Curious Bounds for Floor Function Sums

1 47 6 11 Journal of Integer Sequences, Vol. 1 (018), Article 18.1.8 Curious Bounds for Floor Function Sus Thotsaporn Thanatipanonda and Elaine Wong 1 Science Division Mahidol University International College Nakhon Patho, 7170 Thailand thotsaporn@gail.co wongey@gail.co Abstract Jacobsthal was the first to study a certain kind of su of floor functions. His work was followed by that of Carlitz, Grison, and Tverberg. More recently, Onphaeng and Pongsriia proved soe sharp upper and lower bounds for the sus of Jacobsthal and Tverberg. In this paper, we devise concise forulas for these sus, which we then use to prove the upper and lower bounds claied by Tverberg. Furtherore, we present conjectural lower and upper bounds for these sus. 1 Introduction In 1957, Jacobsthal [] defined and studied a function of the for a1 +a +k a1 +k a +k f ({a 1,a },k) = + for fixed Z + with a 1,a,k Z. He also defined the functions k S ({a 1,a },K) = f ({a 1,a },k), k=0 0 a 1,a,K 1. 1 Correspondence should be addressed to Elaine Wong: wongey@gail.co. 1

It is iportant to note that we can take advantage of the -periodicity of f, and so we restrict our a 1,a, and K values accordingly for the su. Jacobsthal, and then later Carlitz [1], Grison [], and Tverberg [5], proved S ({a 1,a },K) 0. In 01, Tverberg [5] proposed a generalized notation for these su functions for any set A = {a 1,...,a n } with 0 a 1,...,a n,k 1 and n = A, that is, S ({a 1,...,a n },K) = k=0 T [1,n] ( 1) n T k + i T a i He also claied without proof the other upper and lower bounds of S for sets {a 1,a } and {a 1,a,a } (i.e., n =,). In 017, Onphaeng and Pongsriia [4] furnished a proof for the upper bounds when n is even and 4 and the lower bounds when n is odd and. In this paper, we investigate the bounds for S for all n Z + and supply the issing proofs of Tverberg s upper bounds. Furtherore, we conjecture all the bounds for S not previously entioned and suarize the findings in Table 1. Authors who claied their bounds without proof are denoted with an asterisk (*). Otherwise, a proof is given in their corresponding paper.. n Lower Bound Lower Bound Credit Upper Bound Upper Bound Credit 1 0 Trivial 1 Trivial Jacobsthal; 0 Carlitz; Tverberg*; Grison; Corollary 8 Tverberg; Theore 9 Tverberg*; Tverberg*; Onphaeng, Corollary 14 Pongsriia 4 4 Onphaeng, Conjecture 17* (Conjecture) Pongsriia odd n Onphaeng, ( 5) Pongsriia (Conjectures) Conjecture 17* even (Conjectures) Conjecture 17* n Onphaeng, ( 5) Pongsriia Table 1: Bounds for S ({a 1,...,a n },K)

We discuss the case of n = 1 in this section as it sets the foundation for the ain strategy that we use to prove higher cases. We begin with an explicit definition of Jacobsthal s su. Definition 1. For any Z and a 1,K Z + {0}, S ({a 1 },K) = ( a1 +k k=0 ) k. The su can be written concisely without using the suation sybol, which we show below. Note that the periodicity that existed in the n case does not exist here. However, we only prove the following proposition for 0 K 1 because that is all that is needed for higher values of n. We let a od denote the inial non-negative representative in the Z/-equivalence class. Proposition. For 0 K 1 and a 1 Z + {0}, a1 S ({a 1 },K) = (K +1)+ax(0,(a 1 od )+K +1). Proof. We observe a1 +k = a1 k + + {a1 } + { } k, where {x} = x x, the fractional part of x. This notation is distinguished fro the usual set notation according to context. Furtherore, since 0 k 1, {a1 } { } k a1 od + = + k. The result is derived as follows: ( ) a1 +k k ( a1 = + k=0 = k=0 a1 (K +1)+ k + {a1 } + a1 od + k k=0 { } k k )

