Repdigits As Sums of Three Balancing Numbers Mahadi Ddamulira
Repdigits as sums of three balancing numbers Mahadi Ddamulira
To cite this version:
Mahadi Ddamulira. Repdigits as sums of three balancing numbers. Mathematica Slovaca, 2020, 70 (3), pp.557-566. 10.1515/ms-2017-0371. hal-02405969v2
HAL Id: hal-02405969 https://hal.archives-ouvertes.fr/hal-02405969v2 Submitted on 8 Jun 2020
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AUTHOR degree of number algebraic an be COPY p with | q , 2.1 Mathematica } h olwn r oeo h rpriso h oaihi egtfunction height logarithmic the of properties the of some are following The . of . n 1 l o-eaieitgrsolutions integer non-negative All η ≥ sgvnby given is η n = 2 ≥ . p/q ,n N, n ,` d, h N a 3 ( 0 η x ≥ a sartoa ubrwt gcd( with number rational a is = 1 := ) ih1 with h d 0 n , 0 h ( B + spstv n the and positive is η , ( 2 N .Peiiayresults Preliminary 3. n 1 η n , ` a d 1 1 1 ± h 1 ≥ hthsol n itntdgtwe rte nbs 0 That 10. base in written when digit distinct one only has that η + 3 x ( 2 ± ≤ η ,` d, , η N log d 1 B 2 1 s N − .Mi result Main 2. and , ) ) = ) d 1 n = a ≤ ≤ { ∈ 2 .DDAMULIRA M. with ) ≤ + 0 d d + h h | + · · · .Tesqec frpiisi sequence is repdigits of sequence The 9. s ihmnmlpiiieplnma vrteintegers the over polynomial primitive minimal with 1 ( ( B 1 | 10 η η , h X i =1 2 1 1 n ≤ + d ( ` + ) + ) , 3 η n 9 3 − ( ) a d log = 1 , d 6 ≤ ≥ 1 h h η d , = ( ( ( ( ,n N, 7 i 9 η η n , ) max , a rs from arise , 2 2 saetecnuae of conjugates the are ’s 2 8 10 0 o 2 log + ) ) } , ≥ i Y 1 . ` =1 d n , 9 {| − n s ( η 3 2 x ,q p, 1 ( ∈ n , ≥ i − ) | Z , 3 , and 1 = ) 0, , η ,` d, , 1 ) . ( } i ` ) ) ≥ , ) . ,ad1 and 1, fteDohnieequa- Diophantine the of > q η ≤ hnthe Then . ,then 0, d ≤ A 9. 175on 010785 h ( η (2.1) (1.3) (3.1) loga- h = ) ( · ),
ic log Since number irrational the of K npstv integers positive in paper. this in tools main our of one is which [8], Matveev Theorem of result the of version modified and snneo Then nonzero. is Lemma > B u oDjlaadPt˝ se[:Lma5].Frara number fractions. real continued a For of 5a]). theory Lemma the Peth˝o [3: from and (see Dujella result to some due use we so, do To them. reduce to procedure Reduction 3.2. where and hr euse we where Lemma n ∈ ⊂ ercl h euto ueu,Mgot,adSke [:Term94 p 8],wihi a is which 989]), pp. 9.4, Theorem ([2: Siksek and Mignotte, Bugeaud, of result the recall We ial,tefloigLmai loueu.I s[:Lma7]. Lemma [4: is It useful. also is Lemma following the Finally, easm that assume We o ohmgnoslna omi w nee aibe,w s lgtvraino result a of variation slight a use we variables, integer need two we in thus form large, linear too nonhomogeneous are which a variables For our on bounds upper get we calculations, the During Z R 1 } utemr,let Furthermore, . fdegree of o h itnefrom distance the for 3.2 AUTHOR COPY3.1 γ/ 3.1 . . o 0=0 = 10 log If > γ Let . log r Let D 10 ≥ M .Thus, 3. K | Λ n ,v u, γ ` , A 1 − η 1 n | eapstv integer, positive a be , b i . 