1991 Mathematics Subject Classification. Primary 11A25; Secondary 11B83.
Some properties of the pseudo-Smarandache function
Richard Pinch
2 Eldon Road, Cheltenham, Glos GL52 6TU, U.K. E-mail address : rgep@chalcedon.demon.co.uk
Abstract. Charles Ashbacher [1] has posed a number of questions relating to the pseudo-Smarandache function Z ( n ) . In this note we show that the ratio of consecutive values Z ( n + 1 ) / Z ( n ) and Z ( n − 1 ) / Z ( n ) are unbounded; that Z ( 2 n ) / Z ( n ) is unbounded; that n / Z ( n ) takes every integer value infinitely often; and that the series ∑ n 1 / Z ( n ) α is convergent for any α > 1 .
We define the m -th triangular number T ( m ) = m ( m + 1 ) 2 . Kashihara [2] has defined the pseudo-Smarandache function Z ( n ) by Z ( n ) = min { m : n | T ( m ) } . Charles Ashbacher [1] has posed a number of questions relating to the pseudo-Smarandache function Z ( n ) . In this note we show that the ratio of consecutive values Z ( n ) / Z ( n − 1 ) and Z ( n ) / Z ( n + 1 ) are unbounded; that Z ( 2 n ) / Z ( n ) is unbounded; and that n / Z ( n ) takes every integer value infinitely often. He notes that the series ∑ n 1 / Z ( n ) α is divergent for α = 1 and asks whether it is convergent for α = 2 . He further suggests that the least value of α for which the series converges "may never be known" . We resolve this problem by showing that the series converges for all α > 1 .
2 Some properties of the pseudo-Smarandache function
We record some elementary properties of the function Z .
Lemma 1.
(1) If n ≥ T ( m ) then Z ( n ) ≥ m . Z ( T ( m ) ) = m .
(2) For all n we have n < Z ( n ) .
(3) Z ( n ) ≤ 2 n − 1 , and if n is odd then Z ( n ) ≤ n − 1 .
(4) If p is an odd prime dividing n then Z ( n ) ≥ p − 1 .
(5) Z ( 2 k ) = 2 k + 1 − 1 .
(6) If p is an odd prime then Z ( p k ) = p k − 1 and Z ( 2 p k ) = p k − 1 or p k according as p k ≡ 1 or 3 m o d 4 .
We shall make use of Dirichlet's Theorem on primes in arithmetic progression in the following form.
Lemma 2. Let a , b be coprime integers. Then the arithmetic progression a + b t is prime for infinitely many values of t .
3 Successive values of the pseudo-Smarandache function
Using properties (3) and (5), Ashbacher observed that | Z ( 2 k ) − Z ( 2 k − 1 ) | > 2 k and so the difference between the conecutive values of Z is unbounded. He asks about the ratio of consecutive values.
Theorem 1. For any given L > 0 there are infinitely many values of n such that Z ( n + 1 ) / Z ( n ) > L , and there are infinitely many values of n such that Z ( n − 1 ) / Z ( n ) > L .
Proof. Choose k ≡ 3 m o d 4 , so that T ( k ) is even and k divides T ( k ) . We consider the conditions k | m and ( k + 1 ) | ( m + 1 ) . These are satisfied if m ≡ k m o d k ( k + 1 ) , that is, m = k + k ( k + 1 ) t for some t . We have m ( m + 1 ) = k ( 1 + ( k + 1 ) t ) ⋅ ( k + 1 ) ( 1 + k t ) , so that if n = k ( k + 1 ) ( 1 + k t ) / 2 we have n | T ( m ) . Now consider n + 1 = T ( k ) + 1 + k T ( k ) t . We have k | T ( k ) , so T ( k ) + 1 is coprime to both k and T ( k ) . Thus the arithmetic progression T ( k ) + 1 + k T ( k ) t has initial term coprime to its increment and by Dirichlet's Theorem contains infinitely many primes. We find that there are thus infinitely many values of t for which n + 1 is prime and so Z ( n ) ≤ m = k + k ( k + 1 ) t and Z ( n + 1 ) = n = T ( k ) ( 1 + k t ) . Hence Z ( n + 1 ) Z ( n ) ≥ n m = T ( k ) + k T ( k ) t k + 2 T ( k ) t > k 3 . A similar argument holds if we consider the arithmetic progression T ( k ) − 1 + k T ( k ) t . We then find infinitely many values of t for which n − 1 is prime and Z ( n − 1 ) Z ( n ) ≥ n − 2 m = T ( k ) − 2 + k T ( k ) t k + 2 T ( k ) t > k 4 . The Theorem follows by taking k > 4 L . □
We note that this Theorem, combined with Lemma 1(2), gives another proof of the result that the difference of consecutive values is unbounded.
4 Divisibility of the pseudo-Smarandache function
Theorem 2. For any integer k ≥ 2 , the equation n / Z ( n ) = k has infinitely many solutions n .
Proof. Fix an integer k ≥ 2 . Let p be a prime ≡ − 1 m o d 2 k and put p + 1 = 2 k t .