= + = = a1 (K +1)+ k= (a 1 od ) a1 a1 (a 1 od ) 1 k=0 (a1 od )+k (a 1 od ) 1 (K +1)+ 0+ k=0 (a1 od )+k k= (a 1 od ) (K +1)+ax(0,(a 1 od )+K +1). 1 The bounds for S ({a 1 },K) are then easily attained fro Proposition. Corollary. For 0 a 1,K 1, 0 S ({a 1 },K) 1. In particular, the axiu occurs precisely when a 1 = K = 1. Proof. The result follows fro the fact that S ({a 1 },K) = ax(0,a 1 +K +1). In the following sections, we use siilar ethods to provide bounds when n > 1 for the sus S ({a 1,...,a n },K). Lower and upper bounds for n = Carlitz [1], Grison [], Jacobsthal [], and Tverberg [5] all proved the lower bound for n =, while Tverberg was first to ention the upper bound for this case. In this section, we prove the upper bound by introducing a new for for the su just as before. We also use this for to give a new proof for its lower bound. We do this by first writing out Jacobsthal s su explicitly, and then aking a generalization of Proposition. Definition 4. For any Z + and any a 1,a,K Z + {0}, S ({a 1,a },K) = ( a1 +a +k a1 +k a +k + k=0 ) k. 4

We show that the su can be written concisely without using the suation sybol, siilar to Proposition. Proposition 5. For 0 K 1, and any a 1,a Z + {0}, ( a1 +a a1 a ) S ({a 1,a },K) = (K +1) +ax(0,((a 1 +a ) od )+K +1) ax(0,(a 1 od )+K +1) ax(0,(a od )+K +1). Proof. Rewriting the two-variable su in Definition 4 as a series of one-variable sus, S ({a 1,a },K) = S ({a 1 +a },K) S ({a 1 },K) S ({a },K), allows us to apply Proposition to each su to get our result. A syetry exists in Proposition 5. We outline the pattern in the lea below. This, along with a partial result in Theore 7, gives us the desired upper and lower bounds in Corollary 8 and Theore 9, respectively. Lea 6. (Mirrored Sus) For 0 a 1,a 1 and 0 K, S ({a 1,a },K) = S ({ a 1, a }, K). Proof. It is enough to show the clai for 0 a 1 + a. Otherwise, we have that ( a 1 )+( a ) <, in which case we can use a siilar arguent by substituting a 1 with a 1 and a with a. For the case a 1 = a = 0, the result trivially holds by Definition 4. For the case 0 < a 1 +a <, Proposition 5 siplifies to S ({a 1,a },K) = ax(0,a 1 +a +K +1) ax(0,a 1 +K +1) ax(0,a +K +1). (1) Furtherore, we note that < (a 1 +a ) <, which gives S ({ a 1, a }, K) = 1 K +ax(0, (a 1 +a ) K 1) Now consider the following equations that use the fact that ax(0, a 1 K 1) ax(0, a K 1). () ax(0,x) ax(0, x) = x, for all x R. 5

ax(0,a 1 +a +K +1) ax(0, a 1 a K + 1) = a 1 +a +K +1, ax(0, a 1 K 1) ax(0,a 1 +K +1) = a 1 K 1, ax(0, a 1 K 1) ax(0,a +K +1) = a K 1. () (4) (5) Then, ()+(4)+(5) confirs (1) = (). Finally, we consider a 1 +a =. Here, Proposition 5 gives S ({a 1,a },K) = K +1 ax(0,a 1 +K +1) ax(0,a +K +1), S ({ a 1, a }, K) = (K +1) ax(0, a 1 K 1) ax(0, a K 1). (6) (7) In this case, (4)+(5) confirs (6) = (7). This concludes the proof. We show the upper bound for half the range of K using differences. Theore 7. For 0 a 1,a 1 and 0 K 1, S ({a 1,a },K). Proof. We show the stronger result, that for 0 K 1, S ({a 1,a },K) K +1. For the case K = 0, a1 +a S ({a 1,a },0) = +ax(0,((a 1 +a ) od ) +1) ax(0,a 1 +1) ax(0,a +1) a1 +a = +0 0 0 1. For the case 1 K 1, it is enough to show that := S ({a 1,a },K) S ({a 1 +1,a },K 1) 1. 6

