Some identities involving the Smarandache ceil function

Scientia Magna Vol. 2 (2006), No. 1, 45-49

Some identities involving the Smarandache ceil function Wang Yongxing School of Science, Xidian University, Xi’an, Shaanxi, P.R.China Abstract For any fixed positive integer n, the Smarandache ceil function of order k is denoted by N ∗ → N and has the following definition: Sk (n) = min{x ∈ N : n | xk } (∀n ∈ N ∗ ) .

In this paper, we use the elementary methods to study the arithmetical properties of Sk (n), and give some identities involving the Smarandache ceil function.

Keywords

Smarandache ceil function; Arithmetical properties; Identity.

§1. Introduction For any fixed positive integer n, the Smarandache ceil function of order k is denoted by N ∗ → N and has the following definition: Sk (n) = min{x ∈ N : n | xk } (∀n ∈ N ∗ ) .

For example, S2 (1) = 1, S2 (2) = 2, S2 (3) = 3, S2 (4) = 2, S2 (5) = 5, S2 (6) = 6, S2 (7) = 7, S2 (8) = 4, S2 (9) = 3, · · · · · · . S3 (1) = 1, S3 (2) = 2, S3 (3) = 3, S3 (4) = 2, S3 (5) = 5, S3 (6) = 6, S3 (7) = 7, S3 (8) = 2, · · · · · · .

The dual function of Sk (n) is defined as Sk (n) = max{x ∈ N : xk | n} (∀n ∈ N ∗ ) .

For example, S2 (1) = 1, S2 (2) = 1, S2 (3) = 1, S2 (4) = 2, · · · . For any primes p and q with p 6= q, S2 (p2 ) = p, S2 (p2m+1 ) = pm and S2 (pm q n ) = S2 (pm )S2 (q n ).

These functions were introduced by F.Smarandache who proposed many problems in [1].

There are many papers on the Smarandache ceil function and its dual. For example, Ibstedt [2] and [3] studied these functions both theoretically and computationally, and got the following conclusions: (∀a, b ∈ N ∗ )(a, b) = 1 ⇒ Sk (ab) = Sk (a)Sk (b), α1 αr αr 1 α2 Sk (pα 1 p2 . · · · .pr ) = Sk (p1 ). · · · .Sk (pr ).

Ding Liping [4] studied the mean value properties of the Smarandache ceil function, and obtained a sharp asymptotic formula for it. That is, she proved the following conclusion:

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Wang Yongxing

No. 1

Let real number x ≥ 2, then for any fixed positive integer k ≥ 2, we have the asymptotic formula µ ¶¸ ³ 3 ´ X Y· x2 1 1 Sk (n) = ζ(2k − 1) 1− 1 + 2k−3 + O x 2 +² , 2 p(p + 1) p p n≤x

where ζ(s) is the Riemann zeta-function,

Y

denotes the product over all prime p, and ² denotes

p

any fixed positive number.

Lu Yaming [7] studied the hybrid mean value involving Sk (n) and d(n), and obtained the following asymptotic formula: µ ¶ ³ 1 ´ X 1 d(Sk (n)) = ζ(k)x + ζ + O x k+1 , k n≤x

where ζ(s) is the Riemann zeta-function and d(n) is the Dirichlet divisor function.

In this paper, we use the elementary methods to study the arithmetical properties of Smarandache ceil function and its dual, and give some interesting identities involving these functions. That is, we shall prove the following: Theorem 1. For any real number α > 1 and integer k ≥ 2, we have the identity: µ ¶ ∞ X (−1)n−1 2α − k − 1 Y k = α 1+ α , Skα (n) 2 +k−1 p p −1 n=1 where

Y

denotes the product over all prime p.

p

Theorem 2.

For any real number α > 1 and integer k ≥ 2, we also have the identities: ∞ X Sk (n) ζ(α)ζ(kα − 1) = α n ζ(kα) n=1

and

¸ · ∞ X (−1)n−1 Sk (n) ζ(α)ζ(kα − 1) (2α − 1)(2kα−1 − 1) = −1 , nα ζ(kα) 2α−2 (2kα − 1) n=1

where ζ(s) is the Riemann zeta-function.

Taking k = 2, α = 2 and 4, from our Theorems we may immediately deduce the following: Corollary 1. Let Sk (n) denotes the Smarandache ceil function, then we have the identities: ∞ X 1 (−1)n−1 2 (n) = 2 S 2 n=1

and

∞ X (−1)n−1 91 4 (n) = 102 .

S 2 n=1

Corollary 2. Let Sk (n) denotes the dual function of the Smarandache ceil function, then we have the identities: ∞ X 15 S2 (n) = 2 · ζ(3) 2 n π n=1

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47

Some identities involving the Smarandache ceil function

and

∞ X (−1)n−1 S2 (n) 6 = 2 · ζ(3).

2 n π n=1

§2. Proof of the theorems In this section, we shall complete the proof of Theorems. First we prove Theorem 1. For any real number α with α > 1, let f (α) =

∞ X (−1)n−1 .

Skα (n) n=1

Then from the multiplicative property of Sk (n) we may get ∞ X ∞ X 1 1 − α α S (2n − 1) t=1 n=1 Sk (2n − 1)Skα (2t ) n=1 k Ã∞ !à ! ∞ X X 1 1 = 1− .

