proof of Theorems 4 and 5 [Sch67] W. Schwarz, Remarks on the rratonalty and transcendence of certan seres, Math. Scand. 20 (1967) [Tan05] T. Tanaka, Algebrac ndependence of the values of power seres, Lambert seres, and nfnte products generated by lnear recurrences, Osaka J. Math. 42 (2005) Peter Bundschuh Kejo Väänänen Mathematsches Insttut Department of Mathematcal Scences Unverstät zu Köln Unversty of Oulu Weyertal P. O. Box Köln, Germany Oulu, Fnland e-mal: pb@math.un-koeln.de e-mal: kejo.vaananen@.oulu.f