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Solution of a problem by J. Rodriguez 
ee er OY Jy HOQNSUEZ 
by Pal Grønås 


Problem: “Find the largest strictly increasing series of integer numbers for which the 
Smarandacne Function is strictly decreasing”. 

My intention is to prove that there exists series of arbitrary finite length with the prop- 
erties described above. 

To begin with, define p, = 2, p2 = 3, p = 5 and more generaily, p, = the nth prime. 


Now we have the following lemma: 


Lemma: pe < Dewi < 2% forall KEN. (A). 

Proof: A theorem conjectured by Bertrand Russell and proven by Tchebychef states that 
for all natural numbers n > 2, there exists a prime p such that n < p < 2n. Using this 
theorem for n = py, we get pe < p < 2p; (x) for at least one prime p. The smallest prime 
> Pk IS Deroy. SO D> pkpi. But then it is obvious that (*) is satisfied by p = peai. Hence 


lame] 


Pe < Pk+i < 2py. T 


This lemma plays an important role in the proof of the following theorem: 

Theorem: Let n be a natural number > 2 and define the series {z;}}Zå of length n 
by r; = 2* pon, for ke GO ga eck n — l}. Then zę < zg4, and OSES S(Tk41) for all 
ke {0..... n — 2}. (N) l 

Proof: For k € {0,...,n —2} we have the following equivalences: zy < 241 © 

ae ee ee a, eee Pon- < 2P2n-~-1 according to Lemma (A). 
Futhermore pe,z_, > Pan—(n-1) = Pnzi 2 Ps = 5 > 2, so (pon_z, 2) = 1 forall ke 
{0.... n — 1}. Hence $(z,) = S(2* penx) = max{ 5(2*), S(pan-z) } = max{ 5(2*), Pon- }. 
Consequently pan-s < S(zz) < max{ 2k, pan-s } (*) since S(2*) < 2k. 

Moreover we know that pei; — pe > 2 for all k >2 because both Pe and p41 are 


odd integers. This inequality gives us the following result: 
n=l n=-1 
2 Pett ~ Pi)= Pn — Po = Pa —-3 > D0 2 = An —2), 
k=2 k=2 


SO pa 2 2n — 1 foralln > 3. In other words, paz; > 2n +1 > 2(n — 1) for n > 2, ie. 
Pon- > 2k fork =n — 1. The fact that pən-; increases and 2k decreases as k decreases 


from n — 1l to 0 implies that pm- > 2k forall k€ {0,... n — 1}. From this last 
inequality and (=) it follows that S(T) = Pon-z. This formula brings us to the conclusion: 
S(z;) = Pen-k > Pen-e-1 = S(T) for all & = {0,. 23% — 2}. m 


Example: For n =10 Theorem (N) generates the following series: 





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