By using Proposition 5 we explicitly write out the two sus a1 +a S ({a 1,a },K) = (K +1) +ax(0,((a 1 +a ) od )+K +1) ax(0,a 1 +K +1) ax(0,a +K +1), a1 +a +1 S ({a 1 +1,a },K 1) = K +ax(0,((a 1 +a +1) od )+K ) ax(0,a 1 +K +1) ax(0,a +K ). We deterine according to the possible values of a 1 +a and a +K +1. Case 1: (a 1 +a < and a +K +1 0) If a 1 +a < 1, then both sus are the sae. S ({a 1,a },K) = 0+ax(0,a 1 +a +K +1) ax(0,a 1 +K +1) 0, S ({a 1 +1,a },K 1) = 0+ax(0,(a 1 +1)+a +K ) ax(0,(a 1 +1)+K ) 0. If a 1 +a = 1, then both sus evaluate to K. Therefore, we get = 0. Case : (a 1 +a < and a +K +1 > 0) Again, we assue a 1 +a < 1. S ({a 1,a },K) = 0+ax(0,a 1 +a +K +1) ax(0,a 1 +K +1) (a +K +1), S ({a 1 +1,a },K 1) = 0+ax(0,(a 1 +1)+a +K +1) ax(0,(a 1 +1)+K ) (a +K ). Therefore, = S ({a 1,a },K) S ({a 1 +1,a },K 1) = 1. A siilar arguent holds for a 1 +a = 1. Case : (a 1 +a and a +K +1 0) S ({a 1,a },K) = (K +1)+ax(0,(a 1 +a )+K +1) ax(0,a 1 +K +1) 0, S ({a 1 +1,a },K 1) = K +ax(0,(a 1 +1+a )+K ) ax(0,(a 1 +1)+K ) 0. 7

Therefore, = S ({a 1,a },K) S ({a 1 +1,a },K 1) = +1. Case 4: (a 1 +a and a +K +1 > 0) = 0 using siilar reasoning to Case and Case. We suarize the four cases for 0 < K 1 in the table below. Case a 1 +a a +K +1 1 < 0 0 < > 0 1 0 +1 4 > 0 0 Table : Suary of values. This shows that 1 1. Thus, we have shown that S ({a 1,a },K) S ({a 1,a },K 1)+1 K +1 for all 0 K 1. We apply Lea 6 and Theore 7 to show the upper bound for S. Corollary 8. For 0 a 1,a,K 1, S ({a 1,a },K). Proof. For 0 K 1, Theore 7 gives the result. For K, S ({a 1,a },K) = S ({ a 1, a }, K) by Lea 6. Fro there, we apply Theore 7 to the right hand side and get the result. Finally, for K = 1, it is easily seen that S ({a 1,a },K) = 0. This copletes the proof. The lower bound is now easy to show using. Theore 9. For 0 a 1,a,K 1, 0 S ({a 1,a },K). Proof. Without loss of generality, we assue that 0 a a 1 1. Consider 0 K 1. Thus, the conditions of Case (i.e., a1 +a < and a +K +1 > 0) cannot be et because if a 1 + a <, then a. So, a + K + 1 0. This shows that 1, which eans = 0 or 1. This, along with S ({a 1,a },0) 0, gives us the lower bound as desired. Next, we consider K, and apply Lea 6 to coplete the arguent. Lastly, S ({a 1,a }, 1) = 0. We now have the coplete result. 8,

Upper bound for n = Onphaeng and Pongsriia [4] already proved the lower bound for this case. In this section, we follow the style of the previous sections by rewriting the su, observing its syetry, and using a difference to prove the upper bound. This tie, we use Tverberg s forulation to write out the su explicitly. Definition 10. For any Z + and a 1,a,a,K Z + {0}, ( a1 +a +a +k S ({a 1,a,a },K) = k=0 a1 +a +k a +a +k a1 +a +k ) a1 +k a +k a +k k + + +. Like before, we show that the su can be written concisely without using the suation sybol. Proposition 11. For 0 K 1 and a 1,a,a Z + {0}, ( a1 +a +a a1 +a a +a S ({a 1,a,a },K) = a1 +a a1 a a ) + + + (K +1) +ax(0,((a 1 +a +a ) od )+K +1) ax(0,((a 1 +a ) od )+K +1) ax(0,((a +a ) od )+K +1) ax(0,((a 1 +a ) od )+K +1) +ax(0,(a 1 od )+K +1) +ax(0,(a od )+K +1) +ax(0,(a od )+K +1). Proof. We can rewrite our three-variable su in Definition 10 in ters of two-variable sus, that is, S ({a 1,a,a },K) = S ({a 1,a +a },K) S ({a 1,a },K) S ({a 1,a },K). (8) We then apply Proposition 5 to each su to get our result. A syetry exists for S ({a 1,a,a },K), siilar to Lea 6. 9

