# [seqfan] Re: binomial transform in A001519

Vladimir Shevelev shevelev at bgu.ac.il
Tue Feb 4 14:42:13 CET 2014

```Sequence {0,1,0,5,0,25,..} is definitely connected with A001519.
I know nothing about the second binomial transform, but every other term of the first
one {1,8,80,832,8704,...} gives A001519 in a simple way
1/2^0=a(1), 8/2^2=a(2), 80/2^4=a(3), 832/2^6=a(4), 8704/2^8=a(5), etc.,
such that ( probably, one can prove by induction) A001519(n)=4^(- (n-1))*sum{k=1,3,...,2*n-1} 5^((k-1)/2)*C(2*n-1,k). I think that such a formula for F_(2*n-1)
could be known.

Regards,

________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Richard J. Mathar [mathar at mpia-hd.mpg.de]
Sent: 03 February 2014 20:22
To: seqfan at seqfan.eu
Subject: [seqfan] binomial transform in A001519

A comment in A001519 by Witula says
"Moreover, a(n) is the second binomial transform of (0,1,0,5,0,25,...) ..."

However, if I start from [0,1,0,5,0,25...] and take the binomial transform
I get [0,1,2,8,24,80,...]
and after taking again the binomial transform I get
[0,1,4,17,72,305,..]
Can someone explain why the comment says that these operations result in
A001519, which is [1,1,2,5,13,34..]?

R. Mathar

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