1 1 1 > 1 ≥ 76555 − > H η , . . . , and , EDGT SSM FTREBLNIGNUMBERS BALANCING THREE OF SUMS AS REPDIGITS ≥ b , . . . , ≤ ( 1 − ε n n τ max ≤ d 1 2 1 := uhthat such . − ≥ (4 4 B · · · 10 w t X t k × { 1) r n n µq D u epstv elagbacnmesi elagbacnme field number algebraic real a in numbers algebraic real positive be 3 5 ` .Budn h variables the Bounding 4. enneoitgr,adasm that assume and integers, nonzero be 2 1 with 3 9 < 30 otenaetinteger. nearest the to ) o 10 log ( − rm(.)ad(.) ehave we (2.1), and (1.2) From . K ≤ log ≤ r − k n and , h 4 t 1 1 +3 B M / ( 0 γ − η ,w a ocuefo h bv that above the from conclude can we 5, B n M > q < i × 1 = ) 1) ≤ := Λ , ≥ k + | L/ > H | t uτ B τq 4 log < ` < ` 6 max B . n and 5 M k − p q 1 n × If . η η and 2 + and , 1 b eacnegn ftecniudfato expansion fraction continued the of convergent a be i v {| + D 1 | , B + + 1 · · · b (log 0 K 2 > ε B 1 n . 1+log + (1 µ 16 | 2 n , . . . , ` w η | ,B µ B, A, 3 + 4 5 t − b L } AB < 0 t ≥ = , ( hnteei oslto oteinequality the to solution no is there then , ) B n − 1 r d 1 then , | n log( ≤ b 1 3 1) + t , D |} log − 10 ( ≤ o all for esm elnmeswith numbers real some be n Aq/ε K w , ` 1 3 ( log + )(1 , 9 . B − B < L 1) + n X 1 ) 1 . i ewrite we , γ < 2 o 10 log 1 = log ≤ r H B 10 n γ t. , . . . , (log 1 ) +1 ` . A 1 , H A . . . k ) X r . k t =min := , > A {| X 0 − (4.1) (3.2) and 559 n | :
aigaslt auso ohsdso h bv qain eoti that obtain we equation, above the of sides both on values absolute Taking unn a Running hn erwie(.)i he asadapylwrbud o ierfrsi oaihsi three 1 in logarithms Step in forms 4.1. linear for bounds lower apply follows: and ways as three steps in (4.2) rewrite we Then, n so and hsi qiaetto equivalent is This edvd hog 43 by (4.3) through divide We hs ehave we Thus, hs ehave we Thus, ycnuaigteaoerlto in relation above the conjugating By eput We nodrt pl hoe . ene ocekta Λ that check to need we 3.1 Theorem apply to order In eoti nytesltoslse nTerm21 rmnwowrs easm that assume we onwards, now From 2.1. Theorem in listed solutions the only obtain we hc sflefor false is which 560 ehv that have We qain(.)cnb rte as written be can (2.1) Equation
4 γ √ AUTHOR COPYn 1 2 Mathematica
10 + 1 ` 4 γ γ √ n γ − 1 ` 2 n n 2 ≥ 3 − 10 + 1 n γ and 1 ( + 9(1 1 n ` γ
