|
Search: id:A130668
|
|
| |
|
| 0, 0, 1, -2, 5, -11, 23, -48, 102, -220, 476, -1024, 2184, -4624, 9744, -20480, 42976, -90048, 188352, -393216, 819328, -1704192, 3539200, -7340032, 15203840, -31456256, 65010688, -134217728, 276826112, -570429440, 1174409216
(list; graph; listen)
|
|
|
OFFSET
|
0,4
|
|
|
COMMENT
|
Inverse Binomial Transform of A129819. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Feb 25 2009]
|
|
FORMULA
|
A124072 must be consulted . We change the sign of every negative number and consider the differences of every line. Hence for the second line and followers the four terms periodic sequences:
0 1 -1 1 0
1 0 0 1 1
1 0 1 2 1
1 1 3 3 1
2 4 6 4 2
6 10 10 6 6
16 20 16 12 16
36 36 28 28 36
72 64 56 64 72
136 120 120 136 136
256 240 256 272 256
They are linked together : 272=136+136, 256=120+136, 240=120+120, 256=136+120 . The 4 columns are almost known (must the first line be suppressed?) :A038503 (without the first 1),A000749 (without the first 0),A038505, A038504 G. Adamson (two days ago,I submited 1 1 0 0 without knowing Adamson comment . Like the present one, every sequence of A124072 begining by a negative number (-2,-11..) is a "twisted" sequence (see A129339 comments,A129961 and the present 4 columns) . Periodic 2^n .
G.f.: (1+x)*(1+3*x+4*x^2+3*x^3)*x^2/((1+2*x+2*x^2)*(1+2*x)^2). a(n) = (-1)^n*A001787(n+1)/32-A108520(n)/8+A122803(n)/8, n>2. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Feb 25 2009]
|
|
CROSSREFS
|
Sequence in context: A059411 A126017 A034468 this_sequence A083380 A018112 A067149
Adjacent sequences: A130665 A130666 A130667 this_sequence A130669 A130670 A130671
|
|
KEYWORD
|
sign,uned
|
|
AUTHOR
|
Paul Curtz (bpcrtz(AT)free.fr), Jun 27 2007
|
|
EXTENSIONS
|
Extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Feb 25 2009
|
|
|
Search completed in 0.002 seconds
|