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A166920
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2^n -(1+(-1)^n)/2 .
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4
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0, 2, 3, 8, 15, 32, 63, 128, 255, 512, 1023, 2048, 4095, 8192, 16383, 32768, 65535, 131072, 262143, 524288, 1048575, 2097152, 4194303, 8388608, 16777215, 33554432, 67108863, 134217728, 268435455, 536870912, 1073741823, 2147483648, 4294967295
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| Partial sums of A014551. The inverse binomial transform yields a sequence 0,2,-1,5,-7,17,...: zero followed by a sign alternating A014551.
The table of a(n) plus higher order differences in successive rows shows A131577 on the main diagonal.
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LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index to sequences with homogeneous linear recurrences with constant coefficients, signature (2,1,-2)
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FORMULA
| G.f.: x*(2-x)/((1-x)*(1-2*x)*(1+x)).
a(n) = 2^n - (1+(-1)^n)/2.
a(2*n) = A024036(n); a(2*n+1)= A004171(n).
a(n) = 2*a(n-1)+a(n-2)-2*a(n-3).
a(n+1)-2*a(n) = A168361(n).
a(n) = A000225(n+1)-A051049(n) = A014551(n)-A168361(n).
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PROG
| (MAGMA) [2^n -(1+(-1)^n)/2: n in [0..30]]; // Vincenzo Librandi, May 16 2011
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CROSSREFS
| Sequence in context: A128022 A011946 A195095 * A080206 A132862 A055543
Adjacent sequences: A166917 A166918 A166919 * A166921 A166922 A166923
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KEYWORD
| nonn,easy
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AUTHOR
| Paul Curtz (bpcrtz(AT)free.fr), Oct 23 2009
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EXTENSIONS
| Edited and extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 02 2010
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