1 + 4 · n 10 rga nterne0 range the in program γ 9 √ 4 − 1
4 Λ n γ d 4 ` √ γ 2 − 2 1 n √ γ δ √ − d α n γ 3 2 n − =10 := δ n 1 2 n − 2 n ≥ 1 n n 1 1 1 n 4 3 1 = + − 1+ (1 + 1 .Teeoe Λ Therefore, 1. + 1 d δ n + γ √ | γ n γ 3 ` ( + 9(1 δ − n γ 1 n n 2 n + γ 1 γ − γ 3 2 1 + 4 n d Q − + n n n √ γ − 2 + 4 2 · 2 n 3 δ ( n √ 9 − − − 2 γ 1 10 √ n γ δ 2 δ n n n 2 − 2 n n n .DDAMULIRA M. ( + 9(1 δ ) eget we 2), 1 2 ` 1 − 2 n 1 2 + n 4
− + + + 3 2 + + d ≤ ) δ n d √ γ + γ γ n 3 δ · γ n 3 n ≤ n 1 d 9 9 2 n + γ 10 − n 3 3 3 γ 3 = 0. 6= n − 3 − + 4 | ≤ | n n γ = ` 1 1 √ n n − 4 3 3 4 − n −
1 1 = 4 d √ 2 − 2 n 10 ≤ ) < δ √ √ − 10 1 3 / δ 2 n − − n δ (4 2 ` 2 | n + 3 1 ` + 1 n 3 n d · 9 ( 2 ) 1 d 3 √ · .Spoeta Λ that Suppose 0. 6= = 9 δ γ 4 9 4 ≤ + n + · d − n )t get to 2) d 9 1 γ d 10 √ 2 − √ n | − n 4 + d δ 2 1 2 ` √ n 2 2 | 1 1 10 − · n
. δ 3 γ
≤ 9 . 2 10 2 n ) n < ` 1 γ < ( 9 1 2 + − 0,1 100, − δ ` + + n 2 n − | = 1 √ 1 1 δ γ 2 γ δ − | 1 + 2 n n n − n . . 3 3 3 . 1 δ ≤ − ) d 9 < . n n . < 2 d 1 3 + , ≤ 4 γ < 1 √ δ 1 ,ad1 and 9, ,te ehave we then 0, = n 2 3 2 (4 ) − . n √ 1 3) + 2 . ≤ n 1 ` > , ≤ (4.2) (4.3) (4.4) 100. 100
..Se 2 Step 4.2. hs ecntake can we Thus, ycmaigteaoeieult ihtergthn ieo 44 egtthat get we (4.4) of side right-hand the with inequality above the comparing By hs ehave we Thus, by below bounded is (4.4) of side left-hand the that us tells hs ecncnld that conclude can we Thus, to equivalent is This iiigtruh(.)by (4.6) through Dividing n Also, 5 Since 1 < γ < etake we , ehv that have We ow pl hoe . ihtedata the with 3.1 Theorem apply we So η 1 h η ,
AUTHOR COPY 0 and 6 ( η 2
3 η , 4 γ ) B √ 2 n ≤ ≤ ≤ ≤ 1 t b 2 := ∈ 1 4 γ h h h =3 := 2 9 √ < δ < Q := n (4 4 + (4) 4 + (4) + 1 n o o + 3 log 4 + 2 log 1 2 ( 1 A d
√ 10 η , ,b `, h iia oyoilof polynomial minimal The . √ 1 log + 1 γ ) etk h field the take we 2), )+ 2) ` n γ =2lg10, log 2 := EDGT SSM FTREBLNIGNUMBERS BALANCING THREE OF SUMS AS REPDIGITS h h γ .Tu,w a take can we Thus, 1. 2 n ( ( | − − Λ d d 1 γ 4 n n + ) + ) α 1 √ n − 1 2 h 1 | 2 2 n 2 9( + (9((1 =10 := −
> > 2 δ := 4 − 1+ (1 h h n γ n ( + 9(1 √ 1 1 ( ( n − × − √ √ n d − 1 2 + 1 − η , · )+ 2) + 2) 1 2lg10)(log log (2 4 n 4 2 1 A γ 9 10 ≤ . . d 3 γ ( + 1 4 37 γ n γ d 2 b , √ n n 1 n ` n × 2 2 1 · − 4 =log := 1 2
1 × 9 . h h 10 √ − 48 − − n − γ 30 ≤ ≤ 9 + (9) + (9) 2 2 n 10 n n 2 ` K ) δ n 2 3 6 2 := × + 1 / d 9 n ) − 3 3 = 14 + × (4 := 2 log ) n 10 + γ =1 := ( ,η γ, h − 1 √ − γ 3 h h and , n − 4 14 ( Q γ 4 n 4 γ 1 (( + (1) + ((1 1 γ ) eget we 2), d 9 γ √ . n d 2 √ 5 1 (6+( + )(16 ( ( γ − n . − = ) − 3 n − √
