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Search: id:A007909
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| A007909 |
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Expansion of (1-x)/(1-2*x+x^2-2*x^3). |
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+0
7
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| 1, 1, 1, 3, 7, 13, 25, 51, 103, 205, 409, 819, 1639, 3277, 6553, 13107, 26215, 52429, 104857, 209715, 419431, 838861, 1677721, 3355443, 6710887, 13421773, 26843545, 53687091, 107374183, 214748365, 429496729, 858993459, 1717986919, 3435973837, 6871947673
(list; graph; listen)
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OFFSET
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1,4
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COMMENT
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Equals INVERT transform of (1, 0, 2, 2, 2,...). [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Apr 28 2009]
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REFERENCES
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I. Gessel, Problem 10424, Amer. Math. Monthly, 102 (1995), 70.
M. E. Larsen, Summa Summarum, A. K. Peters, Wellesley, MA, 2007; see p. 38. [From N. J. A. Sloane (njas(AT)research.att.com), Jan 29 2009]
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LINKS
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INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 444
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FORMULA
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(1/5)*(2^{n+1}+3cos(n*pi/2)+sin(n*pi/2))
a(n)=sum{k=0..floor((n-1)/3), binomial(n-k-1, 2k)2^k} - Paul Barry (pbarry(AT)wit.ie), Sep 16 2004
(1/5) {2^n + (-1)^[n/2] + 2(-1)^[(n-1)/2] }. - Ralf Stephan, Jun 09 2005
a(n) = 2a(n-1)-a(n-2)+2a(n-3). Sequence is identical to its half second differences from the second term; a(n)+a(n+2)=2^(n+1). - Paul Curtz (bpcrtz(AT)free.fr), Dec 17 2007
a(n)=(2^n)*sum(((-1)^(floor(k/2)))/(2^k),k=1..n) - Aktar Yalcin (aktaryalcin(AT)msn.com), Jul 20 2008
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MAPLE
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U:=n->(1/5)*(2^(n+1)+3*cos(n*Pi/2)+sin(n*Pi/2)); [seq(U(n), n=0..50)];
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CROSSREFS
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Cf. A005251, A007679, A007910.
Sequence in context: A092463 A017994 A078000 this_sequence A099810 A125898 A146928
Adjacent sequences: A007906 A007907 A007908 this_sequence A007910 A007911 A007912
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KEYWORD
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nonn,easy
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AUTHOR
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Mogens Esrom Larsen [ mel(AT)math.ku.dk ]
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