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A108981
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a(n) = 3a(n-1) + 4a(n-2), a(0) = 1, a(1) = 5.
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5
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1, 5, 19, 77, 307, 1229, 4915, 19661, 78643, 314573, 1258291, 5033165, 20132659, 80530637, 322122547, 1288490189, 5153960755, 20615843021, 82463372083, 329853488333, 1319413953331, 5277655813325, 21110623253299
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| The Hankel transform of this sequence is [1,-6,0,0,0,0,0,0,0,0,...]. - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Apr 15 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=charpoly(A,2). [From Milan R. Janjic (agnus(AT)blic.net), Jan 26 2010]
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LINKS
| Index to sequences with linear recurrences with constant coefficients, signature (3,4).
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FORMULA
| Inverse binomial transform of A003948.
a(2n) = 4a(2n-1) - 1; a(2n+1) = 4a(2n) + 1.
a(n) = 3*2^(2*n-1)-a(n-1), with a(0) = 1; also a(n) = abs{3*sum[i = 1..n,(-1)^i*2^(2*i-1)]+1}, with a(0) = 1. - Paolo P. Lava & Giorgio Balzarotti (paoloplava(AT)gmail.com), May 28 2007
O.g.f.: (1+2*x)/[(1+x)(1-4*x)]. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Apr 02 2008
Sum_{k, 0<=k<=n}a(k)=A037481(n+1). - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Apr 15 2008
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MAPLE
| P:=proc(n, k) local a, i, j; a:=1; print(a); for i from 1 by 1 to n do j:=(k+1)*k^(2*i-1)-a; print(j); a:=j; od; end: P(100, 2); - Paolo P. Lava & Giorgio Balzarotti (paoloplava(AT)gmail.com), May 28 2007
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PROG
| (PARI) Vec((1+2*x)/(1+x)/(1-4*x)+O(x^99)) \\ Charles R Greathouse IV, Jan 11 2012
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CROSSREFS
| Sequence in context: A149768 A149769 A149770 * A149771 A149772 A149773
Adjacent sequences: A108978 A108979 A108980 * A108982 A108983 A108984
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KEYWORD
| nonn,easy
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AUTHOR
| Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Jul 23 2005
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EXTENSIONS
| Corrected by T. D. Noe (noe(AT)sspectra.com), Nov 07 2006
Edited by N. J. A. Sloane (njas(AT)research.att.com) at the suggestion of R. J. Mathar, Apr 14 2008
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