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Elementary properties of the numbers I1(n) are described in Section 2. These include a second order recurrence, an exponential generating function as well as an explicit finite sum. These are generalized to the involution polynomials I1(n; t) in Section 3 which are intimately linked to the (probabilistic) Hermite polynomials defined by ⌊n/2⌋ ( 1)j tn−2j (1.4) H (t)= n! − n j!(n 2j)! 2j j=0 X − with generating function ∞ xn (1.5) H (t) = exp xt 1 x2 . n n! − 2 n=0 X The involution polynomials have a combinatorial interpretation as the generating function for fixed points of permutation in Sn. Arithmetic properties of I1(n) are presented in Section 4. Particular emphasis is given to the 2-adic valuation of I1(n). Recall that, for x N and p prime, the p-adic valuation of x, denoted by νp(x), is the highest power∈ of p that divides x. An odd prime p is called efficient if p does not divide I (j) for 0 j p 1. Otherwise it is called inefficient. The 1 ≤ ≤ − prime p = 3 is efficient and p = 5 is inefficient since I1(4) = 10. A periodicity argument is used to show that νp(I1(n) = 0; i.e., p is efficient. Morever, for a prime p, it is shown that either p divides I1(n) infinitely often or never. In the case of an inefficient prime, it is conjecture that the p-adic valuation of the sequence I1(n) can be given in terms of a tree Tp. This phenomena is illustrated for the prime p = 5. It is an open question to characterize efficient (or inefficient) primes. The partial sums of I1(n), denoted by an, are discussed in Section 5. Their arithmetic properties are presented in Section 6. For instance, an explicit expression for their 2-adic valuation is given there. The valuations for odd primes are also conjectured to have a tree structure. This is illustrated in the case p = 5. Section 7 considers the statistics of the sequence dn,ℓ in (1.1). This is a generalization of I1(n)= Cn,2. Finally, the asymptotic behavior of d = C is given in Section 8. n,ℓ | n,ℓ| 2. Basic properties of the involution numbers
This section discusses fundamental properties of I1(n). Some of them are well- known but proofs are included here for the convenience of the reader.
Theorem 2.1. The sequence I1(n) satisfies the recurrence (2.1) I (n)= I (n 1)+(n 1)I (n 2), for n 2, 1 1 − − 1 − ≥ with initial conditions I1(0) = I1(1) = 1.
Proof. There are I1(n 1) involutions that fix n. The number of involutions that contain a cycle (jn), with− 1 j n 1 is n 1 times the number of involutions containing the cycle (n 1, n≤). This≤ is− (n 1)−I (n 2). − − 1 − The recurrence above generates the values n 0 1 2 3 4 5 6 7 8 9 10 I1(n) 1 1 2 4 10 26 76 232 764 2620 9496 This is sequence A000085 in OEIS. The recurrence (2.1) now enables to write a generating function for I (n) . { 1 } INVOLUTIONSANDTHEIRPROGENIES 3
Theorem 2.2. The exponential generating function for I1(n) is ∞ I (n) (2.2) 1 xn = exp(x + 1 x2). n! 2 n=0 X Proof. On the basis of (2.1) verify that both sides of (2.2) satisfy f ′(x)=(1+ x)f(x) and the value f(0) = 1.
x x2/2 Cauchy’s product formula on e and e allows to express I1(n) as a finite sum.
Corollary 2.3. The involution numbers I1(n) are given by
⌊n/2⌋ n 2j j! (2.3) I (n)= . 1 2j j 2j j=0 X 2j j! The numbers appearing in (2.3) are now shown to be of the same parity. j 2j Corollary 2.4. For j N, the numbers (2j)!/(j!2j) are odd integers. ∈ Proof. The identity (2j)! (2j)(2j 1) (j + 1) (2.4) = − ··· j!2j 2j shows that the denominator is a power of 2. To compute this power, use Legendre’s formula (2.5) ν (n!) = n s (n), 2 − 2 where s2(n) is the sum of the digits of n in its binary expansion. Therefore, (2j)! (2.6) ν = (2j s (2j)) (j s (j)) j =0, 2 j!2j − 2 − − 2 − in view of s2(2j)= s2(j).