2 · 3 × γ 2 n ≤ 9 1 < 10 n d )wt degree with 2) + 3 A 4 + 2 over − γ 3 ))) γ · + 8 3 2 1 √ ` 2 log ) + 1 9 3 n 4 10 n 1+lg2( log + 2)(1 log + (1 γ n 4 n log := 3 2 = √ 1 =1 ( + 16 := 3 n √ 1 3 1 n − ` 3 + log ) 2 Z 1 2 − − 1
γ n 2 − γ ( + 9(1 n ( 3 n < − ( = n , is 1 n d 4 9 n δ 1 + 2 . √ h 1 − n 1 2 +1 n − γ n 4 + 3 x 1)=lg1,and 10, log = (10) ( + ) 1 n 4 − γ 2 3 n 1 2 2 √ + log ) n ( 3 . γ γ n − n δ +1 2 γ < n n 2 − δ n 3 D 1 1 3 4 n √ 4 6 log ) n 4 1 n − . − √ K √ − 2 z 2 d γ 3 + + 2 n n )) √ ) − + 2 a roots has 1 + 2 3 =2 ic max Since 2. := 3 n δ δ − 2 n + 3 γ. n δ n n 3 log ) n γ n 2 3 2 γ . n 1 3 )) +1 + n ) 3 ) h 2 δ . < − γ ( n γ n o hoe 3.1 Theorem So, . 3 o 2 log 2 + ) 3 γ ) 4 ) h n . √ , 3 ( √ +1 2 γ )= 2) . { and 1 ,n `, , 2 1 δ o 2. log (4.5) (4.7) (4.6) 3 with ≤ } 561
eput We aigaslt auso ohsdso h bv qain eoti that obtain we equation, above the of sides both on values absolute Taking etake We 562 to equivalent is This o hoe .,tlsu that us tells 3.1, Theorem So, in relation above the conjugating By Λ that check to need we 3.1 Theorem apply to order in before, As hs ecncnld that conclude can we Thus, hnw have we then hc sflefor false is which ..Se 3 Step 4.3. ycmaigteaoeiegaiywt 47,w ocuethat conclude we (4.7), with ineqguality above the comparing By A ehv that have We 3 t =3 := =2 :=
AUTHOR COPYD η , h ( =2, :=
η 4 γ 3 √ 1 2( = ) n 1 2 ≤ =10 := log B ` − 2( 4 γ ≥ := √ | d h h n Λ η , 1 (4 4 + (4) and 1 2 · 2 9 n 10 γ | d − 1 n > > √ , ` 1 2 4 d
A )+ 2) √ − 10 h := − × − ≤ ≤ · d 1 ( 9 2 10 δ ` d ≥ =2lg10, log 2 := n 1 (log 2 + 1 n ,η γ, d 9 + ) · Λ . . ` 2 h 1 9 4 48 4 .Teeoe Λ Therefore, 1. 2 + 9 + (9) − d = − × =10 := γ √ h γ × 4 n γ (4+( + )(14 − ( 30 n d 3 δ 2 √ 3 √ n γ 10 2 4 n Q · d 9 := 2 6 2 n < = 9 √ 1 + )+ 2) h 10 14 1 ( ` × − 1+ (1 + + √ 2 γ 1 | γ + ` ( ( + 9(1 δ A − . 3 .DDAMULIRA M. n n ) eget we 2), 4 δ 42 n γ 4 = 4 γ γ n 3 2 √ n n h 1 1 √ n . 2 n n 2 5 9 ( + (9) 1 γ =log := − 2 + + 2 × − 4 3 2 2 × n = − d ( + 9(1 γ − 1 = n d 9 δ 10 4 2 + √ − 2 n n − n 2 √ 3 − .S eapyTerm31wt h data the with 3.1 Theorem apply we So 0. 6= 2 n 1 2 log ) 2 4 14 10 2 γ 4 1+lg2( log + 2)(1 log + (1 − 10 2 | ≤ | 2 √ log ) 4 γ n √ n )) 1 ( γ n ` d 3 n 1 = 2 ` γ 2 n 2 and √ · 1 ) − n n · 2 9 4 ( + 4 δ 9 b , 4 + 3 1 2 4 γ − δ 1 √ − d | − n d . n n ) √ 4 2 n √ 1 1 n 2 δ √ 1 2 ) 2 n 2 + + 1 2 h ) log ) 2 2 . √ . ( := δ | ( γ − δ + 2 δ n − o 2) log + ) | 4 n 2 n ,b `, γ √ n 1 2 1 + n 1 γ . + 2 < 3 . 4 n δ 2 √ − δ 2 n n 2 2 n 3 .Spoeta Λ that Suppose 0. 6= 1 + , 2 ) 2 δ := ( o 10) log )(2 n + γ ≤ 3 n − . δ 3 4+( + 14 n n < 3 2 ) b , . γ 4 n n √ 2 3 1 +2 2 − =1 := . n 2 . log ) 2 (4.8) 0, = γ.
