A191748 Sequence of all m in {1,2,3,...} such that A191747(m) = 1.

1, 2, 5, 6, 10, 14, 15, 20, 25, 30, 31, 37, 43, 49, 55, 56, 63, 70, 77, 84, 91, 92, 100, 108, 116, 124, 132, 140, 141, 150, 159, 168, 177, 186, 195, 204, 205, 215, 225, 235, 245, 255, 265, 275, 285, 286, 297, 308, 319, 330, 341, 352, 363, 374, 385, 386, 398

Offset: 0

Author

L. Edson Jeffery, Jun 29 2011

Keywords

• nonn

Comments

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.

Crossrefs

Cf. A000330, A056520, A058372, A058373, A191747.

Formula

For the table: T(n,k) = (n+k+2)*(2*(n+k)^2-n+5*k+3)/6, n,k=0,1,2,....