Lea 1. (Mirrored Sus) For 0 a 1,a,a 1 and 0 K, S ({a 1,a,a },K) = S ({ a 1, a, a }, K). Proof. As above, we rewrite the three-variable su as a series of two-variable sus and then reason as follows: S ({a 1,a,a },K) = S ({a 1,a +a },K) S ({a 1,a },K) S ({a 1,a },K) = S ({a 1,(a +a ) od },K) S ({a 1,a },K) S ({a 1,a },K) = S ({ a 1, (a +a ) od }, K) S ({ a 1, a }, K) S ({ a 1, a }, K) = S ({ a 1,( a )+( a )}, K) S ({ a 1, a }, K) S ({ a 1, a }, K) = S ({ a 1, a, a }, K). The first and fifth equalities coe fro (8). We take advantage of the a i -periodicity of the susinthesecondandfourthequalities. Lastly, weapplylea6inthethirdequality. Next, we show the upper bound for half the range of K using. Theore 1. For 0 a 1,a,a 1 and 0 K 1, S ({a 1,a,a },K). Proof. Without loss of generality, we assue that 0 a a a 1 1. We break down the proof into two cases of K, that is, we would like to show { K +1, if 0 K 1 (Case A); S ({a 1,a,a },K), if K 1 (Case B). As above, we define a difference of sus, := S ({a 1,a,a },K) S ({a 1 +1,a,a },K 1), and note that can be converted to via (8) as follows: = S ({a 1,a,a },K) S ({a 1 +1,a,a },K 1) = (S ({a 1,a +a },K) S ({a 1 +1,a +a },K 1)) (S ({a 1,a },K) S ({a 1 +1,a },K 1)) (S ({a 1,a },K) S ({a 1 +1,a },K 1)) = ({a 1,a +a },K) ({a 1,a },K) ({a 1,a },K). 10

Case A: Fro the proof of Theore 9, we know that ({a 1,a },K) 0 and ({a 1,a },K) 0. This, coupled with the fact that ({a 1,a + a },K) 1 (by Table ), gives 1. Furtherore, with S ({a 1,a,a },0) 1, we get S K +1 as in the proof of Theore 7. Case B: If 0, then we can use Case A (i.e., S K +1 ) to show S ({a 1,a,a },K) S ({a 1 +1,a,a },K 1). If = 1, we show S directly. Observe that the only way to obtain = 1 is when ({a 1,a +a },K) = +1, ({a 1,a },K) = 0, ({a 1,a },K) = 0. This arrangeent is achieved when our assued a 1,a,a,K also satisfies all conditions fro Table. So, in Table, we organize each row according to the restrictions that ust be applied. The type refers to the logical operator on the conditions in the sae row in order to achieve that particular value. Keeping these conditions in ind, we bound S ({a 1,a,a },K) according to whether a +a < or not. Condition a Type Condition b +1 (1a) a 1 +(a +a ) od AND (1b) ((a +a ) od )+K +1 0 0 (a) a 1 +a XOR (b) a +K +1 0 0 (a) a 1 +a XOR (b) a +K +1 0 Table : Conditions on values for Case B Case B1: a +a <. Conditions (1a) and (1b) siplify to a 1 +a +a < AND a +a +K +1 0, which therefore satisfies (b) and (b), iplying that it also satisfies (a) and (a) by the XOR condition. Thus, we get a 1 +a < AND a +K +1 0, a 1 +a < AND a +K +1 0. 11

Applying these conditions to Proposition 11 results in the following forula: S ({a 1,a,a },K) = K +1 ax(0,a 1 +a +K +1) ax(0,a 1 +a +K +1)+ax(0,a 1 +K +1), which can be broken down into the following four subcases. Case B1.1: (a 1 +K +1 > 0) S ({a 1,a,a },K) = a 1 a a + + = 0. Case B1.: (a 1 +K +1 0 and a 1 +a +K +1 > 0) S ({a 1,a,a },K) = K +1 (a 1 +a +K +1) (a 1 +a +K +1) = ( a 1 a a ) (a 1 +K)+( 1) 0 +( 1), because a 1 +a +a (1a) and a 1 a a gives a 1. This, along with K, gives us the second to last line. Case B1.: (a 1 +a +K +1 0 and a 1 +a +K +1 > 0) S ({a 1,a,a },K) = a 1 a, because a 1 +a. Case B1.4: (a 1 +a +K +1 0) This case cannot happen because a 1 +a +a gives that a 1 +a which eans a1 +a +K +1, contradicting the condition. We suarize these four subcases in Table 4. Subcase a 1 +a +K +1 a 1 +a +K +1 a 1 +K +1 S 1 > 0 > 0 > 0 0 > 0 > 0 0 > 0 0 0 4 0 0 0 N/A Case B: a +a <. Table 4: Subcases for B1 1