sthat us eput We etake We iiigtruh(.)by (4.9) through Dividing aigaslt auso ohsdso h bv qain eoti that obtain we equation, above the of sides both on values absolute Taking have we Thus, ycnuaigteaoerlto in relation above the conjugating By ycmaigti ih(.0,w banthat obtain we (4.10), with this comparing By si h rvoscss nodrt pl hoe . ene ocekta Λ that check to need we 3.1 Theorem apply to order in Λ cases, that previous the in As hc sflefor false is which n 10 o,w pl em . nteaoeieult 41)wt h data: the with (4.12) inequality above the on 3.2 Lemma apply we Now, Lemma 1 42 o ecmieteieulte 45,(.)ad(.1 ooti h on on bound the obtain to (4.11) and (4.8) (4.5), inequalities the combine we Now ≥ L , n 3 2 := ,te ehave we then 0, = t ≥ 4.1
AUTHORD COPY =3 := n n log =2, := . 3 1 eoti that obtain We . Let η , ≥ | Λ 0 3 , B ` ( | 1 ,n N, n 1 ≥ > > 1 =10 := := ≤ and 1 < < − − 1 EDGT SSM FTREBLNIGNUMBERS BALANCING THREE OF SUMS AS REPDIGITS d n 1 1 n , 2 3 1 . . ≤ . . η , 4 73 , 48 5 2 α × A n , 9 × d n × and , × 1 1 30 3 ≥ 10 2 / 10 =2lg10, log 2 := ,` d, ,
10 n (4 := 10 6 42 .Tu,Λ Thus, 1. 1 14 × 14 √ 10 ` (log < ,η γ, ` α ) Λ log n ) eget we 2), 3
` ≥ − etenneaieitgrsltost h equation the to solutions integer nonnegative the be 1 3 1 4 2 4 · γ Q . n . . 5 − √ n 10 = 9 42 n 6 1 4 n 1 δ ( × 1 hnw have we Then . d 1 1 γ 2 √ n × 3 n ) . √ × n 1 3 n < ` 2 4 2 := − ) eget we 2), 1 A 10 . 2 d ` 3 = 2 < 10 α 9 = 1+lg2( log + 2)(1 log + (1 √ 2 d .S,w pl hoe . ihtedata the with 3.1 Theorem apply we So, 0. 6= 49 = 4 − − 14 9 =log := 2 · d 10 n ercr htw aeproved. have we what record We . . 9 1 | 9 10 10 8 √ 1 δ < ` n 9 × 2 − ` ` 1 . · 4 8
b , · 10 | ≤ | 9 4 10 1 d 9 4 < γ × d 9
√ d 50 13 √ α < and √ 10 γ 2 1 . δ 2 4 log n 2 | := √ 13 . n 2 . +2 n 1 − 2 A log 2 n ,b `, < − 3 1 1 . n . . n 1 =2 := n 1 , 1 +2 1 ( o 10)(log log )(2 2 . log h := ( η − n 3 ) 1 n ≤ 2 log b , 1 hoe .,tells 3.1, Theorem 11. r n 3 1 γ =3 := =1 := ) × 3 n H , 11 .Suppose 0. 6= . 1 sfollows: as (2.1) =3 := (4.10) (4.11) (4.12) (4.9) . with 5 563 ×
