Smarandache Function Journal, Vol. 6, No. 1-2-3, 1995, pp. 16-20. SOME ?ROPERTI ES OF SMARANDACHE FUNCT! ONS OF TnE TYPE ! by Ion and Seleacu Vasile aalaceno~u Depar~men~ of ma~hema~ics of Cralova Universi~y We consider ~he cons~ruc~ion of Smarandache func~ions - of ~he ~ype I S C peiN • p pr i m) which are defined in (1) P and [2] as follows: S n !N-~lN- S 1. (Ie) "" 1 S 1"1 (Ie) = 1m.~ajx~r { S p J( i J .Ie)} 1. , i i for ~ = P, Pz Z Pr r In ~his .paper ~here are present.ed some pi"' oper t.i es of t.hese functions. fie shall s~udy ~he monot.oniclt.y of each function and also the mono~onicity of some subsequences of t.he sequence ( S ) IN-' 1"\ 1"1 • 1. Propos! tion. The function Sn is monotonous increasing for every positiv integer n. Proof. The func~ion S , is abviously monot.onous increasing. Ie < Ie where 1e,Ie e tN-, Supposing that n is a prime number Le~ 1 Z 1 Z lc lc and t.aking accont. t.ha~ CS(n1 e z ))! = mult.iple n 1 - multiple n Z 16 It. result.s t.hat. S (k )!S S (k ), t.herefore S is monot.onous increa- l"I 1 n Z n sing. Let. so: ) = { S (i . k )} = S (i . k ) n 1 1~1~1c P. J 1 P m 1 J m SO: ) = { S (i .k )} = S (i .k ) n Z 1~~r P j J Z P t t Z Because S (i .k) !S S (i . k ) ~ S (i.k) Pm m 1 Pm m Z P t t Z it. result..s t.hat. S (k) !S S (k) so Sis monot.onous i ncr easi ng. l"I 1 l"I Z n 2. Proposition. The sequence of funct..ions (Sp~ )i~- is monot.onous increasing, for every prime number p. Proof. For any t..wo nombers i ,i e IN i < i and for any n~- 1 Z - 2. Z we have = S P i 1 ( n) = S p (i 2. . n) ~ S p Ci Z . n) S p 1. Z( n) t.herefore SP.1 t !SSP.1 z Hence t..he sequence { Sp~ }i~- is monot.onous increasing for every pr i me number p. 3. Proposition. Let p and q t.wo given prime numbers. If p<q t..hen - < S (k) S (k) k E IN P q <p> <p> < p) Proof. Let.. t..he sequence of coefficient.s (see (2]) a ,a ,. ,a , . 2. Z 8 • Every k e IN can be uniquely writ..t..en as + t. a (P> (1) 8 2. 17 where 0 S t .. S p-l,for i = 1,s-1 ,and o ::s t. S p, s The p~ocedu~e of passing from k t.o k+1 in formule (1) is t. • is increasing wit.h a un~t.y. C~I.) if t. • can not. lncrease wit.h a unit.y,t.hen t. .-1 is increaslng wit.h a unit.y and t. • == 0 (~,,~) 1 f nei t.he t. , nor t. are not. i ncreasi ng wi t.h a uni t. y S .-1 = = t.hen t. .-z is increaslng wit.h a unit.y and t. • t. 0 8-1 The procedure is cont.inued in t.he same way unt.il w. obt.ain t.he expresion of k+l. = Denot.ing S Ck+l) - S (Ie) t.he leap of t.he funct.ion S p p p when we pass from Ie t.o Ie +1 corresponding t.o t.h. procedure described above. W. find t.hat. -' in t.h. case (,,) A~Sp) ~ I' .. in t.h. cas. (;, ) A~Sp) :a 0 in t.h. case e i. i.i.> AIcCSp) • 0 It. is abviously seen t.hat.: Sen) == E,\Cn) + SCi) P lc=1 P P = Analogously we writ.e Sen) + S (1) q q Taking int.o account. t.hat. S (1) == P < q s S C1) and using t.he p q procedure of passing from k t.o k+l we deduce t.hat. t.he number of leaps wit.h zero value of S is great.er t.hen t.h. of n~ p leaps wit.h zero value of S • respect.ively t.he number of' l.aps wit.h q val ue p of Sis 1 ess t.hen t.he number of leaps of S wi t.h val ue p q 18 q ~~ resul~ ~ha~ E ~ CS ) + SCi) < E ~ CS ) + SCi) (2) lc p p lc q q lc=1 k=1 • Hence S Cn) < S Cn) n e LN. p q As an example we give a table ... i~h S z and S a for 0<n<21 k 1 2 3 4 9 6 7 8 9 10 11 12 13 14 19 16 17 18 19 20 leap 2 0 2 2 0 0 2 2 0 2 2 0 0 0 2 0 2 2 2 ~he s CJc) 2 4 4 6 8 9 8 10 12 12 14 16 16 16 16 18 18 20 22 24 z the leap 330 3 3 303 330 0 3 3 3 0 3 3 3 S Ck) 3 6 9 9 12 19 18 18 21 24 27 27 27 30 33 36 36 39 42 49 3 Hence S Ck) < S Ck) for k = 1, 2, . . . ,20. z a 4. Remark. For any mono~onous increasing sequence of prime numbers i~ resul~s ~ha~ < S < ... Pz If n = P\i. PzI. P\l. and Pi < Pz < .. <pl then S Ck) = ( S 1.0:) } = S i.Ck) = S Cik) n 1~~l P P P j l t 5. Proposition. If P and q are prime numbers and p.i < q then S i.<S . P q Proof. Because p.i < q i~ results < S .. C 1) ~ p. i q .. S (1) (3) p q = and S .. C k) SCi k) ~ i S Ck) P P P From (3) passing from k to k+l, we deduce S i.) (4) P Taking into account the proposition 3. from (4) it results that ... hen ... 61 pass from k to k+l "'61 ob~ain 19 ., '" ~(Spi ) ~ i ~(Sp ) ~ i. P < q and iE 41lc (Sp ) ~ E ~(SqJ (5) k:1 1e:1 ..... Because have n n S p \.Cn) = S \.(1) + 1: "\: CS i.) ~ S \.(1) + ~ r ,,\:CS ) P 1e=1 P P Ie:! P and n S q (n) = S q (1) + E .1Jc (S ) lc=1 q from (3) and (5) ~~ resul~s S p t.Cn) ~ S q (n) n • [N- 6. Propos! tion. I~ P is a prima number t.han S ,., < S p ~or every n < p n a prime number from n < P. using t.he proposit.ion 3 Pr~~f.lf ~s - it. result.s S (Ie) < S (Ie) for Ie e IN • If n is a. composed, ~ha~ 1'\ p is ~hen S (Ie) = max < S ~.C Ie) } = S i.Ck). 1'\ P J PJ J Because n < P i~ result.s P r < p and using ~h. proposit.ion S r i. and knowing t.hat. i P ~ P r < pit. :-esul ~s t.ha.t. S i. C Ie) ~ S C Ie) r r r pI" r p t.herefore ~or Ie • IN- S (Ie) < S (Ie) n p R_j_r.n.c.s (13 Balacenoiu I , Smarandache Numerical Funct.ions in Smarandacha Func~ion Journal nr. 4 / lQQ4. (c:) Smaranda.che F .• A funct.ion in ~he number t.heory. .. An. Un1v. Ti mi soar a" vol XVI I I , f asc 1. pp -rg-88.