Conditions (1a) and (1b) siplify to a 1 +a +a AND a +a +K +1 0, which therefore satisfies (a) and (a), iplying that it also satisfies (b) and (b) by the XOR condition. Thus, we get a 1 +a AND a +K +1 > 0, a 1 +a AND a +K +1 > 0. Applying these conditions to Proposition 11 results in the following forula: S ({a 1,a,a },K) = (K +1) ax(0,a 1 +a +K +1) ax(0,a 1 +a +K +1)+(a 1 +K +1) +(a +K +1)+(a +K +1), which can be broken down into the following three subcases. B.1: (a 1 +a +K +1 > 0 and a 1 +a +K +1 > 0) S ({a 1,a,a },K) = (a 1 +a +K +1) (a 1 +a +K +1) +(a 1 +a +K +1)+(a +K +1) = a 1, because a 1 +a +a and a 1 a a gives a 1. B.: (a 1 +a +K +1 > 0 and a 1 +a +K +1 0) S ({a 1,a,a },K) = ( a 1 )+(a 1 +a +K +1) B.: (a 1 +a +K +1 0) a 1. S ({a 1,a,a },K) = ( a 1 )+(a 1 +a +K +1) We suarize the three cases in Table 5. +(a 1 +a +K +1) ( a 1 ). 1

Subcase a 1 +a +K +1 a 1 +a +K +1 S 1 > 0 > 0 > 0 0 0 0 Table 5: Subcases for B The results fro Tables 4 and 5 coplete the proof of Case B. Hence, we have proven the theore. Lea 1 and Theore 1 give the ain result, which we state as a corollary. Corollary 14. For 0 a 1,a,a,K 1, S ({a 1,a,a },K). Proof. For 0 K 1, Theore 1 gives us our result. For K, S ({a 1,a,a },K) = S ({ a 1, a, a }, K) by Lea 1. Fro there, we apply Theore 1 to the right hand side and get the result. Finally, S ({a 1,a,a }, 1) = 0. This copletes the proof. 4 (Not so sharp) lower bound for n = 4 Given that 0 a 1,a,a,a 4,K 1, the pattern of the axiu and iniu values of S ({a 1,a,a,a 4 },K) is less clear, as evidenced by soe results of the coputer progra: Maxiu Values of Sus Miniu Values of Sus (S 1,S,...) = (0,4,,8,7,1,11,16,15,0,19,4,8,7,,1,6,5,40,9,44,...). (S 1,S,...) = (0,0,,,, 6, 5, 6, 9, 8, 9, 1, 11, 1, 15, 14, 15, 18, 17, 18, 1, 0,...). Onphaeng and Pongsriia [4] showed the upper bound S ({a 1,a,a,a 4 },K) 4 14.

We conjecture the lower bound S ({a 1,a,a,a 4 },K). In an attept to prove this lower bound, we found that writing a difference of sus (like or ) is not an efficient way to approach the proble. Accordingly, we use another ethod to obtain the following partial result. Theore 15. For 0 a 1,a,a,a 4,K 1, S ({a 1,a,a,a 4 },K). Proof. We can cobine the following bounds fro n =,, naely, along with the identity 0 S ({a 1 +a +a,a 4 },K), S ({a 1 +a,a 4 },K), S ({a 1 +a,a 4 },K), S ({a,a,a 4 },K), 0 S ({a 1,a 4 },K), S ({a 1,a,a,a 4 },K) = S ({a 1 +a +a,a 4 },K) S ({a 1 +a,a 4 },K) to obtain the claied result. S ({a 1 +a,a 4 },K) S ({a,a,a 4 },K)+S ({a 1,a 4 },K), 5 Conjectures In order to coplete the analysis on this type of floor function proble, we want to show all the upper bounds and lower bounds for any nuber of variables, n. Onphaeng and Pongsriia [4] were able to show the upper bound when n is even and the lower bound when n is odd. Theore 16 (Onphaeng, Pongsriia). When n is even and is even, S n. When n is odd and is even, n S. The bounds on both cases are obtained exactly at A = {/,/,...,/}, K = / 1. 15

