# [seqfan] Re: Tabl formula?

Richard Mathar mathar at strw.leidenuniv.nl
Fri Apr 1 19:31:04 CEST 2011

```http://list.seqfan.eu/pipermail/seqfan/2011-April/014747.html

With some experimentation, it seems that the n-th row can be characterized
by having an inverse binomial transform (note this is with an offet 0
for the term that is actually defined with offset 1!) which is the
integers counted backwards. So in the binomial base, the polynomials
might have simple coefficients:
"n=", 1, "terms=", [2, 3, 4]
"inverse binomial transform
[2, 1, 0]

"n=", 2, "terms=", [3, 5, 8, 12]
"inverse binomial transform
[3, 2, 1, 0]

"n=", 3, "terms=", [4, 7, 12, 20, 32]
"inverse binomial transform
[4, 3, 2, 1, 0]

"n=", 4, "terms=", [5, 9, 16, 28, 48, 80]
"inverse binomial transform
[5, 4, 3, 2, 1, 0]

"n=", 5, "terms=", [6, 11, 20, 36, 64, 112, 192]
"inverse binomial transform
[6, 5, 4, 3, 2, 1, 0]

"n=", 6, "terms=", [7, 13, 24, 44, 80, 144, 256, 448]
"inverse binomial transform
[7, 6, 5, 4, 3, 2, 1, 0]

"n=", 7, "terms=", [8, 15, 28, 52, 96, 176, 320, 576, 1024]
"inverse binomial transform
[8, 7, 6, 5, 4, 3, 2, 1, 0]

"n=", 8, "terms=", [9, 17, 32, 60, 112, 208, 384, 704, 1280, 2304]
"inverse binomial transform
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

"n=", 9, "terms=", [10, 19, 36, 68, 128, 240, 448, 832, 1536, 2816, 5120]
"inverse binomial transform
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

"n=", 10, "terms=",
[11, 21, 40, 76, 144, 272, 512, 960, 1792, 3328, 6144, 11264]
"inverse binomial transform
[11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

```

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