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A099375
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Sequence matrix for odd numbers.
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10
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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, 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, 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
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OFFSET
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0,2
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COMMENTS
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Riordan array ((1+x)/(1-x)^2, x).
Row sums yield (n+1)^2.
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.
1 + 1/3 - 4/45 + 44/945 - 428/14175 =1/(1 -1/3 +1/5 -1/7 ..= Pi/4)=4/Pi.
For c(0)=-1, c(1)=1/3, c(2)=4/45, c(3)=44/945, c(4)=428/14175,
c(0)/3 + c(1)=0,
c(0)/5 + c(1)/3 + c(2)=0,
c(0)/7 + c(1)/5 + c(2)/3 + c(3)=0.
Hence a(n+1). Numbers are
-1/3 + 1/3, 1=1,
-1/5 + 1/9 + 4/45, 4=9-5,
-1/7 + 1/15 + 4/135 + 44/945 44=135-63-28. (End)
Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
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)
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LINKS
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FORMULA
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Number triangle T(n, k)=if(k<=n, 2(n-k)+1, 0)=binomial(2(n-k)+1, 2(n-k))
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
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EXAMPLE
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Rows start
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;
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PROG
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(Haskell)
a099375 n k = a099375_row n !! k
a099375_row n = a099375_tabl !! n
a099375_tabl = iterate (\xs -> (head xs + 2) : xs) [1]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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