We conjecture the issing bounds, naely, the lower bounds when n is even and the upper bounds when n is odd. To ake these conjectures, we wrote a Maple progra to calculate the values of S (A,K) for specific,a, and K. For each n, the sets A and K that give extree values of S for interesting patterns, which depend on. Once we deterine such values in A and K, we can quickly copute the extree values of S (A,K) for each n and then use the resulting data to for a holonoic ansatz. The resulting recurrence becoes a ninth order recurrence with polynoial coefficients of degree at ost. We suarize our findings in the following conjecture. For interested readers, this Maple code can be found on Thanatipanonda s website (www.thotsaporn.co). Conjecture 17. Let A := {a 1,a,...,a n } and define ax/in S (A,K) to be the axiu/iniu over all choices of a i and K with 0 a 1,a,...,a n,k 1 and fixed. Furtherore, we let the function S be as defined in the first section. Suppose now { ax S (A,K), n odd; M(n) := in S (A,K), n even. Then for n 4, we conjecture the following result in two parts. 1. n = 4k 1,4k or 4k +1, where k Z + : Under the condition that is a ultiple of k +1, the values of M(n) occur exactly at { } k A = k +1, k k +1,..., k,k = k k +1 k +1 1, { } (k +1) (k +1) (k +1) (k +1) or A =,,...,,K = k +1 k +1 k +1 k +1 1.. n = 4k +, where k Z + : Under the condition that is a ultiple of k + 1 and k +, the values of M(n) occur at (aong other places) { } k A = k +1, k k +1,..., k,k = k k +1 k +1 1, { } (k +1) (k +1) (k +1) (k +1) or A =,,...,,K = k +1 k +1 k +1 k +1 1, { } (k +1) (k +1) (k +1) (k +1) or A =,,...,,K = k + k + k + k + 1, { } (k +) (k +) (k +) (k +) or A =,,...,,K = k + k + k + k + 1. Moreover, M(n) can be calculated directly fro a forula siilar to the equations fro Propositions 5 and 11, or by M(n) = f(n), 16

where f(n) satisfies the recurrence relation 5(n+)(n )f(n) = 10(n +n 8)f(n 1) 4(n 10n+)f(n ) 4(n 11)f(n ) (n 10n 1)f(n 4) 19(n 1)(n 5)f(n 5)+64(n n+51)f(n 6) +84(n 1)f(n 7) 56(n )(n 8)f(n 8) +51(n 9)(n 8)f(n 9) for n 11, with the initial conditions f() = 0,f() = 1/,f(4) = 1,f(5) =,f(6) =,f(7) = 8,f(8) = 18, f(9) = 6, and f(10) = 65. For convenience, we give exaples of soe of the bounds (and the set A for which the values of those bounds occur) produced fro the conjectures above. Suppose n = 4,5 and is a ultiple of. Case n = 4: Case n = 5: S. S 6. For these cases, M(n) occurs at A = {/,/,...,/},K = / 1 or A = {/,/,...,/},K = / 1. Suppose n = 6. If is a ultiple of, then 9 S, with the iniu at (aong other places) A = {/,/,...,/},K = / 1 or A = {/,/,...,/},K = / 1. 17

If is a ultiple of 5, then 15 S, 5 with the iniu at (aong other places) Suppose n = 7,8,9 and is a ultiple of 5. A = {/5,/5,...,/5},K = /5 1 or A = {/5,/5,...,/5},K = /5 1. Case n = 7: Case n = 8: Case n = 9: S 40 5. 90 S. 5 S 180 5. For these cases, M(n) occurs at A = {/5,/5,...,/5},K = /5 1 or A = {/5,/5,...,/5},K = /5 1. 6 Acknowledgeents The authors would like to thank Harry Richan and an anonyous referee, for their helpful coents and suggestions to iprove this anuscript. References [1] L. Carlitz, Soe arithetic sus connected with the greatest integer function, Math Scand. 8 (1960), 59 64. [] R. C. Grison, The evaluation of a su of Jacobsthal, Norske Vid. Selsk. Skr. Trondhei (1974), No. 4. [] E. Jacobsthal, Über eine zahlentheoretische Sue, Norske Vid. Selsk. Forh. Trondhei 0 (1957), 5 41. 18

[4] K. Onphaeng and P. Pongsriia, Jacobsthal and Jacobsthal-Lucas nubers and sus introduced by Jacobsthal and Tverberg, J. Integer Sequences 0 (017), Article 17..6. [5] H. Tverberg, On soe nuber-theoretic sus introduced by Jacobsthal, Acta Arith., 155 (01), 49 51. 010 Matheatics Subject Classification: Priary 11A5; Secondary 05D99. Keywords: floor function, optiization, su. Received Septeber 19 017; revised versions received February 14 018; February 16 018; February 18 018. Published in Journal of Integer Sequences, February 018. Return to Journal of Integer Sequences hoe page. 19