Put n = T ( p ) / t = p ( p + 1 ) / 2 t = p k . Then n | T ( p ) so that Z ( n ) ≤ p . We have p | n , so Z ( n ) ≥ p − 1 : that is, Z ( n ) must be either p or p − 1 . Suppose, if possible, that it is the latter. In this case we have 2 n | p ( p + 1 ) and 2 n | ( p − 1 ) p , so 2 n divides p ( p + 1 ) − ( p − 1 ) p = 2 p : but this is impossible since k > 1 and so n > p . We conclude that Z ( n ) = p and n / Z ( n ) = k as required. Further, for any given value of k there are infinitely many prime values of p satisfying the congruence condition and hence infinitely many values of n = T ( p ) such z / Z ( n ) = k . □
5 Another divisibility question
Theorem 3. The ratio Z ( 2 n ) / Z ( n ) is not bounded above.
Proof. Fix an integer k . Let p ≡ − 1 m o d 2 k be prime and put n = T ( p ) . Then Z ( n ) = p . Consider Z ( 2 n ) = m . We have 2 k p | p ( p + 1 ) = 2 n and this divides m ( m + 1 ) / 2 . We have m ≡ ε m o d p and m ≡ δ m o d 2 k + 1 where each of ε , δ can be either 0 or − 1 .
Let m = p t + ε . Then m ≡ ε − t ≡ δ m o d 2 k : that is, t ≡ ε − δ m o d 2 k .
This implies that either t = 1 or t ≥ 2 k − 1 . Now if t = 1 then m ≤ p and T ( m ) ≤ T ( p ) = n , which is impossible since 2 n ≤ T ( m ) . Hence t ≥ 2 k − 1 .
Since Z ( 2 n ) / Z ( n ) = m / p > t / 2 , we see that the ratio Z ( 2 n ) / Z ( n ) can be made as large as desired. □
6 Convergence of a series
Ashbacher observes that the series ∑ n 1 / Z ( n ) α diverges for α = 1 and asks whether it converges for α = 2 .
In this section we prove convergence for all α > 1 .
Lemma 3. log n ≤ ∑ m = 1 n 1 m ≤ 1 + log n ; 1 2 ( log n ) 2 − 0.257 ≤ ∑ m = 1 n log m m ≤ 1 2 ( log n ) 2 + 0.110 for n ≥ 4 .
Proof. For the first part, we have 1 / m ≤ 1 / t ≤ 1 / ( m − 1 ) for t ∈ [ m − 1 , m ] .
Integrating, 1 m ≤ ∫ m − 1 m 1 t d t ≤ 1 m − 1 . Summing, ∑ 2 n 1 m ≤ ∫ 1 n 1 t d t ≤ ∑ 2 n 1 m − 1 , that is, ∑ 1 n 1 m ≤ 1 + log n and log n ≤ ∑ 1 n − 1 1 m . The result follows.
For the second part, we similarly have log m / m ≤ log t / t ≤ log ( m − 1 ) / ( m − 1 ) for t ∈ [ m − 1 , m ] when m ≥ 4 , since log x / x is monotonic decreasing for x > e .
Integrating, log m m ≤ ∫ m − 1 m log t t d t ≤ log ( m − 1 ) m − 1 . Summing, ∑ 4 n log m m ≤ ∫ 3 n log t t d t ≤ ∑ 4 n log ( m − 1 ) m − 1 , that is,
∑ 1 n log m m − log 2 2 − log 3 3
≤ 1 2 ( log n ) 2 − 1 2 ( log 3 ) 2
≤ ∑ 1 n log m m − log n n − log 2 2 .
We approximate the numerical values log 2 2 + log 3 3 − 1 2 ( log 3 ) 2 < 0.110 and log 2 2 − 1 2 ( log 3 ) 2 > − 0.257 . to obtain the result. □
Lemma 4. Let d ( m ) be the function which counts the divisors of m .
For n ≥ 2 we have ∑ m = 1 n d ( m ) / m < 7 ( log n ) 2 .
Proof. We verify the assertion numerically for n ≤ 6 . Now assume that n ≥ 8 > e 2 . We have
∑ m = 1 n d ( m ) m = ∑ m = 1 n ∑ d e = m 1 m = ∑ d ≤ n ∑ d e ≤ n 1 d e
= ∑ d ≤ n 1 d ∑ e < n / d 1 e ≤ ∑ d ≤ n 1 d ( 1 + log ( n / d ) )
≤ ( 1 + log n ) 2 − 1 2 ( log n ) 2 + 0.257
= 1.257 + 2 log n + 1 2 ( log n ) 2
< 4 3 ( log n 2 ) 2 + 2 log n ( log n 2 ) + 1 2 ( log n ) 2
< 2 ( log n ) 2 .
Lemma 5. Fix an integer t ≥ 5 . Let e t > Y > e ( t − 1 ) / 2 . The number of integers n with e t − 1 < n ≤ e t such that Z ( n ) ≤ Y is at most 196 Y t 2 .
