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One-half of the antidiagonal sums of array A220508.

To prove Paul D. Hanna's formula for the row n o.g.f. A(x,n) = Sum_{m >= 1} T(n,m)*x^m, we use Leibniz's rule for the k-th derivative of a product of functions: dx^k(exp(k^n*x) * (1 - A(x,n)))/dx = Sum_{s=0..k} binomial(k,s) * d^s(exp(k^n*x))/dx^s * d^(k-s) (1 - A(x,n))/dx^(k-s) = k^(n*k) * exp(k^n*x) * (1 - Sum_{m>=1} T(n,m) * x^m) - Sum_{s=0..k-1} binomial(k,s) * k^(n*s) * exp(k^n*x) * (Sum_{m>=1} (m!/(m-(k-s))!) * T(n,m) * x^(m-(k-s))). The coefficient of x^k for exp(k^n*x) * (1 - A(x,n)) is obtained by setting x = 0 in the k-the derivative, and it is equal to k^(n*k) - Sum_{s=0..k-1} binomial(k,s) * k^(n*s) * (k-s)! * T(n,k-s) = k! * (k^(n*k)/k! - Sum_{s=0..k-1} k^(n*s)/s! * T(n,k-s)) = 0 because of the recurrence that T(n,k) satisfies.

To prove the formula below for T(n,k) that involves the compositions of k, we use mathematical induction on k. For k = 1, it is obvious. Assume it is true for all n and all m < k. Consider the compositions of k.

There is only one of size r = 1, namely k, and corresponds to the term k^(n*k)/k! in the recurrence T(n,k) = k^(n*k)/k! - Sum_{s=1..k-1} k^(n*s)/s! * T(n,k-s).

For the other compositions (s_1, ..., s_r) of k (of any size r >= 2), we group them according to the their last element s_r = s in {1, 2, ..., k - 1}, which gives rise to the factor k^(n*s)/s! = (Sum_{i=1..r} s_i)^(n*s_r)/s_r!. Using the inductive hypothesis, we substitute the expression for T(n,k-s) in the recurrence T(n,k) = k^(n*k)/k! - Sum_{s=1..k-1} k^(n*s)/s! * T(n,k-s). Each term in the expression for T(n,k-s) corresponds to a composition of k - s and is postmultiplied by k^(n*s)/s! = (Sum_{i=1..r} s_i)^(n*s_r)/s_r!. We thus get a term in the expression for T(n,k) that corresponds to a composition of the form (composition of k - s) + s, and the sign of this term is (-1)^((size of composition of k - s) + 1). The rest of the proof follows easily.

Antidiagonal sums of square array A342354.

This is a square array defined by J. M. Bergot in A005917 (originally by mistake in A047926). Here is the edited description of the array by this contributor.

Construct an array M with M(0,n) = 2*n^2 + 4*n + 1 = A056220(n+1), M(n,0) = 2*n^2 + 1 = A058331(n) and M(n,n) = 2*n*(n+1) + 1 = A001844(n). Row(n) begins with all the increasing odd numbers from A058331(n) to A001844(n) and column(n) begins with all the decreasing odd numbers from A056220(n+1) to A001844(n). The sum of the terms in row(n) plus those in column(n) minus M(n,n) equals A005917(n+1).

For n >= 1, a(n) is the number of deterministic, completely-defined, initially-connected finite automata with n inputs and 3 unlabeled states. A020522 counts similar automata with n inputs and 2 unlabeled states.

According to a comment by Nelma Moreira in A006689 and A006690, the number of such automata with N inputs and M unlabeled states is Sum (Product_{i=1..M-1} i^(f_i - f_{i-1} - 1)) * M^(M*N - f_{M-1} - 1), where the sum is taken over integers f_1, ..., f_{M-1} satisfying 0 <= f_1 < N and f_{i-1} < f_{i} < i*N for i = 2..M-1. (See Theorem 8 in Almeida, Moreira, and Reis (2007). The value of f_0 is not relevant.) For this sequence we have M = 3 unlabeled states, for A020522 we have M = 2 unlabeled states, for A006689 we have N = 2 inputs, and for A006690 we have N = 3 inputs.

A similar formula for the number of such automata with N inputs and M unlabeled states was given by Robinson (1985, Eq. (2.3) upon division by (p-1)!). It is Sum_{r=1..M} (-1)^(r-1) * Sum_{k_1,...,k_r} (k_1/(Product_{i=1..r} k_i!)) * Product_{j=1..r} (Sum_{i=1..j} k_i)^(N*k_j), where the second sum is over all compositions (k_1,...,k_r) of length r of M. (A composition of length r of M is an ordered partition (k_1,...,k_r) with k_i >= 1 for i = 1..r and Sum_{i=1..r} k_i = M.)

Both formulas with N = n and M = 3 give a(n) = (27^n - 9^n)/2 - 12^n + 6^n.

In Liskovets (2006, Eq. (11), p. 546), a(n) equals H_N(M) = h_N(M)/(M-1)! with N = n and M = 3. The sequence h_N(M) satisfies the recurrence h_N(M) = M^(N*M) - Sum_{t=1..M-1} binomial(M-1, t-1) * M^(N*(M-t)) * h_N(t) with h_N(1) = 1. A recurrence for H_N(M) = h_N(M)/(M-1)! originally appeared in Liskovets (1969, Eq. (3), p. 17, denoted by h_{n,r}).

Because A323833(n,n/2) = 0 for n even (if A323833 is read as a triangle), we also have a(n) = Sum_{k=0..ceiling((n-1)/2)} A323833(n,k).

Even graphs are colloquially known as Euler graphs (even though, strictly speaking, that is not correct).

Robinson (1969, Section 4, p. 153) actually calculates the o.g.f. of row n=6 of this irregular triangle T(n,k), but he is discouraged by the asymmetry of the coefficients of the polynomial. (The n-th row of this triangle T(n,k) is symmetric only when n is odd). He states: "The processes of Section 1 can be extended brutally to accommodate the line [edge] parameter, but the result does not promise to be pleasing."

The value T(0,0) = 0 has no physical meaning. It is there because it makes the formula for the bivariate e.g.f.-o.g.f. (shown below) work.

Since T(n,k) counts even connected graphs with n vertices and k edges, for n >= 2, each vertex must have at least two edges, so k >= n. Hence, T(n,k) = 0 for 0 <= k < n.

We have T(n,n) = (n-1)!/2 for n >= 3 because T(n,n) counts the different labelings of cyclic graphs with n vertices and n edges, and we have (n-1)! cyclic permutations of the numbers 1, 2, ..., n. We divide by 2 because we get the same labeling if we flip the cyclic graph over (like a bracelet).

We have T(n, n*(n-1)/2) = 1, if n is odd, because the complete graph on n vertices is even (each vertex has degree n-1) and has only one non-isomorphic labeling.

We have T(n, n*(n-1)/2 - s) = 0 for s = 0, 1, 2, ..., (n/2)-1, if n is even, because in an even graph with n vertices we cannot have more than n*(n-2)/2 = n*(n-1)/2 - n/2 edges.

Finally, we have T(n, n*(n-2)/2) = A001147(n/2), if n is even >= 4, because any labeling in an even graph with n vertices and n*(n-2)/2 edges corresponds to a perfect matching in a complete graph with n vertices (by considering the pairs of vertices that are not connected).

See the comments for A058878 about the different (and sometimes confusing) terminology regarding even and (connected or not) Euler graphs.