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A049652
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a(n) = (F(3*n+2) - 1)/4, where F=A000045 (the Fibonacci sequence).
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12
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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
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OFFSET
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0,3
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COMMENTS
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"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." (End)
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. - Wolfdieter Lang, Oct 18 2010
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LINKS
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FORMULA
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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, 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, May 06 2003
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MAPLE
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a:= n-> add(fibonacci(i, 4), i=0..n): seq(a(n), n=0..22); # Zerinvary Lajos, Mar 20 2008
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MATHEMATICA
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s = 0; lst = {s}; Do[s += Fibonacci[n, 4]; AppendTo[lst, s], {n, 1, 22, 1}]; lst (* Zerinvary Lajos, Jul 14 2009 *)
LinearRecurrence[{5, -3, -1}, {0, 1, 5}, 30] (* or *) Table[(Fibonacci[ 3*n+2] - 1)/4, {n, 0, 30}] (* G. C. Greubel, Dec 05 2017 *)
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PROG
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(Magma) [(Fibonacci(3*n+2) - 1)/4: n in [0..30]]; // G. C. Greubel, Dec 05 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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