hrfr,i ohcss ehv that have we cases, both in Therefore, of Γ acltdec au flog(39 of value each calculated optr hs ene ordc hs ons od o ertr o(.) 47,ad(4.10) and (4.7), (4.4), to return we so, procedure. do following To the bounds. via these 3.1 reduce Lemma apply to and need we Thus, computer. iiigtruhteaoeieult ylog by inequality above the through Dividing ics o ogtrdo hscniinltr oeta e that Note later. condition this of rid get that to assume how discuss we reasons, technical For ildsushwt e i fti odto ae.Nt hte that Note later. condition this of rid get to how discuss will then aethat have o ehia esn,a eoew suethat assume we before as reasons, technical For fw put we If ecnrwie(.)as (5.1) rewrite can we fΓ If 564 Thus, eel htteconvergent the that reveals enwapyLma31o 52.W put We (5.2). on 3.1 Lemma apply now We 2 h onsotie nLma41aetolret ar u ennflcmuain nthe on put computations we meaningful First, out carry to large too are 4.1 Lemma in ontained bounds The et eput we Next, τ < 3 ssc that such is > ,then 0, n ,te ehv that have we then 0, AUTHOR1 COPY − n n 1 2 − < p q n 108 108 0hlsi ohcases. both in holds 70 2 q Γ 108 ≤ 2 = 0 τ Γ 70 := < > := 2703843740443108411802421359516257223259008405220106 2069931281589203990595364033267574277731383243231951 3 0 . := ` | < 6 ntecs that case the In Γ o 10 log o 10 log M log 3
` | ` 0 o 10 log 0 0 | = o 10 log and e γ log < q Γ < < 0 108
3 − ` .Rdcn h bounds the Reducing 5. Γ < | | γ − o 10 log Γ Γ and 3 ε /ε n − | 3 2 d 1 2 − `τ < | | d | log ≥ n < < ) n < e / − 1 Γ 1 n − 0 e e log µ log 3 γ 1 n | | 1 + . d Γ Γ / .DDAMULIRA M. 0623392 − n 1 log + n 3 2 − .Hnee Hence 2. := γ γ 1 | | log((4 1 + e = 1 M − − log n − log + n on htalo hmaea ot7.Tu,we Thus, 70. most at are them of all that found and γ µ 2 log((4 eget we , = 1 = 1 d =10 := n n γ log ≥ | 2 2 d Γ < ( + 9(1 log + 3 > √ < − 0frtemmn n ot 41) ewill We (4.10). to go and moment the for 20 | | log | γ Λ Λ 39 Λ d 2) 50 4 n .Teeoe with Therefore, 0. 0 ewudhv that have would we 20, 3 2 Γ √ 3 d 4 | | 3 3 / γ uc optrsac in search computer quick A . | · 9 d √ 2) < < 9) γ < γ < ≥ √ Γ n 2 − / 4 3
1 2 γ γ .Tu,w e that get we Thus, 2. 0frtemmn n ot 47.We (4.7). to go and moment the for 20 9) d γ ( − − < n n n 9 √ n , n 1 1 2 , 1 Λ = 1 − 2 − − 2 γ − 1 1 Γ ) 2 n n n n n 2 2 2 3 1 2
) − − − − , − 1 . 2 < 1 2 1 n 2 3 γ Λ = 1 ≤ . . 2 . ≤ 2 .Tu,Γ Thus, 0. 6= 1 γ log d n ≤ d 2 1 ≤ γ − ≤ γ A d 2 2 n 9 ≤ 2 9 .Tu,Γ Thus, 0. 6= . =4 and 46 := , . 9 n . 1 − 3 n .I Γ If 0. 6= 2 Mathematica < B 2 20 := .If 0. 6= < 3 γ (5.1) (5.2) < 70. we 0
