%I #35 Jul 23 2014 11:00:58
%S 1,3,1,5,3,1,7,5,3,1,9,7,5,3,1,11,9,7,5,3,1,13,11,9,7,5,3,1,15,13,11,
%T 9,7,5,3,1,17,15,13,11,9,7,5,3,1,19,17,15,13,11,9,7,5,3,1,21,19,17,15,
%U 13,11,9,7,5,3,1,23,21,19,17,15,13,11,9,7,5,3,1,25,23,21,19,17,15,13,11,9
%N Sequence matrix for odd numbers.
%C Riordan array ((1+x)/(1-x)^2, x).
%C Inverse matrix is A101038.
%C Row sums yield (n+1)^2.
%C Diagonal sums yield sum{k=0..floor(n/2),2(n-2k)+1}=C(n+2,2)=A000217(n+1). Note that sum{k=0..n,2(n-2k)+1}=n+1.
%C From _Paul Curtz_, Sep 25 2011. (Start)
%C Consider from A187870(n-2) and A171080(n)
%C 1 + 1/3 - 4/45 + 44/945 - 428/14175 =1/(1 -1/3 +1/5 -1/7 ..= Pi/4)=4/Pi.
%C For c(0)=-1, c(1)=1/3, c(2)=4/45, c(3)=44/945, c(4)=428/14175,
%C c(0)/3 + c(1)=0,
%C c(0)/5 + c(1)/3 + c(2)=0,
%C c(0)/7 + c(1)/5 + c(2)/3 + c(3)=0.
%C Hence a(n+1). Numbers are
%C -1/3 + 1/3, 1=1,
%C -1/5 + 1/9 + 4/45, 4=9-5,
%C -1/7 + 1/15 + 4/135 + 44/945 44=135-63-28. (End)
%C T(n,k) = A158405(n+1,n+1-k), 1<=k<=n. [_Reinhard Zumkeller_, Mar 31 2012]
%C From _Peter Bala_, Jul 22 2014: (Start)
%C Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
%C /I_k 0\
%C \ 0 M/
%C having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A208904. (End)
%H Reinhard Zumkeller, <a href="/A099375/b099375.txt">Rows n=0..150 of triangle, flattened</a>
%F Number triangle T(n, k)=if(k<=n, 2(n-k)+1, 0)=binomial(2(n-k)+1, 2(n-k))
%F a(n)=2*A004736(n)-1; a(n)=2*((t*t+3*t+4)/2-n)-1, where t=floor((-1+sqrt(8*n-7))/2). - _Boris Putievskiy_, Feb 08 2013
%e Rows start
%e 1;
%e 3,1;
%e 5,3,1;
%e 7,5,3,1;
%e 9,7,5,3,1;
%e 11,9,7,5,3,1;
%e 13,11,9,7,5,3,1;
%o (Haskell)
%o a099375 n k = a099375_row n !! k
%o a099375_row n = a099375_tabl !! n
%o a099375_tabl = iterate (\xs -> (head xs + 2) : xs) [1]
%o -- _Reinhard Zumkeller_, Mar 31 2012
%Y Cf. A005408, A004736. A208904.
%K nonn,easy,tabl
%O 0,2
%A _Paul Barry_, Jan 22 2005