cp's OEIS Frontend

Original name: (1,3,5,7,9,..) = A005408 convolved with (1,0,2,3,4,..) = 1 followed by A087156.

Number of linear forests of planted planar trees with n nodes (Christian G. Bower).

Number of ordered trees with n+2 edges and having no branches of length 1 starting from the root. Example: a(1)=1 because the only ordered tree with 3 edges having no branch of length 1 starting from the root is the path tree of length 3. a(n) = A127158(n+2,0). - Emeric Deutsch, Mar 01 2007

Hankel transform is A056520. - Paul Barry, Oct 16 2007

For each n >= 1 the n X n matrix Z(n) is constructed as follows. The i-th row of Z(n) is obtained by generating a hexagonal array of numbers with 2*n+1 rows, 2*n numbers in the odd numbered rows and 2*n+1 numbers in the even numbered rows. The first row is all 0's except for two 1's in the i-th and the (2*n+1-i)th positions. The remaining rows are generated using the same rule for generating Pascal's triangle. The i-th row of Z(n) then consists of the first n numbers in the bottom row of our array.

For example the top row of Z(2) is [5,5], found from the array:

. 1 0 0 1

1 1 0 1 1

. 2 1 1 2

2 3 2 3 2

. 5 5 5 5

Zigzag matrices have remarkable properties. Here is a selection:

1) Z(n) is symmetric.

2) det(Z(n)) = A085527(n).

3) tr(Z(n)) = A033876(n-1).

4) If 2*n+1 is a power of a prime p then all entries of Z(n) are multiples of p.

5) If 4*n+1 is a power of a prime p then the dot product of any two distinct rows of Z(n) is a multiple of p.

6) It is always possible to move from the bottom left entry of Z(n) to the top right entry using only rightward and upward moves and visiting only odd numbers.

A001700(n) = last term of last row of Z(n): a(A000330(n-1)) = A001700(n); A230585(n) = first term of first row of Z(n): a(A056520(n-1)) = A230585(n); A051417(n) = greatest common divisor of entries of Z(n). - Reinhard Zumkeller, Oct 25 2013

Note that A191747={1,1,0,0,1,1,0,0,0,1,0,0,0,1,...} is the sequence formed by concatenation of the row entries of successive N X N identity matrices, N=1,2,....

This sequence is read from the antidiagonals of the table

T(n,k)=

1, 5, 14, 30, 55, ..

2, 10, 25, 49, 84, ..

6, 20, 43, 77, 124, ..

15, 37, 70, 116, 177, ..

31, 63, 100, 168, 245, ..

...

in which the n-th row is found from the n-th generating function (-n+(2*n+1)*x-(n-1)*x^2)/(1-x)^4, n in {0,1,2,...}, by taking the (n+1)-th term on, and, similarly, the k-th column is found from the k-th generating function (2*k+1-(5*k+2)*x+4*(k+1)*x^2-(k+1)*x^3)/(1-x)^4, k in {0,1,2,...}, by taking the k-th term on. For the first three rows, n=0 gives the core sequence A000330, n=1 gives essentially A058373, ignoring the two initial zeros, and n=2 gives -A058372. The first column, for k=0, is A056520, where it is known that A056520(m)=A000330(m)+1. Thus a trivial relation, A191748(m,j)=A056520(m)+j*(m+2)=A000330(m)+j*(m+2)+1, j in {0,...,m}, m>0, with A191748(0,0)=1, gives the triangle

1

2, 5,

6, 10, 14,

...

However, the j-th row R_j of the table is given by R_j(n)=(n+1)*(2*n^2+n-6*j)/6, n=j+1,j+2,j+3,..., and the k-th column C_k by C_k(n)=(n+2)*(2*n^2-n+6*k+3)/6, n=k,k+1,k+2,..., with j,k in {0,1,...}. Substituting n+k for n in the second formula (to account for varying offsets) gives the formula for T(n,k) below.

Row n starts with n and ends with F(n), where F=A000045 (Fibonacci numbers).

Column 2: A056520

For a discussion and guide to related arrays, see A208510.

a(n) = A088961(A056520(n-1));

a(n) = A000108(n+1) for n > 1.