fΓ If hrfr,i ohcss ehv that have we cases, both in Therefore, that have we cases, both in Therefore, fΓ If iiigtruhteaoeieult ylog by inequality above the through Dividing eput We log by inequality above the through Dividing eput We where eueteoiia supinthat assumption original the use We Γ eel htte18t ovretof convergent 108-th the that reveals enwapyLma31o 54.W put We (5.4). on 3.1 Lemma apply now We hrfr,with Therefore, ewudhv that have would we htalo hmaea ot7.Tu,w aethat have we Thus, 72. most at are them of all that 1 aty eput we Lastly, .I Γ If 0. 6= 1 2 > > k ,te ehv that have we then 0, ,te ehv that have we then 0,
AUTHOR COPY:= τ := n 1 Γ τ 1 0 0 < 1 − o 10 log := log < < := ,then 0, 0 A 0 n |
o 10 log 2 Γ < γ log < ` ` =7and 7 := ecnrwie(.)as (5.3) rewrite can We . 1 o 10 log o 10 log log n | |
Γ ` 2 γ = and 2 o 10 log EDGT SSM FTREBLNIGNUMBERS BALANCING THREE OF SUMS AS REPDIGITS − log γ |
= ` n and − − o 10 log γ 3
` n n 0 < 0 µ B | | − o 10 log 3 3 e e d,k < 0 < Γ Γ 0 20 log + n := 2 1 µ < − < | Γ 2 := Γ − − d,k,s < log((4 2 + γ Γ 1 n γ | `τ − 1 1 | < 1 3 2 Thus, 72. log + log((4 | | n log((4 < ecluae ahvleo log(7 of value each calculated we log < n := < < 1 e − Γ 1 e τ d > e | 2 1 1 n log √ γ Γ Γ log((4 / / ssc that such is − 2 1 1 0 n ot 44.Nt hte that Note (4.4). to go and 100 d log + 2) .Hnee Hence 2. .Hnee Hence 2. d | + √ − ( + 9(1 √ γ − e = 1 M / log γ γ 2) µ log + 2) e = 1 ( + 9(1 = 1 n eget we , eget we , d,k d =10 := / / 2 log √ ( + 9(1 γ ( + 9(1 Γ − log γ | ( + 9(1 2) 2 | < n Γ Λ γ | n γ Λ 1 1 / log 1 50 4 n γ 3 − n | Γ Γ ( + 9(1 7 ( + 9(1 2 | Λ d 2 1 1 2 n γ ≤ uc optrsac in search computer quick A . | γ < · √ − q 1 3 − k < γ < < γ γ 108 n | n )) 2 2hlsi ohcases. both in holds 72 + 1 n − 4 < 3 γ n − .Tu,w e that get we Thus, 2. .Tu,w e that get we Thus, 2. 1 d γ ( n , + 4 γ 3 γ γ − n n √ n 1 1 > d γ n k n 2 2 n − ≤ 2 γ √ − n 1 2 )) 2 − 2 + 2 − 2 2 − n 1 n 6 n 2 72 + − .
2 n n 3 M 3 γ 1 − ) 2 2 < 3 − . γ s . ) n ) ≤ . 1 )) n ntecs that case the In 3 and γ . 2 )) d ,
, − n 2 <
q n ≤ − 108 3 < 2 ε n ) 9 γ d,k γ 3 , Γ /ε n 1 γ 1 log 2 1 2
n ≤ − γ ≤ d,k 1 − 1 < 2 ≥ n γ log ≤ γ d d 3 Λ = 1 ) 2 2 γ / . . ≤ 0 ≤ γ k n log γ . 2 1 00139047 ≤ 9 9 . . n , . Mathematica 1 γ 2 70 .Thus, 0. 6= − n found and , n 3 < > (5.3) (5.5) (5.4) 565 20, 0.
