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A049652
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a(n)=(F(3n+2)-1)/4, where F=A000045 (the Fibonacci sequence).
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7
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0, 1, 5, 22, 94, 399, 1691, 7164, 30348, 128557, 544577, 2306866, 9772042, 41395035, 175352183, 742803768, 3146567256, 13329072793, 56462858429, 239180506510, 1013184884470, 4291920044391, 18180865062035
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| Comments from Anant P. Godbole (godbolea(AT)etsu.edu), Apr 27 2006: "a(n) equals the number of 2 by 2 determinants that need to be computed while finding the determinant of an n X n matrix using the method discovered by C. L. Dodgson (Lewis Carroll) in the 19th century.
"To evaluate the determinant of an n X n matrix A, set up a 2 by 2 determinant with entries that equal the determinants of the "northwest, northeast, southwest and southeast" (n-1) by (n-1) submatrices of A. Divide this by determinant of the "central" (n-2) by (n-2) submatrix of A. If the latter is zero, the problem can be fixed by row interchanges.
"The Dodgson method does better than the standard method of using cofactors and an expansion in terms of a row/column if n is 6 or larger. By this we mean that a fewer number of 2 by 2 determinants need to be calculated. Of course, the method of choice is diagonalization which can be achieved in polynomial time. Dodgson's method runs in exponential time, whereas the "standard" method requires one to evaluate n!/2 two by two determinants.
"A beautiful combinatorial proof of Dodgson's result was recently given by Zeilberger and an application is presented by Amdeberhan and Ekhad, where a conjecture of Kuperberg and Propp is proved using Dodgson's formula."
This is the sequence A(0,1;4,1;1)of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. [From Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Oct 18 2010]
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REFERENCES
| C. L. Dodgson, "Condensation of determinants," Proceedings of the Royal Society of London, 15 (1866), 150-155.
D. Zeilberger, "Dodgson's determinant-evaluation rule proved by TWO-TIMING MEN and WOMEN," Electron. J. Combin. 4 (no. 2, "The Wilf Festschrift") (1997), #R22, 2 pp. (p. 12).
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LINKS
| T. Amdeberhan and S. Ekhad, A condensed condensation proof of a determinant evaluation conjectured by Greg Kuperberg and Jim Propp.
W. Lang, Notes on certain inhomogeneous three term recurrences. [From Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Oct 18 2010]
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FORMULA
| a(0) = 0, a(1) = 1; a(n) = ceiling(r(4)*a(n-1)) where r(4) = 2+sqrt(5) is the positive root of X^2 = 4*X+1. More generally the sequence a(1) = 1, a(n) = ceiling(r(z)*a(n-1)) where r(z) = (1/2) *(z+sqrt(z^2+4)) is the positive root of X^2 = z*X+1 satisfies the linear recurrence : n>3 a(n) = (z+1)*a(n-1)-(z-1)*a(n-2)-a(n-3) and the closed form formula : a(n) = floor(t(z)*r(z)^n) where t(z) = 1/(2*z)*(1+(z+2)/sqrt(z^2+4)) is the positive root of z*(z^2+4)*X^2 = (z^2+4)*X+1. - Benoit Cloitre (benoit7848c(AT)orange.fr), May 06 2003
a(0) = 0, a(1) = 1, a(2) = 5, a(3) = 22, a(n) = 5a(n-1)-3a(n-2)-a(n-3); a(n) = floor(t(4)*r(4)^n) where t(4) = 1/8*(1+3/sqrt(5)) is the positive root of 80*X^2 = 20*X+1. - Benoit Cloitre (benoit7848c(AT)orange.fr), May 06 2003
a(n+2)=4*a(n+1)+a(n)+1 - Anant P. Godbole (godbolea(AT)etsu.edu), Apr 27 2006
G.f.: x/(x-1)/(x^2+4*x-1) = 1/4/(x-1)+1/4*(-x-1)/(x^2+4*x-1) . - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 23 2007
a(n)=1/2*sum(fibonacci(3k),k=0..n)[From Gary Detlefs (gdetlefs(At)aol.com), Dec 7 2010]
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MAPLE
| a:=n->sum(fibonacci(i, 4), i=0..n): seq(a(n), n=0..22); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 20 2008
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MATHEMATICA
| s = 0; lst = {s}; Do[s += Fibonacci[n, 4]; AppendTo[lst, s], {n, 1, 22, 1}]; lst [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 14 2009]
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CROSSREFS
| Cf. A000071, A048739.
Equals (1/2) A099919.
Sequence in context: A095932 A000346 A026672 * A026877 A128746 A049675
Adjacent sequences: A049649 A049650 A049651 * A049653 A049654 A049655
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KEYWORD
| nonn
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AUTHOR
| Clark Kimberling (ck6(AT)evansville.edu)
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