[seqfan] Re: binomial transform in A001519

Andrew Weimholt andrew.weimholt at gmail.com
Mon Feb 3 21:12:08 CET 2014


Witula's comments seem highly suspect to me.

They contain references to his comments in other sequences which contain
references to his comments in
yet other sequences.
Look at this comment in A014445...

a(n) = 3^n*b(n;2/3) = -b(n;-2), but we have 3^n*a(n;2/3) = F(3n+1) =
A033887 and a(n;-2) = F(3n-1) = A015448, where a(n;d) and b(n;d),
n=0,1,..., d, denote the, so called, delta-Fibonacci numbers (the argument
"d" of a(n;d) and b(n;d) is abbreviation of the symbol "delta") defined by
the following equivalent relations: (1 + d*((sqrt(5) - 1)/2))^n = a(n;d) +
b(n;d)*((sqrt(5) - 1)/2) equiv. a(0;d)=1, b(0;d)=0,
a(n+1;d)=a(n;d)+d*b(n;d), b(n+1;d)=d*a(n;d)+(1-d)b(n;d) equiv.
a(0;d)=a(1;d)=1, b(0;1)=0, b(1;d)=d, and x(n+2;d) + (d-2)*x(n+1;d) +
(1-d-d^2)*x(n;d) = 0 for every n=0,1,...,d, and x=a,b equiv. a(n;d) =
sum_{k=0,..,n} C(n,k)*F(k-1)*(-d)^k, and b(n;d) = sum_{k=0,..,n}
C(n,k)*(-1)^(k-1)*F(k)*d^k equiv. a(n;d) = sum_{k=0,..,n}
C(n,k)*F(k+1)*(1-d)^(n-k)*d^k, and b(n;d) = sum_{k=1,..,n}
C(n;k)*F(k)*(1-d)^(n-k)*d^k. The sequences a(n;d) and b(n;d) for special
values d are connected with many known sequences: A000045, A001519,
A001906, A015448, A020699, A033887, A033889, A074872, A081567, A081568,
A081569, A081574, A081575, A163073 (see also Witula's et al. papers). -
Roman Witula, Jul 12 2012

This appears unintelligible to me. His comments frequently mention
delta-Fibonacci numbers for which I can find no OEIS entry. His comments
contain no examples of delta-Fibonacci numbers - only a hodgepodge of
relations that he claims define them, and numerous references to his own
paper which I cannot access (page does not load on my browser).

Andrew



On Mon, Feb 3, 2014 at 10:22 AM, Richard J. Mathar <mathar at mpia-hd.mpg.de>wrote:

> 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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