On Divisibility of Convolutions of Central Binomial Coefficients

Recently, Z. Sun proved that 2 (2m + 1) ( 2m m ) | ( 6m 3m )( 3m m ) for m ∈ Z>0. In this paper, we consider a generalization of this result by defining bn,k = 2k (n + 2k − 2)!! (n− 2)!! k! . In this notation, Sun’s result may be expressed as 2 (2m + 1) | b(2m+1),(2m+1)−1 for m ∈ Z>0. In this paper, we prove that 2n | bn,un±2r for n ∈ Z>0 and u, r ∈ Z>0 with un ± 2r > 0. In addition, we prove a type of converse. Namely, fix k ∈ Z and u ∈ Z>0 with u > 0 if k < 0. If 2n | bn,un+k for all n ∈ Z>0 with un + k > 0, then there exists a unique r ∈ Z>0 so that either k = 2 or k = −2.


Introduction
There has been much recent work on topics relating to central binomial coefficients ( [1,2,3,5,6,7,8,9,10,11,12]).In particular, Z.Sun in [9] proved interesting results on congruences of sums of products of central binomial coefficients.One such result is that 2 (2m + 1) 2m In this paper, we consider a generalization of Sun's new sequence.Recognizing 6m 3m 3m m / 2m m as the coefficient of x 2m in the (2m + 1)-fold convolution of the central binomial sequence with itself, we define b n,k to be the k th term of the n-fold convolution of the central binomial sequence with itself-which turns out to be In this notation, Sun's result is that 2 In this paper (Theorem 1), we prove that for n ∈ Z >0 and u, r ∈ Z 0 with un ± 2 r > 0. In particular, Sun's result is a special case of the above theorem in which n = 2m + 1, u = 1, r = 0, and the − sign is chosen.In addition, we prove a type of converse (Theorem 2).Namely, fix k ∈ Z and u for all n ∈ Z >0 with un + k > 0, then there exists a unique r ∈ Z 0 so that either

Definition
Write b for the sequence of central binomial coefficients, b j = 2j j , with j ∈ Z 0 .For n, k ∈ Z 0 with n 1, we define the doubly indexed sequence b n,k ∈ Z >0 to be the k th term of the n-fold convolution of b with itself, b n,k = (b * n ) k .In the degenerate case of n = 0, we define b 0,0 = 1 and b 0,k = 0 for k > 0. It follows that the generating function for the sequence k → b n,k is (1 − 4x) −n/2 and a trivial calculation shows that when n 2. For use in Theorem 2 and in order to compare with [9], we note that it is straightforward to verify that the electronic journal of combinatorics 21(1) (2014), #P1.32 for m ∈ Z >0 in the first formula above and m ∈ Z 0 in the second.

Divisibility
Here we present the main result on the divisibility of the sequence b n,k .
Proof.Begin with the convolution definition The set X carries a natural action of the symmetric group, S n , acting by permuting the coordinates.Write O 1 , . . ., O N for the orbits of X under the action of S n .Clearly each O k , 1 k N , has a unique representative of the form As x k ∈ X, it follows that Therefore, and we may write the greatest common divisor of d 1k , d 2k , . . ., d m k k as 2 q k for some q k , 0 q k r.Since it follows that 2 q k | n, we also see that 2 q k n.
Choose w jk ∈ Z, 1 j m k , so that m k j=1 w jk d jk = 2 q k .Write e j for the j th standard basis vector, e j = ( j 0, . . .0, 1, 0, . . .0) ∈ Z m k (suppressing the m k dependence).Then As we are done.Suppose, therefore, that we are in the case of c 1k = 0 (so m k 2).Let Now all that we get is that 2 that q k + 1 s k .So suppose that q k + 1 > s k and write d jk = 2 q k t jk for t jk ∈ Z >0 .Since we must have t jk j − 1 1 when j 2, we get Since it is impossible to obtain q + 1 > 2 q , we arrive at the desired contradiction.
the electronic journal of combinatorics 21(1) (2014), #P1.32 For the case of 2n | b n,un+2 r in the above theorem, n = 0 is ruled out in order to make sure that division by 2n is well defined (note b 0,0 = 1 and b 0,k = 0 for k > 0).For the case of 2n | b n,un−2 r , we require un > 2 r since b n,0 = 1.
We also note that Sun's result . This is then a special case of our equation 2n | b n,un−2 r in which n = 2m + 1, u = 1, and r = 0.This gives the same statement as the above equation, but written as

A Type of Converse
The next result is a type of converse to Theorem 1.
for all n ∈ Z >0 with un + k > 0, then there exists a unique r ∈ Z 0 so that either Proof.First we show that k = 0.For this choose any odd prime p and consider b 2p,2up .Using Theorem 4 of [4] and the Division Algorithm, work mod p to see that Thus p b 2p,2up and so k = 0. Next we consider the case of u = 0 (so k > 0 here).Actually the following argument works whenever k > 0 so that is all we actually assume.If k has an odd prime divisor, p, write k = pk for some k ∈ Z >0 .Consider b 2p,2up+k .Then, working mod p again, Thus p b 2p,2up+k and so k must be a power of 2. Note that the only reason this argument may fail for k < 0 is that we might have 2up + k 0. Finally, consider the case of u = 0. Suppose there exists an odd prime p so p | k.Then write k = pk and fix m 0 ∈ Z 0 so 2 m 0 +1 u + k > 0 and 2 m 0 +1 u is congruent mod p the electronic journal of combinatorics 21(1) (2014), #P1.32 to either 0 or 1 (depending on whether p | u or p u). Now consider n = 2 m 0 +1 p p N + 1 for any sufficiently large N .We will show that p b n,un+k .For this, write For sufficiently large N , un + k can be expanded in base p as Clearly adding un + k to p N +1 + (p − 1) in base p results in no carries so that p b n,un+k .As a result, k has no odd prime divisors and we are done.

Relation to Known Sequences
As a result of Theorem 1, we have the following integer sequences n → B n,u,r,± ≡ b n,un±2 r 2n .
For most choices of parameters u, r, ±, this sequence seems to be new.However, for a few special choices, the sequence is known.Up to a shift and a few initial terms, the sequence B n,0,2,+ is the OEIS integer sequence A077415, B n,1,0,+ is A085614, B n,1,1,+ is A078531, and B n,1,0,− appears as every other term in A089073.In addition, the odd terms of B n,0,2,+ are A162540, the even terms of B n,0,2,+ are A102860 and the negative of A136264, and (by construction) the odd terms of B n,1,0,− are Sun's A176898.

Final Remarks
It would be interesting to find a combinatorial interpretation for the sequences B n,u,r,± .For instance, one is given the case of B n,1,0,− (A089073) or B n,1,1,+ (A078531) as the number of symmetric non-crossing connected graphs on equidistant nodes of a circle and B n,1,0,+ (A085614) is the number of elementary arches of size n.
In addition, information on corresponding generating functions would be of interest.Some are known.For example B n,1,0,+ (A085614) is the series reversion of x − 3x 2 + 2x 3 .

.
d k where we use multinomial notation above and write d k = (d 1k , d 2k , . . ., d m k k ).Using this, we may rewrite the formula for b n,un±2 r as b n,un±2 r = We will prove the theorem by demonstrating that 2n | n d k m k j=1 2c jk c jk d jk for each k. the electronic journal of combinatorics 21(1) (2014), #P1.32

then a similar argument as in the above paragraph shows 2n | n d k m k j=1 2c jk c jk d jk and
we are done.It remains only to show 1, and s N .Now we apply Kummer's theorem to the binomial coefficient in b