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Number of partitions of n into distinct powers of 3.

Trajectory of 1 under the morphism: 1 -> 110, 0 -> 000. Thus 1 -> 110 ->110110000 -> 110110000110110000000000000 -> ... - Philippe Deléham, Jul 09 2005

Also, an example of a d-perfect sequence.

This is a composite of two earlier sequences contributed at different times by N. J. A. Sloane and by Reinhard Zumkeller, Mar 05 2005. Christian G. Bower extended them and found that they agreed for at least 512 terms. The proof that they were identical was found by Ralf Stephan, Jun 13 2005, based on the fact that they were both 3-regular sequences.

"This [constant] arises from the study of a vibrating, homogeneous membrane that is uniformly stretched across the unit disk. [Its square] is the principal frequency of the sound one hears when a kettledrum is struck." - Quoted from the book by Steven R. Finch.

Siegel proves (the Main Theorem) that J_0(z) is transcendental if z is algebraic and nonzero, but since in our case J_0(z) = 0 is not transcendental it follows that z cannot be algebraic. - Charles R Greathouse IV, Oct 20 2020

a(n) is the number of ways to form an ordered pair of n-permutations and then choose a subset of its common descent set. Cf. A192721. - Geoffrey Critzer, Apr 29 2023

In other words, there are no "column rises", where a "column rise" means a pair of adjacent columns where each entry in the left column is strictly less than the adjacent entry in the right column.

This is R(n,k,0) in [Abramson-Promislow].

From Petros Hadjicostas, Sep 09 2019: (Start)

As stated above, in the notation of Abramson and Promislow (1978), we have T(n,k) = R(n, k, t=0).

Let P_k be the set of all lists a = (a_1, a_2, ..., a_k) of integers a_i >= 0, i = 1, ..., k, such that 1*a_1 + 2*a_2 + ... + k*a_k = k; i.e., P_k is the set all integer partitions of k. Then |P_k| = A000041(k).

From Eq. (6), p. 248, in Abramson and Promislow (1978), with t=0, we get T(n,k) = Sum_{a in P_k} (-1)^(k - Sum_{j=1..k} a_j) * (a_1 + a_2 + ... + a_k)!/(a_1! * a_2! * ... * a_k!) * (k! / ((1!)^a_1 * (2!)^a_2 * ... * (k!)^a_k))^n.

The integer partitions of k = 1..10 are listed on pp. 831-832 of Abramowitz and Stegun (1964). We see that, for k = 1..6, the corresponding multinomial coefficients k! / ((1!)^a_1 * (2!)^a_2 * ... * (k!)^a_k) are all distinct; that is, A070289(k) = A000041(k) and A309951(k,s) = A325305(k,s) for s = 0..A000041(k). For 7 <= k <= 10, this is not true anymore; i.e., A070289(k) < A000041(k) for 7 <= k <= 10 (and we conjecture that this is the case for all k >= 7).

From the theory of difference equations, we see that Abramson and Promislow's Eq. (6) on p. 248 (with t=0) implies that Sum_{s = 0..A070289(k)} (-1)^s * A325305(k,s) * T(n-s,k) = 0 for n >= A070289(k) + 1. For k = 1..5, these recurrences give R. H. Hardin's empirical recurrences shown in the Formula section below.

We also have Sum_{s = 0..A000041(k)} (-1)^s * A309951(k,s) * T(n-s,k) = 0 for n >= A000041(k) + 1, but for k >= 7, the recurrence we get (for column k) may not necessarily be minimal.

To derive the recurrence for row n, let y=0 in Eq. (8), p. 249, of Abramson and Promislow (1978). We get 1 + Sum_{k >= 1} T(n,k)*x^k/(k!)^n = 1/f_n(-x), where f_n(x) = Sum_{i >= 0} (x^i/(i!)^n). Matching coefficients, we get Sum_{s = 1..k} T(n,s) * (-1)^(s-1) * binomial(k,s)^n = 1, from which the recurrence in the Formula section follows.

(End)

Let S_n denote the symmetric group on {1,2,...,n}. A permutation p_1p_2...p_n in S_n has a descent at position i (1 <= i <= n-1) if p_i > p_(i+1). The Eulerian numbers A008292 (with an offset of 0 in the column indexing) enumerate permutations by descents. We define a pair of permutations p_1p_2...p_n and q_1q_2...q_n to have a common descent at position i (1 <= i <= n-1) if both p_i > p_(i+1) and q_i > q_(i+1). For example, the permutations (3241) and (4231) in S_4 have common descents at positions i = 1 and i = 3. The table entry T(n,k) gives the number of pairs of permutations in the Cartesian product S_n x S_n with k common descents.

The generalized Stirling numbers associated with this triangle is A061691. See also A192722.

From Petros Hadjicostas, Sep 08 2019: (Start)

We generalize Daniel Suteu's recurrence from A212856. Notice first that, in the notation of Abramson and Promislow (1978), we have a(n) = R(m=4, n, t=0).

Letting y=0 in Eq. (8), p. 249, of Abramson and Promislow (1978), we get 1 + Sum_{n >= 1} R(m,n,t=0)*x^n/(n!)^m = 1/f(-x), where f(x) = Sum_{i >= 0} (x^i/(i!)^m). Matching coefficients, we get Sum_{s = 1..n} R(m, s, t=0) * (-1)^(s-1) * binomial(n,s)^m = 1, from which the recurrence in the Formula section follows.

(End)

From Petros Hadjicostas, Sep 08 2019: (Start)

We generalize Daniel Suteu's recurrence from A212856. Notice first that, in the notation of Abramson and Promislow (1978), we have a(n) = R(m=5, n, t=0).

Letting y=0 in Eq. (8), p. 249, of Abramson and Promislow (1978), we get 1 + Sum_{n >= 1} R(m,n,t=0)*x^n/(n!)^m = 1/f(-x), where f(x) = Sum_{i >= 0} (x^i/(i!)^m). Matching coefficients, we get Sum_{s = 1..n} R(m, s, t=0) * (-1)^(s-1) * binomial(n,s)^m = 1, from which the recurrence in the Formula section follows.

(End)

From Petros Hadjicostas, Sep 08 2019: (Start)

We generalize Daniel Suteu's recurrence from A212856. Notice first that, in the notation of Abramson and Promislow (1978), we have a(n) = R(m=6, n, t=0).

Letting y=0 in Eq. (8), p. 249, of Abramson and Promislow (1978), we get 1 + Sum_{n >= 1} R(m,n,t=0)*x^n/(n!)^m = 1/f(-x), where f(x) = Sum_{i >= 0} (x^i/(i!)^m). Matching coefficients, we get Sum_{s = 1..n} R(m, s, t=0) * (-1)^(s-1) * binomial(n,s)^m = 1, from which the recurrence in the Formula section follows.

(End)