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%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
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