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Search: id:A121724
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| A121724 |
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Generalized central binomial coefficients for k=2. |
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+0
5
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| 1, 1, 5, 9, 45, 97, 485, 1145, 5725, 14289, 71445, 185193, 925965, 2467137, 12335685, 33563481, 167817405, 464221105, 2321105525, 6507351113, 32536755565
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OFFSET
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0,3
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COMMENT
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Hankel transform is 4^binomial(n+1,2)=A053763(n+1). Case k=2 of T(n,k)=2*k^2*(2k)^n*INT(x^n*sqrt(1-x^2)/(1+k^2-2kx),x,-1,1)/pi. T(n,k) has Hankel transform (k^2)^binomial(n+1,2). k=1 corresponds to C(n,floor(n/2)).
Series reversion of x(1+x)/(1+2x+5x^2).
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FORMULA
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G.f.: (sqrt(1-16x^2)+2x-1)/(2x(1-5x))=c(4x^2)/(1-xc(4x^2)), c(x) the g.f. of A000108; a(n)=(1/(n+1))sum{k=0..n+1, sum{j=0..k, C(n,k)C(k,j)C(2n-2k+j,n-2k+j)(-1)^(n-2k+j)*2^j*5^(k-j)}};.
a(n)=8*4^n*INT(x^n*sqrt(1-x^2)/(5-4x),x,-1,1)/pi
a(n) = Sum_{k, 0<=k<=floor(n/2)} A009766(n-k,k)*2^2k . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Aug 18 2006
a(n)=Sum_{k, 0<=k<=n}4^(n-k)*A120730(n,k). [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 16 2008]
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CROSSREFS
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Sequence in context: A110421 A123822 A145031 this_sequence A149496 A149497 A149498
Adjacent sequences: A121721 A121722 A121723 this_sequence A121725 A121726 A121727
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KEYWORD
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easy,nonn
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AUTHOR
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Paul Barry (pbarry(AT)wit.ie), Aug 17 2006, Feb 28 2007
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