A second recurrence for the involution numbers is presented next. Theorem 2.5. For n,m N, the involution numbers satisfy ∈ n m (2.7) I (n + m)= k! I (n k)I (m k). 1 k k 1 − 1 − kX≥0 Proof. Split up the set [n+m] into two disjoint subsets A and B, of n and m letters, respectively. Count the involutions in Inv(n + m) acoording to the number k of cross-permutations that make up a cycle (ab), with a A and b B. The letters n m ∈ ∈ a and b can be chosen in k k ways and k! ways to place the k cycles (ab). The remaining elements in A (respectively B) allow I1(n k) (respectively I1(m k)) n− m − involutions. To complete the argument, summing k! k k I1(n k)I1(m k) over k. − − As a direct consequence of (in fact, equivalent to) Theorem 2.5 the following analytic statement is recorded. This result bypasses the need for an otherwise messy chain rule for derivatives. 4 TEWODROSAMDEBERHANANDVICTORH.MOLL
Corollary 2.6. Higher order derivatives of the function f(x) = exp x + x2/2 are computed by the umbral dm m m (2.8) f(x)= f(x) I (m k)xk := f(x)(x + I )m. dxm k 1 − 1 Xk=0 dm xn Proof. From Theorem 2.2, f(x) = I (n + m) . The right-hand side of dxm 1 n! n Theorem 2.5 implies X n m xn k! I (n k)I (m k) = k k 1 − 1 − n! n X Xk m xn−k I (m k)xk I (n k) k 1 − 1 − (n k)! n Xk X − m = f(x) I (m k)xk. k 1 − Xk The claim follows.
r The recurrence (2.7) is now used to prove periodicity of I1(n) mod p . r Theorem 2.7. Let p be a prime and r N. Then I1 mod p is a periodic sequence of period pr. ∈ Proof. Write n = cpr + t with 0 t
pr pr 2 (2.11) I (2pr)= k! I )(pr k)2 I (pr)2 mod pr 1 k 1 − ≡ 1 Xk=0 and then induction on c gives (2.12) I (cpr) I (pr)c mod pr. 1 ≡ 1 r r The next step is to show that I1(p ) 1 mod p . Then (2.10) and (2.12) imply the required periodicity. Observe first that≡ for m 0 mod p, 6≡ pr pr pr 1 (2.13) = − 0 mod pr, m m m 1 ≡ − so that r−1 pr 2mp (mp)! (2.14) I (pr) mod pr, 1 ≡ 2mp mp 2mp m=0 X where the upper bound arises from ν ((mp)!) m + m/r r if m r. p ≥ ⌊ ⌋≥ ≥ INVOLUTIONSANDTHEIRPROGENIES 5
The final step is to show that pr r 1 if m 0 mod p (2.15) νp = − 6≡ 2mp r 2 if m 0 mod p. ( − ≡ r r This would imply I1(p ) 1 mod p since νp((mp)!) 2 for m 2. The periodicity r ≡ ≥ ≥ of I1(n) mod p follows from here. To prove (2.15), recall Legendre’s formula x s (x) (2.16) ν (x!) = − p p p 1 − where sp(x) is the digit sum of x in base p. This gives pr s (pr)+ s (2mp)+ s (pr 2mp) (2.17) ν = − p p p − p 2mp p 1 − 1+ s (2m)+ s (pr−1 2m) = − p p − . p 1 − r−2 Write 2m = u pi with 0 u p 1. Then i ≤ i ≤ − i=0 X r−2 r−2 pr−1 2m = pr−1 u pi =1+ (p 1 u )pi − − i − − i i=0 i=0 X X r−2 = (p u )+ (p 1 u )pi. − 0 − − i i=1 X If m 0 mod p, then 6≡ r−2 (2.18) s (pr−1 2m) = (p u )+ (p 1 u ). p − − 0 − − i i=1 X On the other hand, if νp(m)= a, a direct calculation leads to r−2 (2.19) s (pr−1 2m) = (p u )+ (p 1 u ). p − − a − − i i=a+1 X The claim (2.15) now follows from (2.17).
Corollary 2.8. Assume I1(n) 0 mod p for 0 n p 1 and p an odd prime. Then ν (I (n)) 0. 6≡ ≤ ≤ − p 1 ≡ Corollary 2.9. A prime p divides the sequence I1(n) infinitely often or never at all. In the process of discovering the previous congruences, the following result was obtained by the authors. Even though it is not related yet to the material that follows, it is of intrinsic interest and thus placed here for future use. In the sequel, λ n means λ is a partition of n. ⊢ Proposition 2.10. Let λ = (λ1, , λk) n with λ1 λk 1. Denote pn = pn . ··· ⊢ ≥ ··· ≥ ≥ pλ pλ1, ··· , pλk a) If p 3 is a prime, then pn n mod p2. ≥ pλ ≡ λ b) If p 5 is a prime, then pn n mod p3. ≥ pλ ≡ λ 6 TEWODROSAMDEBERHANANDVICTORH.MOLL
Proof. The case k = 2 is considered first. Take an n p rectangular grid. Choose pb of these squares and paint them red. One option× is to paint b entire rows red, n call this type-1. This can be done in b different ways. In all other cases, there exist at least two rows each consisting of t red squares, where 0