%I
%S 1,1,1,0,1,1,0,0,1,1,0,0,0,1,1,0,0,0,0,1,1,0,0,0,0,0,1,1,0,0,0,0,0,0,
%T 1,1,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,1,1,0,0,
%U 0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,1,1
%N Riordan array (1+x,1) read by rows.
%C Pair sum operator. Columns have g.f. (1+x)x^k. Row sums are A040000. Diagonal sums are (1,1,1,....). Riordan inverse is (1/(1+x), 1). A097806=B*A059260^(1), where B is the binomial matrix.
%C Triangle T(n,k), 0<=k<=n, read by rows given by [1, 1, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938 .  _Philippe Deléham_, May 01 2007
%C Table T(n,k) read by antidiagonals. T(n,1) = 1, T(n,2) = 1, T(n,k) = 0, k > 2.  Boris Putievskiy, Jan 17 2013
%H Boris Putievskiy, <a href="http://arxiv.org/abs/1212.2732">Transformations [of] Integer Sequences And Pairing Functions</a> arXiv:1212.2732 [math.CO].
%F Number triangle T(n, k)=if(n=k or nk=1, 1, 0).
%F a(n)=A103451(n+1).  _Philippe Deléham_, Oct 16 2007
%F From _Boris Putievskiy_, Jan 17 2013: (Start)
%F a(n) = floor((A002260(n)+2)/(A003056(n)+2)), n > 0.
%F a(n) = floor((i+2)/(t+2)), n > 0,
%F where i=nt*(t+1)/2, t=floor((1+sqrt(8*n7))/2). (End)
%e Rows begin {1}, {1,1}, {0,1,1}, {0,0,1,1}...
%e From _Boris Putievskiy_, Jan 17 2013: (Start)
%e The start of the sequence as table:
%e 1..1..0..0..0..0..0...
%e 1..1..0..0..0..0..0...
%e 1..1..0..0..0..0..0...
%e 1..1..0..0..0..0..0...
%e 1..1..0..0..0..0..0...
%e 1..1..0..0..0..0..0...
%e 1..1..0..0..0..0..0...
%e . . .
%e The start of the sequence as triangle array read by rows:
%e 1;
%e 1,1;
%e 0,1,1;
%e 0,0,1,1;
%e 0,0,0,1,1;
%e 0,0,0,0,1,1;
%e 0,0,0,0,0,1,1;
%e 0,0,0,0,0,0,1,1;
%e . . .
%e Row number r (r>4) contains (r2) times '0' and 2 times '1'. (End)
%K easy,nonn,tabl
%O 0,1
%A _Paul Barry_, Aug 25 2004