hr 1 where 55 as (5.5) eaanapyLma31o 56.W loput also We (5.6). on 3.1 Lemma apply again We Mathematica .Teeoe with Therefore, 0. on htalo hmaea ot7.Tu,w aethat have we Thus, 76. most at are them of all that found that ral mrvdteqaiyo rsnaino h paper. the of presentation of quality the improved greatly Acknowledgement 1]OI ONAININC.: FOUNDATION OEIS [10] cetd1.1.2019 12. 12. 2019 Accepted 7. 8. Received 566 2 UEU,Y—INTE .SKE,S.: M.—SIKSEK, Y.—MIGNOTTE, BUGEAUD, [2] 1 AEA .PNA .K.: G. A.—PANDA, BAHERA, [1] 8 AVE,E M.: E. MATVEEV, [8] A.—PETH DUJELLA, [3] 6 UA F.: LUCA, [6] GARC [5] G [4] 7 UA .NREY,B V.—TOGB B. F.—NORMENYO, LUCA, [7] 9 OMNO .V—UA F.—TOGB V.—LUCA, B. NORMENYO, [9] 98–105. ieeutosI ioac n ua efc powers. perfect Lucas and Fibonacci I. equations tine 156 leri ubr II numbers algebraic sequence gar. Seq. Integer 291–306. (1998), nIvsiaMathematics Izvestiya in n ZA S UZMAN ´ 1 2 21) 255–265. (2019), (2) 77 > AUTHOR COPY ≤ ´ ALOMEL IA 2 21)318–328. (2018) (2) 0.Hne hoe . holds. 2.1 Theorem Hence, 100. k nae ahmatqe uQu´ebec Mathem´antiques du Annales , := eel htte18t ovretof convergent 108-th the that reveals edgt ssm ftreFbncinumbers Fibonacci three of sums as Repdigits NHZ .LC,F.: S.—LUCA, ANCHEZ, 22 ´ n 2 21) r.I 19.2.3. ID Art. (2019), (2) 1 − ´ ,A C.—HERN A. I, A . n z.Rs.Aa.Nu e.Mat. Ser. Nauk Akad. Ross. Izv. , 3 =3 and 39 := nepii oe on o ooeeu ainllna omi h oaihsof logarithms the in form linear rational homogeneous a for bound lower explicit An h uhrtak h eee o h sflcmet n ugsin that suggestions and comments useful the for referee the thanks author The ( = ,A.: O, ˝ 64 n 6 20) 1217–1269. (2000), (6) 1 h nLn nylpdao nee Sequences Integer of Encyclopedia On-Line The − eeaiaino hoe fBkradDavenport and Baker of theorem a of generalization A 0 n ntesur ot ftinua numbers triangular of roots square the On B 2 NE HERN ANDEZ < ´ ( + ) := | `τ iercmiain ffcoil and factorials of combinations Linear n γ − ,A.: E, ´ 2 ,A.: E, ´ ecluae ahvleo log(39 of value each calculated we .DDAMULIRA M. n − REFERENCES 3 n 38 + 3 edgt ssm ftrePl numbers Pell three of sums as Repdigits ) 2 21) 169–188. (2014), (2) NE,S.: ANDEZ, µ ´ edgt ssm ftreLcsnumbers Lucas three of sums as Repdigits lsia n oua prahst xoeta Diophan- exponential to approaches modular and Classical ≤ d,k,s n.o ah (2) Math. of Ann. τ 64 4 n 1 and 142 E-mail AUSTRIA Graz A-8010 24/II Kopernikusgasse Technology of Theory University Number Graz and Analysis of Institute ssc that such is | 6 20) 2–8,(nRsin.Egihtranslation English Russian). (in 125–180, (2000), (6) ah Commun. Math. , < M 39 n dauiatga.t [email protected] [email protected]; : edgt ssm ftoPdvnnumbers Padovan two of sums as Repdigits =10 := 1 · ≤ γ ≤ − 76 n s 1 50 q . . := 108 hscnrdcsorassumption our contradicts This uc optrsac in search computer quick A . 163 n > 17 ioac Quart. Fibonacci , 2 2 20) 969–1018. (2006), (2) 2019, , − 1 21) 1–11. (2012), (1) 6 s M uisi iayrecurrence binary a in -units n 3 and q https://oeis.org ≤ 108 .J Math. J. Q. , 2 ecnrewrite can We 72. ε eid ah Hun- Math. Period. , /ε d,k,s d,k,s olq Math. Colloq. , ≥ ) 37 / 0 log 2 (1999), (2) . 00125 49 . γ (5.6) (195) and J. , >