A049652 a(n) = (F(3*n+2) - 1)/4, where F=A000045 (the Fibonacci sequence).

0, 1, 5, 22, 94, 399, 1691, 7164, 30348, 128557, 544577, 2306866, 9772042, 41395035, 175352183, 742803768, 3146567256, 13329072793, 56462858429, 239180506510, 1013184884470, 4291920044391, 18180865062035, 77015380292532, 326242386232164, 1381984925221189, 5854182087116921

Offset: 0

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

From Anant Godbole, Apr 27 2006: (Start)

"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

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• T. Amdeberhan and S. Ekhad, A condensed condensation proof of a determinant evaluation conjectured by Greg Kuperberg and Jim Propp.
• C. L. Dodgson, Condensation of determinants, Proceedings of the Royal Society of London, 15 (1866), 150-155.
• Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
• Doron 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).
• Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Crossrefs

Cf. A000045, A000071, A048739, A099919.

Programs

Magma

[(Fibonacci(3*n+2) - 1)/4: n in [0..30]]; // G. C. Greubel, Dec 05 2017

Maple

a:= n-> add(fibonacci(i,4), i=0..n): seq(a(n), n=0..22); # Zerinvary Lajos, Mar 20 2008

Mathematica

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 *)

PARI

a(n)=fibonacci(3*n+2)\4 \\ Charles R Greathouse IV, Jun 11 2015

Formula

a(n) = A099919(n)/2.

From Benoit Cloitre, May 06 2003: (Start)

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.

a(0) = 0, a(1) = 1, a(2) = 5, a(3) = 22, a(n) = 5*a(n-1) - 3*a(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. (End)

a(n+2) = 4*a(n+1) + a(n) + 1. - Anant Godbole, Apr 27 2006

G.f.: x/((x-1)*(x^2+4*x-1)). - R. J. Mathar, Nov 23 2007

E.g.f.: exp(x)*(exp(x)*(5*cosh(sqrt(5)*x) + 3*sqrt(5)*sinh(sqrt(5)*x)) - 5)/20. - Stefano Spezia, May 24 2024