S α (2n − 1) S α (2t ) n=1 k t=1 k

f (α) =

∞ X

(1)

For any prime p, note that Sk (p) = p, Sk (p2 ) = p, · · · , Sk (pk ) = p, Sk (pk+1 ) = p2 , Sk (p ) = pt+1 for any integers t ≥ 0 and 1 ≤ r ≤ k. So from the Euler product formula [6] we have Ã∞ ! ¶ X Yµ 1 1 1 1 = + + ··· + α k + ··· 1+ α S α (2n − 1) Sk (p) Skα (p2 ) Sk (p ) p n=1 k tk+r

p6=2

=

Yµ

1+

p p6=2

=

Yµ

p p6=2

=

k k k k + 2α + · · · + nα + (n+1)α + · · · pα p p p

k 1+ α p −1

¶

¶

µ ¶ 2α − 1 Y k 1+ α .

2α + k − 1 p p −1

(2)

Similarly, we also have 1−

∞ X

1

S α (2t ) t=1 k

=1−

2α

k . −1

Combining (1), (2) and (3) we may immediately get the identity µ ¶ ∞ X 2α − k − 1 Y k (−1)n−1 = 1 + .

Skα (n) 2α + k − 1 p pα − 1 n=1 This proves Theorem 1.

(3)

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Wang Yongxing

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Now we prove Theorem 2. For any real number α with α > 1 and integer k ≥ 2, let g(α) =

∞ X Sk (n) . nα n=1

Then from the multiplicative property of Sk (n) and the Euler product formula [6] we have g(α)

=

Yµ

1+

p

Yµ

Sk (p) Sk (p2 ) Sk (p3 ) + + + ··· pα p2α p3α

¶

1 1 1 p p + 2α + · · · + (k−1)α + kα + · · · + (2k−1)α + · · · α p p p p p p à !

1 1 1 Y 1 − pkα p 1 − pkα p2 1 − pkα = + kα + 2kα + ··· p 1 − p1α p 1 − p1α 1 − p1α p à ! ¶ 1 Y 1 − pkα Yµ p p2 = 1 + kα + 2kα + · · · p p 1 − p1α p p à !

1 Y 1 − pkα Y 1 = 1 1 1 − pα 1 − pkα−1 p p =

=

¶

1+

ζ(α)ζ(kα − 1) . ζ(kα)

This proves the first formula of Theorem 2.

Similarly, we can also get ∞ X (−1)n−1 Sk (n) nα n=1

∞ ∞ ∞ X Sk (2n − 1) X X Sk (2n − 1) Sk (2t ) − (2n − 1)α (2n − 1)α 2tα n=1 n=1 t=1 !à ! Ã∞ ∞ X X Sk (2n − 1) Sk (2t ) = 1− (2n − 1)α 2tα t=1 n=1 à ! µ ¶ ∞ X Sk (2t ) Y Sk (p) Sk (p2 ) Sk (p3 ) = 1− 1+ + + + ··· 2tα pα p2α p3α p t=1

=

=

ζ(α)ζ(kα − 1) ζ(kα)

1−

p6=2 ∞ X t=1

1+

∞ X Sk (2t ) t=1 α

=

Sk (2t ) 2tα 2tα

· ¸ ζ(α)ζ(kα − 1) (2 − 1)(2kα−1 − 1) − 1 . ζ(kα) 2α−2 (2kα − 1)

This completes the proof of Theorem 2.

Taking k = 2, then from Theorem 1 we have µ ¶ ∞ X (−1)n−1 2α − 3 Y 2 2α − 3 Y pα + 1 2α − 3 ζ 2 (α) = 1 + = α = α , α α α α S2 (n) 2 +1 p p −1 2 +1 p p −1 2 + 1 ζ(2α) n=1

(4)

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Some identities involving the Smarandache ceil function

49

where ζ(α) is the Riemann zeta-function.

Note that ζ(2) = π 2 /6, ζ(4) = π 4 /90 and ζ(8) = π 8 /9450, from (4) we may immediately deduce Corollary 1.

Corollary 2 follows from Theorem 2 with k = 2.

References [1] F. Smarandache, Only problems, Not solutions, Chicago, Xiquan Publ. House, 1993.

[2] Ibstedt, Surfing on the Ocean of Numbers-A Few Smarandache Notions and Similar Topics, Erhus University press, New Mexico, 1997.

[3] Ibstedt, Computational Aspects of Number Sequences, American Research Press, Lupton, 1999.

[4] S.Tabirca and T. Tabirca, Some new results concerning the Smarandache ceil function, Smarandache Notions Journal, 13(2002), 30-36.

[5] Ding Liping, On the mean value of Smarandache ceil function, Scientia Magna, 1(2005), No. 2, 74-77.

[6] Tom M.Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, 1976.

[7] Lu Yaming, On a dual function of the Smarandache ceil function, Research on Smarandache problems in number theory (edited by Zhang Wenpeng, Li Junzhuang and Liu Duansen), Hexis, Phoenix AZ, 2005, 55-57.

[8] Ji Yongqiang, An equation involving the Smarandache ceil function, Scientia Magna, 1(2005), No. 2, 149-151.