Proof. Consider such an n with m = Z ( n ) ≤ Y . Now n | m ( m + 1 ) , say k 1 n 1 = m and k 2 n 2 = m + 1 , with n = n 1 n 2 . Thus k = k 1 k 2 = m ( m + 1 ) / n and k 1 n 1 ≤ Y . The value of k is bounded below by 2 and above by m ( m + 1 ) / n ≤ 2 Y 2 / e t − 1 = K , say. Given a pair ( k 1 , k 2 ) , the possible values of n 1 are bounded above by Y / k 1 and must satisfy the congruence condition k 1 n 1 + 1 ≡ 0 modulo k 2 : there are therefore at most Y / k 1 k 2 + 1 such values.
Since Y / k ≥ Y / K = e t − 1 / 2 Y > 1 / 2 e , we have Y / k + 1 < ( 2 e + 1 ) Y / k < 7 Y / k .
Given values for k 1 , k 2 and n 1 , the value of n 2 is fixed as n 2 = ( k 1 n 1 + 1 ) / k 2 .
There are thus at most ∑ k ≤ K d ( k ) possible pairs ( k 1 , k 2 ) and hence at most ∑ k ≤ K 7 Y d ( k ) / k possible quadruples ( k 1 , k 2 , n 1 , n 2 ) . We have K > 2 so that the previous Lemma applies and we can deduce that the number of values of n satisfying the given conditions is at most 49 Y ( log K ) 2 . Now K = 2 Y 2 / e t − 1 < 2 e t + 1 so log K < t + 1 + log 2 < 2 t . This establishes the claimed upper bound of 196 Y t 2 . □
Theorem 4. Fix 1 2 < β < 1 and an integer t ≥ 5 . The number of integers n with e t − 1 < n ≤ e t such that Z ( n ) < n β is at most 196 t 2 e β t .
Proof. We apply the previous result with Y = e β t . The conditions of β ensure that the previous lemma is applicable and the upper bound on the number of such n is 196 e β t t 2 as claimed. □
Theorem 5. The series ∑ n = 1 ∞ 1 Z ( n ) α is convergent for any α > 2 .
Proof. We note that if α > 2 then 1 / Z ( n ) α < 1 / n α / 2 and the series is convergent.
So we may assume 2 < α ≤ 2 . Fix β with 1 / α < β < α / 2 . We have 1 2 < β < 1 2 < α / 2 .
We split the positive integers n > e 4 into two classes A and B . We let class A be the union of the A t where, for positive integer t ≥ 5 we put into class A t those integers n such that e t − 1 < n ≤ e t for integer t and Z ( n ) ≤ n β . All values of n with Z ( n ) > n β we put into class B . We consider the sum of 1 / Z ( n ) α over each of the two classes. Since all terms are positive, it is sufficient to prove that each series separately is convergent.
Firstly we observe that for n ∈ B , we have 1 / Z ( n ) α < 1 / n α β and since α β > 1 the series summed over the class B is convergent.
Consider the elements n of A t : so for such n we have e t − 1 < n ≤ e t and Z ( n ) < n β . By the previous result, the number of values of n satisfying these conditions is at most 196 t 2 e β t . For n ∈ A t , we have Z ( n ) ≥ n , so 1 / Z ( n ) α ≤ 1 / n α / 2 < 1 / e α ( t − 1 ) / 2 . Hence the sum of the subseries ∑ n ∈ A t 1 / Z ( n ) α is at most 196 e α / 2 t 2 e ( β − α / 2 ) t . Since β < α / 2 for α > 2 , the sum over all t of these terms is finite.
We conclude that ∑ n = 1 ∞ 1 / Z ( n ) α is convergent for α > 2 □
Proof. We fix β 0 = 1 > β 1 > ⋯ > β r = 1 2 with β j < α β j + 1 for 0 ≤ j ≤ r − 1 .
We define a partition of the integers e t − 1 < n < e t into classes B t and C t ( j ) , 1 ≤ j ≤ r − 1 . Into B t place those n with Z ( n ) > n β 1 . Into C t ( j ) place those n with n β j + 1 < Z ( n ) < n β j . Since β r = 1 2 we see that every n with e t − 1 < n < e t is placed into one of the classes.
The number of elements in C t ( j ) is at most 196 t 2 e β j t and so ∑ n ∈ C t ( j ) 1 Z ( n ) α < 196 t 2 e β j t e − β j + 1 α ( t − 1 ) = 196 e β j + 1 α t 2 e ( β j − α β j + 1 ) t . For each j we have β j < α β j + 1 so each sum over t converges.
The sum over the union of the B t is bounded above by ∑ n 1 n α β 1 , which is convergent since α β 1 > β 0 = 1 .
We conclude that ∑ n = 1 ∞ 1 / Z ( n ) α is convergent. □
Charles Ashbacher, Pluckings from the tree of Smarandache sequences and functions, American Research Press, 1998, http://www.gallup.unm.edu/ ∼ smarandache/Ashbacher-pluckings.pdf.
K. Kashihara, Comments and topics on Smarandache notions and problems, Erhus University Press, Vail, AZ, USA, 1996.