A191748 - OEIS

%I #23 May 09 2022 11:01:45

%S 1,2,5,6,10,14,15,20,25,30,31,37,43,49,55,56,63,70,77,84,91,92,100,

%T 108,116,124,132,140,141,150,159,168,177,186,195,204,205,215,225,235,

%U 245,255,265,275,285,286,297,308,319,330,341,352,363,374,385,386,398

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

%C 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,....

%C This sequence is read from the antidiagonals of the table

%C T(n,k)=

%C    1,   5,  14,   30,   55,  ..

%C    2,  10,  25,   49,   84,  ..

%C    6,  20,  43,   77,  124,  ..

%C   15,  37,  70,  116,  177,  ..

%C   31,  63, 100,  168,  245,  ..

%C   ...

%C   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

%C    1

%C    2,  5,

%C    6, 10, 14,

%C    ...

%C 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.

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

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

%K nonn

%O 0,2

%A _L. Edson Jeffery_, Jun 29 2011